problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
value | source stringclasses 8
values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
4.94 Solve the equation $\operatorname{arctg} x+\operatorname{arctg}(1-x)=2 \operatorname{arctg} \sqrt{x-x^{2}}$. | [Solution] $\because x(1-x) \geqslant 0, \therefore \quad x \in[0,1]$ and $x(1-x) \leqslant \frac{1}{4}$. Taking the tangent on both sides of the equation, we get
$$\frac{x+(1-x)}{1-x(1-x)}=\frac{2 \sqrt{x(1-x)}}{1-x(1-x)}$$
Since $x(1-x) \neq 1$, we get the equation
$$\sqrt{x(1-x)}=\frac{1}{2}$$
This yields $x=\frac... | x=\frac{1}{2} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,509 |
4.95 Solve the equation $\cos \cos \cos \cos x=\sin \sin \sin \sin x$.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | [Solution] The equation has no solution.
We will prove that for all $x \in \mathbb{R}$, we have
$\cos \cos \cos \cos x > \sin \sin \sin \sin x$
Indeed, if $x \in [\pi, 2\pi]$, then
$\cos \cos \cos \cos x > 0, \sin \sin \sin \sin x \leq 0$, so (1) holds.
If $x \in \left[0, \frac{\pi}{2}\right]$, then
$\cos x, \sin x, \c... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | inequalities | false | 735,510 |
$1 \cdot 51$ Let $n$ be even, and select four different numbers $a, b, c, d$ from the integers $1, 2, \cdots, n$ such that $a+c=b+d$. Prove: The number of ways to select (without considering the order of $a, b, c, d$) is $\frac{n(n-2)(2 n-5)}{24}$. | [Solution] Without loss of generality, let $a>b>d$, then $d>c$. Consider selecting three numbers $a>b>c$ from $1,2, \cdots, n$ such that $a+c-b \neq b$: There are $C_{n}^{3}$ ways to choose three numbers $a>b>c$. Among these, the number of ways that satisfy $a+c=2 b$ is when $a$ and $c$ have the same parity, which tota... | \frac{n(n-2)(2 n-5)}{24} | Combinatorics | proof | Yes | Yes | inequalities | false | 735,511 |
4.97 Real number $p \geqslant \frac{1}{4}$. Find all positive real solutions $x$ of the following equation:
$$\log _{\sqrt{2}}^{2} x+2 \log _{\sqrt{2}} x+2 \log _{\sqrt{2}}\left(x^{2}+p\right)+p+\frac{15}{4}=0$$ | ［Solution］ When $p>\frac{1}{4}$,
$$\left(x-\frac{1}{2}\right)^{2}+p-\frac{1}{4}>0$$
That is,
$$x^{2}+p>x$$
Since $\sqrt{2}>1, x>0$, we have
$$\log _{\sqrt{2}}\left(x^{2}+p\right)>\log _{\sqrt{2}} x$$
Thus,
$$\begin{aligned}
\text { the left side of the original equation } & >\log _{\sqrt{2}}^{2} x+2 \log _{\sqrt{2}}... | x=\frac{1}{2} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,512 |
4. 98 Find all integer solutions of the following equation:
$$\cos \left(\frac{\pi}{8}\left(3 x-\sqrt{9 x^{2}+160 x+800}\right)\right)=1 .$$ | [Solution] Let $x$ be an integer solution of the original equation, then there must exist an integer $n$ such that
$$\frac{\pi}{8}\left(3 x-\sqrt{9 x^{2}+160 x+800}\right)=2 n \pi$$
i.e., $\square$
$$\begin{array}{l}
9 x^{2}+160 x+800=(3 x-16 n)^{2} \\
x(3 n+5)=8 n^{2}-25
\end{array}$$
However,
Therefore,
$$8 n^{2}-2... | x=-7 \text{ or } x=-31 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,513 |
4.99 Solve the system of equations $\left\{\begin{array}{l}x+y=4, \\ \left(x^{2}+y^{2}\right)\left(x^{3}+y^{3}\right)=280 .\end{array}\right.$ | [Solution] Substituting (1) into (2) yields
$$\left(x^{2}+y^{2}\right)\left(x^{2}+y^{2}-x y\right)-70=0$$
From (1), we get $x^{2}+y^{2}=16-2 x y$
Substituting (4) into (3) and simplifying, we get $3 x^{2} y^{2}-40 x y+93=0$, so $x y=3$ or $x y=\frac{31}{3}$.
Combining with (1), we have
$$\left\{\begin{array} { l }
{ ... | null | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,514 |
$4 \cdot 100$ Solve the system of equations $\left\{\begin{array}{l}x+y+z=0, \\ x^{3}+y^{3}+z^{3}=-18 .\end{array}\right.$ for integer solutions. | [Solution] From (1), we get $z=-(x+y)$
(3)
Substituting into (2) yields $x^{3}+y^{3}-(x+y)^{3}=-18$, simplifying to $xy(x+y)=6$.
From (3), we get $xyz=-6$,
Therefore, $x, y, z$ must be divisors of 6.
From (1) and (2), we know that among $x, y, z$, there is exactly one negative number, and the absolute value of this neg... | not found | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,515 |
4-101 Proof: The system of equations with three unknowns and two equations
$$\left\{\begin{array}{l}
x-y=2, \\
x y+z^{2}+1=0
\end{array}\right.$$
has no real solutions for $z \neq 0$. | [Proof] Transform the original system of equations into
$$\left\{\begin{array}{l}
x+(-y)=2 \\
x(-y)=1+z^{2}
\end{array}\right.$$
By the inverse of Vieta's formulas, the numbers $x$ and $-y$ are the roots of the equation
$$m^{2}-2 m+\left(1+z^{2}\right)=0$$
But \quad $\Delta=4-4\left(1+z^{2}\right)=-z^{2}$
When $z \n... | proof | Algebra | proof | Yes | Yes | inequalities | false | 735,516 |
4・102 Solve the system of equations
$$\left\{\begin{array}{l}
x(x+1)(3 x+5 y)=144, \\
x^{2}+4 x+5 y=24 .
\end{array}\right.$$ | [Solution] From (2), we get $x(x+1)+(3 x+5 y)=24$. Treating $x(x+1)$ and $(3 x+5 y)$ as the roots of the equation $a^{2}-24 a+144=0$, we obtain $x(x+1)=12$ and $3 x+5 y=12$.
Solving $x(x+1)=12$ yields $x_{1}=-4, x_{2}=3$.
Substituting into $3 x+5 y=12$, we get $\quad y_{1}=\frac{24}{5}, y_{2}=\frac{3}{5}$.
$\therefore ... | \left(-4, \frac{24}{5}\right),\left(3, \frac{3}{5}\right) | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,517 |
4. 103 Real numbers $\alpha, \beta$ satisfy the system of equations
$$\left\{\begin{array}{l}
\alpha^{3}-3 \alpha^{2}+5 \alpha-17=0 \\
\beta^{3}-3 \beta^{2}+5 \beta+11=0
\end{array}\right.$$
Find $\alpha+\beta$. | [Solution] The system of equations is transformed into
$$\begin{array}{l}
(\alpha-1)^{3}+2(\alpha-1)-14=0 \\
(\beta-1)^{3}+2(\beta-1)+14=0
\end{array}$$
Adding the two equations, we get $(\alpha-1)^{3}+(\beta-1)^{3}+2(\alpha+\beta-2)=0$,
which simplifies to $(\alpha+\beta-2)\left[(\alpha-1)^{2}+(\beta-1)^{2}-(\alpha-1... | 2 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,518 |
4. 104 Find the integer solutions of the system of equations
$$\left\{\begin{array}{l}
x z-2 y t=3 \\
x t+y z=1
\end{array}\right.$$ | ［Solution］From the system of equations, we get
$$(x z-2 y t)^{2}+2(x t+y z)^{2}=\left(x^{2}+2 y^{2}\right)\left(z^{2}+2 t^{2}\right)=11 .$$
Therefore, we have $x^{2}+2 y^{2}=1$ or $z^{2}+2 t^{2}=1$.
If $x^{2}+2 y^{2}=1$, then we get $y=0, x= \pm 1$, and from the second equation, we get $t= \pm 1$, and from the first e... | (1,0,3,1),(-1,0,-3,-1),(3,1,1,0),(-3,-1,-1,0) | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,519 |
4. 106 Solve the system of equations
$$\left\{\begin{array}{l}
x y^{2}-2 y+3 x^{2}=0 \\
y^{2}+x^{2} y+2 x=0
\end{array}\right.$$ | ［Solution］ When $x=0$, the system of equations has a unique solution:
$$\left\{\begin{array}{l}
x=0 \\
y=0
\end{array}\right.$$
When $x \neq 0$, by (1) - (2) $\times x$ we get
$$-2 y-x^{3} y+x^{2}=0$$
which simplifies to $\left(2+x^{3}\right) y=x^{2}$,
from which we know $2+x^{3} \neq 0$, thus $y=\frac{x^{2}}{2+x^{3}... | null | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,520 |
4・107 Assume $x, y, z$ are all real numbers, and it is known that they satisfy
$$\left\{\begin{array}{l}
x+y+z=a \\
x^{2}+y^{2}+z^{2}=\frac{a^{2}}{2}(a>0) .
\end{array}\right.$$
Prove that $x, y, z$ cannot be negative, and they cannot be greater than $\frac{2}{3} a$. | [Proof] Substituting $z=a-(x+y)$ into (2), we get
$$x^{2}+y^{2}+(x+y)^{2}-2 a(x+y)+a^{2}=\frac{a^{2}}{2}$$
which simplifies to $y^{2}+(x-a) y+\left(x-\frac{a}{2}\right)^{2}=0$.
Since $y$ is a real number, we have
$$(x-a)^{2}-4\left(x-\frac{a}{2}\right)^{2} \geqslant 0$$
which simplifies to $x(2 a-3 x) \geqslant 0$,
t... | proof | Algebra | proof | Yes | Yes | inequalities | false | 735,521 |
4-108 Find the real solutions of the system of equations
$$\left\{\begin{array}{l}
x+y=3 \\
x y-y^{2}=1
\end{array}\right.$$ | [Solution] From (1) $x=3-y$,
(3)
substituting into (2) $2 y^{2}-3 y+1=0$,
which is $(2 y-1)(y-1)=0$,
thus
$$\begin{array}{l}
y_{1}=\frac{1}{2}, x_{1}=\frac{5}{2} \\
y_{2}=1, x_{2}=2
\end{array}$$
we get two sets of solutions
$$\left\{\begin{array} { l }
{ x _ { 1 } = \frac { 5 } { 2 } , } \\
{ y _ { 1 } = \frac { 1 }... | null | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,522 |
4- 109 Prove that for any integers $a$ and $b$, the system of equations
$$\left\{\begin{array}{l}
x+y+2 z+2 t=a, \\
2 x-2 y+z-t=b .
\end{array}\right.$$
has integer solutions. | [Proof] Solving the given system of equations for $x$ and $z$, we get
$$\begin{aligned}
x=\frac{1}{3} & (-a+2 b+5 y+4 t) \\
& =b+2 y+t-\frac{1}{3}(a+b+y-t) \\
z & =\frac{1}{3}(2 a-b-4 y-5 t) \\
& =a-y-2 t-\frac{1}{3}(a+b+y-t)
\end{aligned}$$
For any integers $a$ and $b$, choose integers $y$ and $t$ such that the numbe... | proof | Algebra | proof | Yes | Yes | inequalities | false | 735,523 |
4. 110 Solve the system of equations
$$\left\{\begin{array}{l}
y^{2}=x^{3}-3 x^{2}+2 x, \\
x^{2}=y^{3}-3 y^{2}+2 y .
\end{array}\right.$$
for all solutions. | [Solution] (1) - (2) gives
That is, $y^{3}-2 y^{2}+2 y=x^{3}-2 x^{2}+2 x$.
$$y^{2}-x^{2}=\left(x^{3}-3 x^{2}+2 x\right)-\left(y^{3}-3 y^{2}+2 y\right)$$
Let $f(x)=x^{3}-2 x^{2}+2 x$, then equation (3) becomes
$$f(y)=f(x)$$
Since
$$\begin{array}{l}
f^{\prime}(x)=3 x^{2}-4 x+2 \\
\Delta=(-4)^{2}-4 \cdot 3 \cdot 2=-80$... | null | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,524 |
4・111 Let $a, b, c, d$ be integers such that the system of equations
$$\left\{\begin{array}{l}
a x+b y=m, \\
c x+d y=n .
\end{array}\right.$$
has integer solutions for all integers $m, n$. Prove that
$$a d-b c= \pm 1$$ | [Proof] Clearly, among the four numbers $a, b, c, d$, it is impossible for all to be 0. Thus, without loss of generality, assume $a \neq 0$.
Let $D = ad - bc$. Certainly, $D \neq 0$, otherwise, let $\lambda = \frac{c}{a} = \frac{d}{b}$, then it can be deduced that the original system of equations has integer solutions... | D = \pm 1 | Algebra | proof | Yes | Yes | inequalities | false | 735,525 |
4・112 Given the system of equations
$$\left\{\begin{array}{l}
y-2 x-a=0 \\
y^{2}-x y+x^{2}-b=0
\end{array}\right.$$
where $a, b$ are integers, and $x$ and $y$ are unknowns. Prove that if there is a rational solution to this system of equations, then it must be an integer. | [Proof] From equation (1), we get $x=\frac{y-a}{2}$. Substituting (3) into (2), we obtain
$$y^{2}-y \cdot \frac{y-a}{2}+\left(\frac{y-a}{2}\right)^{2}-b=0,$$
Simplifying, we get $\quad 3 y^{2}=4 b-a^{2}$,
which can be rewritten as $(3 y)^{2}=3\left(4 b-a^{2}\right)$.
If $x$ and $y$ are rational numbers satisfying the ... | proof | Algebra | proof | Yes | Yes | inequalities | false | 735,526 |
4・113 Consider the system of equations
$$\left\{\begin{array}{l}
x + p y = n, \\
x + y = p^{z}.
\end{array}\right.$$
(Where $n$ and $p$ are given natural numbers) What is the necessary and sufficient condition for the system to have positive integer solutions $(x, y, z)$? Prove that the number of such solutions cannot ... | [Solution] If the original system of equations has a positive integer solution $(x, y, z)$, then from equation (2) we know $p>1$, and we have
$$\begin{array}{l}
x=\frac{p^{z+1}-n}{p-1}=\frac{p^{z+1}-1}{p-1}-\frac{n-1}{p-1} \\
y=\frac{n-p^{z}}{p-1}=\frac{n-1}{p-1}-\frac{p^{z}-1}{p-1}
\end{array}$$
Since for all positiv... | proof | Algebra | proof | Yes | Yes | inequalities | false | 735,527 |
4・114 Try to solve the system of equations
$$\left\{\begin{array}{l}
\lg x+\lg y=1 \\
x^{2}+y^{2}-3 x-3 y=8
\end{array}\right.$$ | [Solution] From (1) we get $x y=10$.
(3)
(2) $+2 \times(3)$, and rearranging terms, we get
$$(x+y)^{2}-3(x+y)-28=0$$
So $x+y=7$ and $x+y=-4$.
Combining (4) with (3), we get the following system of equations:
$$\begin{array}{l}
\left\{\begin{array}{l}
x+y=7 \\
x y=10
\end{array}\right. \\
\left\{\begin{array}{l}
x+y=-4... | x_1=2, y_1=5; x_2=5, y_2=2 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,528 |
4・115 Solve the system of equations
$$\left\{\begin{array}{l}
x^{2}+y^{2}+\frac{2 x y}{x+y}=1 \\
\sqrt{x+y}=x^{2}-y
\end{array}\right.$$ | [Solution] From $x^{2}+y^{2}+\frac{2 x y}{x+y}=1$, we get
$$(x+y-1)\left(x^{2}+y^{2}+x+y\right)=0$$
If $x^{2}+y^{2}+x+y=0$, then
$$x+y=-\left(x^{2}+y^{2}\right) \leqslant 0$$
From the original equation, we know $x+y \geqslant 0$.
Therefore, $x+y=0$, but for the original equation, $x+y=0$ cannot be the denominator. Th... | null | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,529 |
4. 116 Consider the system of equations $\left\{\begin{array}{l}a_{11} x_{1}+a_{12} x_{2}+\cdots+a_{1 n} x_{n}=0, \\ a_{21} x_{1}+a_{22} x_{2}+\cdots+a_{2 n} x_{n}=0, \\ \cdots \cdots \cdots \cdots . \\ a_{m 1} x_{1}+a_{m 2} x_{2}+\cdots+a_{m n} x_{n}=0 .\end{array}\right.$
where the coefficients $a_{i j}$ are integer... | [Proof] First, let $n=2 m$, and let $A=\max \left|a_{i j}\right|, B=m A$,
$S=\left\{\left(x_{1}, x_{2}, \cdots, x_{n}\right) \mid x_{j}\right.$ is an integer and
$\left.\left|\cdot x_{j}\right| \leqslant B, 1 \leqslant j \leqslant n\right\}$,
$T=\left\{\left(y_{1}, y_{2}, \cdots, y_{m}\right) \mid y_{j}\right.$ is an i... | proof | Algebra | proof | Yes | Yes | inequalities | false | 735,530 |
4・117 Solve the system of equations
$$\left\{\begin{array}{l}
\frac{x-y \sqrt{x^{2}-y^{2}}}{\sqrt{1-x^{2}+y^{2}}}=a \\
\frac{y-x \sqrt{x^{2}-y^{2}}}{\sqrt{1-x^{2}+y^{2}}}=b
\end{array}\right.$$
where \(a, b\) are known numbers. | [Solution] (1) + (2) gives
$$\frac{(x+y)\left(1-\sqrt{x^{2}-y^{2}}\right)}{\sqrt{1-x^{2}+y^{2}}}=a+b$$
(1) - (2) gives
$$\frac{(x-y)\left(1+\sqrt{x^{2}-y^{2}}\right)}{\sqrt{1-x^{2}+y^{2}}}=a-b$$
(3) $\times$ (4) gives
$$x^{2}-y^{2}=a^{2}-b^{2} .$$
Substituting (5) into (3) and (4) respectively gives
$$\left\{\begin{ar... | \left\{\begin{array}{l}
x=\frac{a+b \sqrt{a^{2}-b^{2}}}{\sqrt{1-a^{2}+b^{2}}} \\
y=\frac{b+a \sqrt{a^{2}-b^{2}}}{\sqrt{1-a^{2}+b^{2}}}
\end{array}\right.
} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,531 |
4・118 Solve the system of equations
$$\left\{\begin{array}{l}
\sqrt{\frac{x}{y}}-\sqrt{\frac{y}{x}}=\frac{7}{\sqrt{x y}} \\
\sqrt[4]{x^{3} y}-\sqrt[4]{x y^{3}}=\sqrt{12}
\end{array}\right.$$ | [Solution] For $x>0$, then $y>0$. From (1) we get $x-y=7$,
(3)
From (2) we get $\sqrt[4]{x y}(\sqrt{x}-\sqrt{y})=\sqrt{12}$,
squaring both sides, we get $\sqrt{x y}(\sqrt{x}-\sqrt{y})^{2}=12$.
From (3) we get, $\sqrt{x}-\sqrt{y}=\frac{7}{\sqrt{x}+\sqrt{y}}$,
substituting into (4) and simplifying, we get
$$12 x+12 y-25 ... | \left\{\begin{array}{l}x=16, \\ y=9\end{array}\right.} 和 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,532 |
4. 119 Solve the system of equations $\left\{\begin{array}{l}\cos x=2 \cos ^{3} y, \\ \sin x=2 \sin ^{3} y .\end{array}\right.$ | [Solution] Squaring and adding the two equations, we get
$$\begin{aligned}
1 & =4\left(\cos ^{6} y+\sin ^{6} y\right) \\
& =4\left(\cos ^{4} y-\cos ^{2} y \sin ^{2} y+\sin ^{4} y\right) \\
& =4\left(1-3 \sin ^{2} y \cos ^{2} y\right)
\end{aligned}$$
So $\sin 2 y= \pm 1, y=\frac{k \pi}{2}+\frac{\pi}{4}(k \in \mathbb{Z}... | \left\{\begin{array}{l}
x=2 l \pi+\frac{k \pi}{2}+\frac{\pi}{4}, \\
y=\frac{k \pi}{2}+\frac{\pi}{4} .
\end{array} \quad(l, k \in \mathbb{Z})\right.} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,533 |
4. 120 If $\operatorname{tg} x+\operatorname{tg} y=25$, and $\operatorname{ctg} x+\operatorname{ctg} y=30$. Find $\operatorname{tg}(x+y)$. | [Solution] From $\operatorname{ctg} x+\operatorname{ctg} y=30$ we get
then
$$\frac{1}{\operatorname{tg} x}+\frac{1}{\operatorname{tg} y}=30$$
which means
so
$$\begin{array}{l}
\frac{\operatorname{tg} x+\operatorname{tg} y}{\operatorname{tg} x \cdot \operatorname{tg} y}=30 \\
\frac{25}{\operatorname{tg} x \cdot \oper... | \frac{5}{6} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,534 |
4. 121 Solve the system of equations
$$\left\{\begin{array}{l}
2^{\lg x}+3^{\lg y}=5, \\
2^{\lg x} \cdot 3^{\lg y}=4 .
\end{array}\right.$$ | [Solution] Let $2^{\lg x}, 3^{\lg y}$ be the roots of the equation $a^{2}-5 a+4=0$, then we have $\left\{\begin{array}{l}2^{\lg x}=4, \\ 3^{\lg y}=1 ;\end{array}\right.$ or $\quad\left\{\begin{array}{l}2^{\lg x}=1, \\ 3^{\lg y}=4 .\end{array}\right.$
Therefore, $\left\{\begin{array}{l}x_{1}=100, \\ y_{1}=1 ;\end{array}... | null | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,535 |
4. 122 Solve the system of equations $\left\{\begin{array}{l}\sqrt{x-1}+\sqrt{y-3}=\sqrt{x+y}, \\ \lg (x-10)+\lg (y-6)=1 .\end{array}\right.$ | [Solution] The original system of equations can be transformed into the system of equations
$$\left\{\begin{array}{l}
x y-3 x-y-1=0 \\
x y-6 x-10 y+50=0
\end{array}\right.$$
Subtracting (4) from (3) and simplifying, we get
$$x=17-3 y .$$
Substituting (5) into (3) and rearranging, we get
$$3 y^{2}-25 y+52=0$$
Thus $\s... | no solution | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,536 |
4-123 Let $a, b, c$ be given positive real numbers, try to determine all positive real numbers $x, y, z$ that satisfy the system of equations
$$\left\{\begin{array}{l}
x+y+z=a+b+c \\
4 x y z-\left(a^{2} x+b^{2} y+c^{2} z\right)=a b c .
\end{array}\right.$$ | ［Solution］ The second equation above is equivalent to
$$\frac{a^{2}}{y z}+\frac{b^{2}}{x z}+\frac{c^{2}}{x y}+\frac{a b c}{x y z}=4$$
Let $x_{1}=\frac{a}{\sqrt{y z}}, y_{1}=\frac{b}{\sqrt{z x}}, z_{1}=\frac{c}{\sqrt{x y}}$, then
$$x_{1}^{2}+y_{1}^{2}+z_{1}^{2}+x_{1} y_{1} z_{1}=4$$
Obviously, $00$, thus
$$z_{1}=-2 \s... | x=\frac{b+c}{2}, y=\frac{c+a}{2}, z=\frac{a+b}{2} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,537 |
4- $124 m, n$ are two distinct positive integers, find the common complex roots of the equations
$$x^{m+1}-x^{n}+1=0$$
and
$$x^{n+1}-x^{m}+1=0$$ | [Solution] Without loss of generality, assume $m>n$ (if $m<n$, just swap $m$ and $n$ in the following proof). Let $x$ be a common complex root of the two equations given in the problem. Subtracting the two equations, we have
that is, $\square$
$$\begin{array}{c}
x^{m+1}-x^{n+1}-x^{n}+x^{m}=0 \\
x^{n}(x+1)\left(x^{m-n}... | x=\frac{1}{2} \pm \frac{\sqrt{3}}{2} i | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,538 |
4・126 Solve the system of equations
$$\left\{\begin{array}{l}
x+y+z=a \\
x^{2}+y^{2}+z^{2}=b^{2} \\
x y=z^{2}
\end{array}\right.$$
where $a, b$ are known numbers. Also, indicate the conditions that $a, b$ must satisfy for $x, y, z$ to be distinct positive numbers. | [Solution] From (1) ${ }^{2}$ - (2) we get $x y+y z+z x=\frac{1}{2}\left(a^{2}-b^{2}\right)$, i.e., $z a=\frac{1}{2}\left(a^{2}-b^{2}\right)$.
Therefore, the solution is
$$\left\{\begin{array}{l}
x=\frac{1}{4 a}\left(a^{2}+b^{2} \pm \sqrt{10 a^{2} b^{2}-3 a^{4}-3 b^{4}}\right) \\
y=\frac{1}{4 a}\left(a^{2}+b^{2} \mp \s... | When\ and\ only\ when\ |b|<a<\sqrt{3}|b|,\ x,\ y,\ and\ z\ are\ distinct\ positive\ numbers. | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,540 |
1. 54 For any positive integer $n$, let $d(n)$ denote the number of positive divisors of $n$ (including 1 and $n$ itself). Determine all possible positive integers $K$ such that there exists a positive integer $n$ satisfying $\frac{d\left(n^{2}\right)}{d(n)}=K$ | [Solution] If $K$ is a possible positive integer such that there exists $n \in N$, satisfying
$$\frac{d\left(n^{2}\right)}{d(n)}=K$$
We assume the prime factorization of $n$ is
$$n=p_{1^{1}}^{\alpha_{1}} \cdot p_{2^{1}}^{\alpha} \cdots \cdots p_{s_{s}}^{\alpha}$$
Thus,
$$\begin{array}{l}
d(n)=\prod_{i=1}^{s}\left(a_{... | proof | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 735,541 |
4. 127 Solve the system of equations
$$\left\{\begin{array}{l}
y z=3 y+2 z-8 \\
z x=4 z+3 x-8 \\
x y=2 x+y-1
\end{array}\right.$$ | \[ \text{Solution:} \] Rewrite the three equations of the system as
\[ \left\{ \begin{array}{l}
(y-2)(z-3)=-2 \\
(z-3)(x-4)=4 \\
(x-1)(y-2)=1
\end{array} \right. \]
From (1) and (3), we get \((z-3)=-2(x-1)\). Substituting into (2), we get \((x-3)(x-2)=0\).
Thus, \(x=2\) or \(x=3\).
Substituting back into (2) and (3), ... | null | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,542 |
4・128 Find the relationship between the coefficients $a, b, c$ for the system of equations
$$\left\{\begin{array}{l}
a x^{2}+b x+c=0, \\
b x^{2}+c x+a=0, \\
c x^{2}+a x+b=0 .
\end{array}\right.$$
to have real solutions. | [Solution] Adding the equations in the system, we get
$$(a+b+c) x^{2}+(a+b+c) x+(a+b+c)=0$$
Thus, $(a+b+c)\left(x^{2}+x+1\right)=0$.
Since for any real number $x, x^{2}+x+1 \neq 0$, the necessary condition for the system of equations to have a solution is $a+b+c=0$.
Conversely, if $a+b+c=0$, then $x=1$ is clearly a s... | a+b+c=0 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,543 |
4. 129 Solve the system of equations $\left\{\begin{array}{l}x^{2}=6+(y-z)^{2}, \\ y^{2}=2+(z-x)^{2}, \\ z^{2}=3+(x-y)^{2} .\end{array}\right.$ | [Solution] Rewrite the original system of equations as
$$\left\{\begin{array}{l}
6=x^{2}-(y-z)^{2}=(x-y+z)(x+y-z), \\
2=y^{2}-(z-x)^{2}=(y-z+x)(y+z-x), \\
3=z^{2}-(x-y)^{2}=(z-x+y)(z+x-y) .
\end{array}\right.$$
(1) $\times$ (2) $\times$ (3) gives
$$36=(x-y+z)^{2}(y-z+x)^{2}(z-x+y)^{2}$$
Thus, we have $\quad(x-y+z)(y-z... | \begin{array}{l}
x_{1}=\frac{5}{2}, y_{1}=\frac{3}{2}, z_{1}=2 \\
x_{2}=\frac{-5}{2}, y_{2}=\frac{-3}{2}, z_{2}=-2
\end{array} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,544 |
4・ 130 Solve the system of equations
$$\left\{\begin{array}{l}
x^{2}=a+(y-z)^{2} \\
y^{2}=b+(z-x)^{2} \\
z^{2}=c+(x-y)^{2}
\end{array}\right.$$ | [Solution] Move all variables in the system of equations to the left and factorize, we get
$$\left\{\begin{array}{l}
(x-y+z)(x+y-z)=a, \\
(x+y-z)(-x+y+z)=b, \\
(-x+y+z)(x-y+z)=c .
\end{array}\right.$$
If \( a b c > 0 \), multiplying the above three equations and taking the square root, we get
$$(x-y+z)(x+y-z)(-x+y+z)=... | \begin{array}{l}
x= \pm \frac{b+c}{2 b c} \sqrt{a b c} \\
y= \pm \frac{a+c}{2 a c} \sqrt{a b c} \\
z= \pm \frac{a+b}{2 a b} \sqrt{a b c}
\end{array} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,545 |
4・131 If
$$\left\{\begin{array}{l}
a^{2}+b^{2}=1, \\
c^{2}+d^{2}=1, \\
a c+b d=0
\end{array}\right.$$
Find the value of $a b+c d$. | [Solution] Multiply both sides of equation (3) by $a d+b c$, we get
$$(a c+b d)(a d+b c)=0$$
which is
$$a b\left(c^{2}+d^{2}\right)+c d\left(a^{2}+b^{2}\right)=0$$
From (1) and (2), we have $a b+c d=0$. | a b+c d=0 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,546 |
4・132 There are three unknowns $x, y, z$ satisfying the following equations
$$\left\{\begin{array}{l}
x+y+z=a \\
x^{2}+y^{2}+z^{2}=b^{2} \\
x^{-1}+y^{-1}+z^{-1}=c^{-1}
\end{array}\right.$$
Determine the value of $x^{3}+y^{3}+z^{3}$. | [Solution] (1) ${ }^{2}-(2)$ gives
$$x y+y z+z x=\frac{1}{2}\left(a^{2}-b^{2}\right) .$$
(3) gives $x y z=c(x y+y z+z x)=\frac{c\left(a^{2}-b^{2}\right)}{2}$.
Therefore, $x^{3}+y^{3}+z^{3}$
$$\begin{array}{l}
=(x+y+z)^{3}-3(x+y+z)(x y+y z+z x)+3 x y z \\
=a^{3}-3 a \cdot \frac{1}{2}\left(a^{2}-b^{2}\right)+3 \cdot \fr... | a^{3}+\frac{3}{2}\left(a^{2}-b^{2}\right)(c-a) | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,547 |
4・133 Let $x_{1}, x_{2}, \cdots, x_{7}$ be real numbers satisfying the following equations:
$$\begin{array}{l}
\sum_{k=1}^{7} k^{2} x_{k}=1 \\
\sum_{k=1}^{7}(k+1)^{2} x_{k}=12 \\
\sum_{k=1}^{7}(k+2)^{2} x_{k}=123
\end{array}$$
Find the value of $\sum_{k=1}^{7}(k+3)^{2} x_{k}$. | [Solution] Let $a \cdot k^{2}+b(k+1)^{2}+c \cdot(k+2)^{2}=(k+3)^{2}$. By equating the corresponding coefficients, we get
$$\left\{\begin{array}{l}
a+b+c=1 \\
2 b+4 c=6 \\
b+4 c=9
\end{array}\right.$$
Solving these equations, we get $a=1, b=-3, c=3$.
Thus, (1) + (2) $\times(-3)+(3) \times 3$ gives
$$\sum_{k=1}^{7}(k+3)... | 334 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,548 |
$4 \cdot 134$ Given the system of equations
$$\left\{\begin{array}{l}
a_{11} x_{1}+a_{12} x_{2}+a_{13} x_{3}=0 \\
a_{21} x_{1}+a_{22} x_{2}+a_{23} x_{3}=0 \\
a_{31} x_{1}+a_{32} x_{2}+a_{33} x_{3}=0
\end{array}\right.$$
The coefficients satisfy the following conditions:
(1) $a_{11}, a_{22}$, and $a_{33}$ are positive;... | [Proof] Let $x_{1}, x_{2}, x_{3}$ be a set of solutions, and $\left|x_{1}\right| \geqslant\left|x_{2}\right| \geqslant\left|x_{3}\right|$, then
$$\begin{aligned}
& \left|a_{11} x_{1}+a_{12} x_{2}+a_{13} x_{3}\right| \\
\geqslant & \left|a_{11} x_{1}\right|-\left|a_{12} x_{2}\right|-\left|a_{13} x_{3}\right| \\
\geqslan... | proof | Algebra | proof | Yes | Yes | inequalities | false | 735,549 |
4・135 Find all real or complex roots of the system of equations
$$\left\{\begin{array}{l}
x+y+z=3 \\
x^{2}+y^{2}+z^{2}=3 \\
x^{5}+y^{5}+z^{5}=3
\end{array}\right.$$ | [Solution] Let $x, y, z$ be the roots of the cubic equation $r^{3}-a r^{2}+b r-c=0$. Then, by Vieta's formulas, we have: $a=3, b=3$, and
$$x^{n+3}-a x^{n+2}+b x^{n+1}-c x^{n}=0 \quad (n \text{ is a natural number})$$
This also holds for $y$ and $z$. Thus, we have:
$$\begin{array}{l}
x^{n+3}-a x^{n+2}+b x^{n+1}-c x^{n}... | x=y=z=1 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,550 |
4. 136 Positive numbers $x, y, \approx$ satisfy the system of equations
$$\left\{\begin{array}{l}
x^{2}+x y+\frac{y^{2}}{3}=25 \\
\frac{y^{2}}{3}+z^{2}=9 \\
z^{2}+z x+x^{2}=16
\end{array}\right.$$
Find the value of $x y+2 y z+3 z x$. | [Solution] Take point $M$, and draw line segments $M A, M B$, $M C$, such that $\angle A M B=90^{\circ}, \angle B M C=150^{\circ}$, $\angle C M A=120^{\circ}$. If $M A=z$,
$$\begin{array}{l}
M B=\frac{\sqrt{3}}{3} y, \\
M C=x
\end{array}$$
Then by the cosine rule we get
$$\begin{aligned}
A B= & \sqrt{M A^{2}+M B^{2}} ... | 24 \sqrt{3} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,551 |
4・140 Solve the system of equations
$$\left\{\begin{array}{l}
x y z=x+y+z, \\
y z t=y+z+t, \\
z t x=z+t+x, \\
t x y=t+x+y .
\end{array}\right.$$ | [Solution] (1) - (2) gives $y z(x-t)=x-t$, so we have $y z=1$ or $x=t$.
If $y z=1$, then from (1), we get $x=x+y+z$, thus $y+z=0$, but the system of equations $y \cdot z=1$ and $y+z=0$ has no real solutions, so it must be that $x=t$.
Substituting $x=t$ into (3), we get $z=\frac{2 x}{x^{2}-1}$,
Substituting $x=t$ into... | (0,0,0,0),(\sqrt{3}, \sqrt{3}, \sqrt{3}, \sqrt{3}),(-\sqrt{3},-\sqrt{3},-\sqrt{3},-\sqrt{3}) | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,553 |
4・141 Solve the system of equations
$$\left\{\begin{array}{l}
x_{5} + x_{2} = y x_{1}, \\
x_{1} + x_{3} = y x_{2}, \\
x_{2} + x_{4} = y x_{3}, \\
x_{3} + x_{5} = y x_{4}, \\
x_{4} + x_{1} = y x_{5}.
\end{array}\right.$$
where $y$ is a parameter. | [Solution]
$$\left\{\begin{array}{l}
x_{5}+x_{2}=y x_{1} \\
x_{1}+x_{3}=y x_{2} \\
x_{2}+x_{4}=y x_{3} \\
x_{3}+x_{5}=y x_{4} \\
x_{4}+x_{1}=y x_{5}
\end{array}\right.$$
Obviously, for any value of $y$,
$$x_{1}=x_{2}=x_{3}=x_{4}=x_{5}=0$$
is a solution to the system of equations.
Next, we seek the "non-trivial soluti... | \begin{array}{l}
\text{(i) When } y^{2}+y-1 \neq 0 \text{ and } y \neq 2, \text{ the system has only the trivial solution} \\
x_{1}=x_{2}=x_{3}=x_{4}=x_{5}=0 \\
\text{(ii) When } y=2, \text{ the system has the solution } x_{1}= | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,554 |
4・142 Solve the system of equations
$$\left\{\begin{array}{l}
x_{1} \cdot x_{2} \cdot x_{3}=x_{1}+x_{2}+x_{3}, \\
x_{2} \cdot x_{3} \cdot x_{4}=x_{2}+x_{3}+x_{4}, \\
x_{3} \cdot x_{4} \cdot x_{5}=x_{3}+x_{4}+x_{5}, \\
\cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \\
x_{1985} \cdot x_{1986} \cdot... | [Solution] (1) - (2), we get $x_{2} \cdot x_{3}\left(x_{1}-x_{4}\right)=x_{1}-x_{4}$. Therefore, $x_{2} x_{3}=1$ or $x_{1}=x_{4}$.
If $x_{2} \cdot x_{3}=1$, then from (1) we get $x_{1}=x_{1}+x_{2}+x_{3}$, thus $x_{2}+x_{3}=0$, but the system of equations $x_{2} \cdot x_{3}=1$ and $x_{2}+x_{3}=0$ has no real solutions.... | x=0, \pm \sqrt{3} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,555 |
4・143 Solve the system of equations
$$\left\{\begin{array}{l}
x-y+z=1, \\
y-z+u=2, \\
z-u+v=3, \\
u-v+x=4, \\
v-x+y=5
\end{array}\right.$$ | [Solution] (2) $+(3)-(5)$ gives $x=0$,
(3) + (4) - (1) gives
$y=6$,
(4) + (5) - (2) gives
$z=7$
(5) + (1) - (3) gives
$u=3$,
(1) + (2) - (4) gives
$v=-1$.
Upon verification, the solution to the original system of equations is
$$\left\{\begin{array}{l}
x=0 \\
y=6 \\
z=7 \\
u=3 \\
v=-1
\end{array}\right.$$ | \left\{\begin{array}{l}
x=0 \\
y=6 \\
z=7 \\
u=3 \\
v=-1
\end{array}\right.} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,556 |
4. 144 If $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$ satisfy the following system of equations
$$\left\{\begin{array}{l}
2 x_{1}+x_{2}+x_{3}+x_{4}+x_{5}=6, \\
x_{1}+2 x_{2}+x_{3}+x_{4}+x_{5}=12, \\
x_{1}+x_{2}+2 x_{3}+x_{4}+x_{5}=24, \\
x_{1}+x_{2}+x_{3}+2 x_{4}+x_{5}=48 \\
x_{1}+x_{2}+x_{3}+x_{4}+2 x_{5}=96 .
\end{array}\rig... | [Solution] Adding up the five equations in the given system and then dividing by 6, we get
$$x_{1}+x_{2}+x_{3}+x_{4}+x_{5}=31 \text {. }$$
Subtracting (1) from the fourth equation in the system yields $x_{4}=17$. Subtracting (1) from the fifth equation in the system yields $x_{5}=65$.
Therefore, $3 x_{4}+2 x_{5}=3 \ti... | 181 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,557 |
4・145 Solve the system of equations
$$\left\{\begin{array}{l}
x_{1}+x_{2}+x_{3}=6, \\
x_{2}+x_{3}+x_{4}=9, \\
x_{3}+x_{4}+x_{5}=3, \\
x_{4}+x_{5}+x_{6}=-3, \\
x_{5}+x_{6}+x_{7}=-9, \\
x_{6}+x_{7}+x_{8}=-6, \\
x_{7}+x_{8}+x_{1}=-2, \\
x_{8}+x_{1}+x_{2}=2 .
\end{array}\right.$$ | [Solution] Add all 8 equations, we get
$$\begin{aligned}
& 3\left(x_{1}+x_{2}+x_{3}+\cdots+x_{8}\right)=0 \\
\therefore \quad & x_{1}+x_{2}+x_{3}+\cdots+x_{8}=0 .
\end{aligned}$$
(1) + (4) + (7) - (9), we get $x_{1}=1$.
Similarly, we can get $x_{2}=2, x_{3}=3, x_{4}=4, x_{5}=-4$,
$$x_{6}=-3, x_{7}=-2, x_{8}=-1 .$$ | x_{1}=1, x_{2}=2, x_{3}=3, x_{4}=4, x_{5}=-4, x_{6}=-3, x_{7}=-2, x_{8}=-1 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,558 |
4. 146 Solve the system of equations
$$\left\{\begin{array}{l}
\frac{y z}{y+z}=a \\
\frac{x z}{x+z}=b \\
\frac{x y}{x+y}=c
\end{array}\right.$$ | [Solution] Let $a \neq 0, b \neq 0, c \neq 0$, note that $x \neq 0, y \neq 0, z \neq 0$. Therefore, the original system of equations can be rewritten as
$$\left\{\begin{array}{l}
\frac{1}{y}+\frac{1}{z}=\frac{1}{a}, \\
\frac{1}{x}+\frac{1}{z}=\frac{1}{b}, \\
\frac{1}{x}+\frac{1}{z}=\frac{1}{c} .
\end{array}\right.$$
A... | \left\{\begin{array}{l}
x=\frac{2 a b c}{a c+a b-b c} \\
y=\frac{2 a b c}{a b+b c-a c} \\
z=\frac{2 a b c}{a c+b c-a b}
\end{array}\right.
} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,559 |
1.56 Find the largest natural number with the following property: its last digit is not 0, and after deleting one of its digits (but not the first digit), the original number is an integer multiple of the resulting number.
Translate the above text into English, please retain the original text's line breaks and format,... | [Solution] Let the original number be $b$ times the obtained number, where $b$ is a positive integer. Then, we have
$$\begin{aligned}
& \overline{a_{k} \cdots a_{l+1}} \cdot 10^{l}+a_{l} \cdot 10^{l-1}+\overline{a_{l-1} \cdots a_{1}} \\
= & b\left(\overline{a_{k} \cdots a_{l+1}} \cdot 10^{l-1}+\overline{a_{l-1} \cdots ... | 180625 | Geometry | math-word-problem | Yes | Yes | inequalities | false | 735,560 |
4・147 Solve the system of equations
$$\left\{\begin{array}{l}
\frac{x_{2} x_{3} x_{4} \cdots x_{n}}{x_{1}}=a_{1} \\
\frac{x_{1} x_{3} x_{4} \cdots x_{n}}{x_{2}}=a_{2} \\
\cdots \cdots \cdots \cdots \\
\frac{x_{1} x_{2} x_{3} \cdots x_{n-1}}{x_{n}}=a_{n}
\end{array}\right.$$ | [Solution] First, consider the case where all $a_{i}>0$:
(1) If all $x_{k}(k=1,2, \cdots, n)$ are greater than 0.
Multiply the left and right sides of all equations, we get
Thus,
$$\left(x_{1} x_{2} x_{3} \cdots x_{n}\right)^{n-2}=a_{1} a_{2} a_{3} \cdots a_{n}$$
Rewrite the $k$-th equation as
$$a_{k} x_{k}^{2}=x_{1}... | x_{k}=\sqrt{\frac{\sqrt[n-2]{a_{1} a_{2} \cdots a_{n}}}{a_{k}}}, k=1,2, \cdots, n | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,561 |
4. 148 Given
$$\left\{\begin{array}{l}
\frac{x^{2}}{2^{2}-1^{2}}+\frac{y^{2}}{2^{2}-3^{2}}+\frac{z^{2}}{2^{2}-5^{2}}+\frac{\omega^{2}}{2^{2}-7^{2}}=1, \\
\frac{x^{2}}{4^{2}-1^{2}}+\frac{y^{2}}{4^{2}-3^{2}}+\frac{z^{2}}{4^{2}-5^{2}}+\frac{\omega^{2}}{4^{2}-7^{2}}=1, \\
\frac{x^{2}}{6^{2}-1^{2}}+\frac{y^{2}}{6^{2}-3^{2}}... | [Solution] First, we notice that: $x, y, z, \omega$ satisfying the given system of equations is equivalent to $t=$ $4,16,36,64$ satisfying the equation
$$\frac{x^{2}}{t-1}+\frac{y^{2}}{t-9}+\frac{z^{2}}{t-25}+\frac{\omega^{2}}{t-49}=1 .$$
Clearing the denominators and rearranging terms, we get
$$\begin{array}{l}
(t-1)... | 36 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,562 |
4. 149 For real numbers $a, b, c$, and $a \neq 0$, consider the system of equations for $x_{1}, x_{2}, \cdots, x_{n}$
$$\left\{\begin{array}{l}
a x_{1}^{2}+b x_{1}+C=x_{2}, \\
a x_{2}^{2}+b x_{2}+C=x_{3}, \\
\cdots \cdots \cdots \cdots \cdots \cdots \\
a x_{n-1}^{2}+b x_{n-1}+C=x_{n}, \\
a x_{n}^{2}+b x_{n}+C=x_{1} .
\... | [Solution] $\because \sum_{i=1}^{n}\left[a x_{i}^{2}+(b-1) x_{i}+C\right]=0$,
i.e., $\square$
$$a \sum_{i=1}^{n}\left[\left(x_{i}+\frac{b-1}{2 a}\right)^{2}-\frac{\Delta}{4 a^{2}}\right]=0 .$$
$\therefore$ When $\Delta > 0$, there are clearly two distinct sets of solutions,
$$\begin{array}{l}
x_{1}=x_{2}=\cdots=x_{n}=\... | proof | Algebra | proof | Yes | Yes | inequalities | false | 735,563 |
4・150 Solve the system of equations
$$\left\{\begin{array}{l}
\left|a_{1}-a_{2}\right| x_{2}+\left|a_{1}-a_{3}\right| x_{3}+\left|a_{1}-a_{4}\right| x_{4}=1, \\
\left|a_{2}-a_{1}\right| x_{1}+\left|a_{2}-a_{3}\right| x_{3}+\left|a_{2}-a_{4}\right| x_{4}=1, \\
\left|a_{3}-a_{1}\right| x_{1}+\left|a_{3}-a_{2}\right| x_{2... | [Solution] We notice that in the system of equations, if we swap the subscript $i$ with $j$, and $j$ with the equations remain unchanged. Therefore, without loss of generality, we can assume $a_{1}>a_{2}>a_{3}>a_{4}$. The system of equations then becomes:
$$\left\{\begin{array}{l}
\left(a_{1}-a_{2}\right) x_{2}+\left(a... | x_{j}=x_{k}=0, x_{i}=x_{l}=\frac{1}{a_{i}-a_{l}} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,564 |
4. $151 n^{2}$ numbers $x_{i j}(i, j=1,2, \cdots, n)$ satisfy the following $n^{3}$ equations:
$$x_{i j}+x_{j k}+x_{k i}=0(i, j, k=1,2, \cdots, n)$$
Prove that there exist numbers $a_{1}, a_{2}, \cdots, a_{n}$ such that for any $i, j$, we have $x_{i j}=a_{i}-a_{j}$. | [Proof] Let $i=j=k$, then we get $3 x_{i i}=0$, hence $\quad x_{i i}=0(i=1,2, \cdots, n)$.
Let $i=j$, then we get $x_{i i}+x_{i k}+x_{k i}=0$, hence $\quad x_{i k}=-x_{k i}(i, k=1,2, \cdots, n)$.
Now fix $i, j$, and add the $n$ equations
$$x_{i j}+x_{j k}+x_{k i}=0,(k=1,2, \cdots, n)$$
We get
$$n x_{i j}=S_{i}-S_{j}$... | proof | Algebra | proof | Yes | Yes | inequalities | false | 735,565 |
4-152 Given a system of $p$ equations with $q=2 p$ unknowns:
$$\left\{\begin{array}{l}
a_{11} x_{1}+a_{12} x_{2}+\cdots+a_{1 q} x_{q}=0 \\
a_{21} x_{1}+a_{22} x_{2}+\cdots+a_{2 q} x_{q}=0 \\
\cdots \cdots \cdots \cdots \cdots \\
a_{p 1} x_{1}+a_{p 2} x_{2}+\cdots+a_{p q} x_{q}=0
\end{array}\right.$$
where $a_{i j} \in... | [Proof] If for every $i$, the value of $x_{i}$ belongs to $\{0, \pm 1, \pm 2, \cdots, \pm p\}$, then there are $2 p+1$ possible choices. By the multiplication principle, the number of $n$-tuples $\left(x_{1}, x_{2}, \cdots, x_{q}\right)$ that satisfy the above conditions is $(2 p+1)^{q}$.
If $a_{i j} \in\{-1,0,1\},\le... | proof | Algebra | proof | Yes | Yes | inequalities | false | 735,566 |
4・153 Solve the system of equations
$$\left\{\begin{array}{l}
\sin x + 2 \sin (x + y + z) = 0 \\
\sin y + 3 \sin (x + y + z) = 0 \\
\sin z + 4 \sin (x + y + z) = 0
\end{array}\right.$$ | [Solution] (1) + (2) - (3) gives
$$\sin x+\sin y-\sin z+\sin (x+y+z)=0$$
Using sum-to-product identities, we get
$$\begin{array}{l}
2 \sin \frac{x+y}{2} \cos \frac{x-y}{2}+2 \cos \left(\frac{x+y}{2}+z\right) \sin \frac{x+y}{2}=0, \\
\sin \frac{x+y}{2} \cos \frac{x+z}{2} \cos \frac{y+z}{2}=0 .
\end{array}$$
Thus, we h... | \left\{\begin{array}{l}
x=n \pi \\
y=l \pi \\
z=m \pi
\end{array}\right.
} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,567 |
4・154 From the system of equations
$$\left\{\begin{array}{l}
a=\cos u+\cos v+\cos w \\
b=\sin u+\sin v+\sin w \\
c=\cos 2 u+\cos 2 v+\cos w \\
d=\sin 2 u+\sin 2 v+\sin 2 w
\end{array}\right.$$
eliminate \( u, v, w \). | [Solution] From the system of equations, we have
$$\begin{array}{l}
a^{2}+b^{2}=3+2[\cos (u-v)+\cos (v-w)+\cos (w-u)], \\
a^{2}-b^{2}=C+2[\cos (u+v)+\cos (v+w)+\cos (w+u)], \\
2 a b=d+2[\sin (u+v)+\sin (v+w)+\sin (w+u)],
\end{array}$$
Therefore, $\left(\frac{a^{2}-b^{2}-c}{2}\right)^{2}+\left(\frac{2 a b-d}{2}\right)^... | \left(a^{2}-b^{2}-c\right)^{2}+(2 a b-d)^{2}=4\left(a^{2}+b^{2}\right) | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,568 |
4・155 To find the minimum value of $n$ for which the following system of equations
$$\left\{\begin{array}{l}
\sin x_{1}+\sin x_{2}+\cdots+\sin x_{n}=0 \\
\sin x_{1}+2 \sin x_{2}+\cdots+n \sin x_{n}=100
\end{array}\right.$$
has a solution. | [Solution] If $n<20$, then we subtract ten times equation (1) from equation (2), obtaining $-9 \sin x_{1}-8 \sin x_{2}-\cdots-\sin x_{9}+\sin x_{11}+2 \sin x_{12}+\cdots+(n-$ 10) $\sin x_{n}$
$$=100 \text {. }$$
The absolute value of the left side of (3) does not exceed $(9+8+\cdots+1) \times 2=90$, so (3) cannot hold.... | 20 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,569 |
4. 156 Solve the system of equations
$$\left\{\begin{array}{l}
\frac{4 x^{2}}{1+4 x^{2}}=y \\
\frac{4 y^{2}}{1+4 y^{2}}=z \\
\frac{4 z^{2}}{1+4 z^{2}}=x
\end{array}\right.$$
for all real solutions, and prove that your solution is correct. | ［Solution］ If $x=y$, then
$$\begin{array}{l}
\frac{4 x^{2}}{1+4 x^{2}}=x, \\
4 x^{2}=x+4 x^{3}, \\
x\left(4 x^{2}-4 x+1\right)=0, \\
x=0 \text { or } \frac{1}{2},
\end{array}$$
Thus, we get $x=y=z=0$ or $x=y=z=\frac{1}{2}$.
If $x>y$, then
$$\begin{array}{l}
\frac{4 z^{2}}{1+4 z^{2}}>\frac{4 x^{2}}{1+4 x^{2}} \\
1-\fra... | (x, y, z)=(0,0,0) \text { and }(x, y, z)=\left(\frac{1}{2}, \frac{1}{2}, \frac{1}{2}\right) | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,570 |
1.57 Find all positive integers $k$, such that the sum of $k$ consecutive positive integers is a cube of a positive integer. | ［Solution］ Let
$$\begin{aligned}
S_{n}(k) & =n+(n+1)+(n+2)+\cdots+(n+k-1) \\
& =\frac{1}{2} k(2 n+k-1)
\end{aligned}$$
Here $n$ is a positive integer. Let
$$t=2 n+k-1$$
Then
$$n=\frac{1}{2}(t-k+1)$$
Thus, we have
$$S_{n}(k)=\frac{1}{2} k t$$
Since $n$ is a positive integer, $t$ and $k$ have different parities, and ... | proof | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 735,571 |
4・157 $w, a, b, c$ are pairwise distinct real numbers, and it is known that there exist real numbers $x, y, z$, satisfying
$$\left\{\begin{array}{l}
x+y+z=1, \\
x a^{2}+y b^{2}+z c^{2}=w^{2}, \\
x a^{3}+y b^{3}+z c^{3}=w^{3}, \\
x a^{4}+y b^{4}+z c^{4}=w^{4} .
\end{array}\right.$$
Express $w$ in terms of $a, b, c$. | [Solution] From the first equation of the system, we know that $x, y, z$ are not all zero. Without loss of generality, assume $z \neq 0$.
From the first equation of the system, we get
$$x=1-(y+z)$$
Substituting the above equation into the last three equations of the system, we have
$$\left\{\begin{array}{l}
y\left(b^... | w=\left\{\begin{array}{l}
-\frac{a b}{a+b}, \text { when } a b+a c+b c=0, \\
-\frac{a b c}{a b+a c+b c}, \text { when } a b+a c+b c \neq 0.
\end{array}\right.} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,572 |
4. 158 Find the integer solutions of the following system of equations
$$\left\{\begin{array}{l}
x^{x+y}=y^{60} \\
y^{x+y}=x^{15}
\end{array}\right.$$ | [Solution] Since $y^{(x+y)^{2}}=x^{15(x+y)}=\left(y^{60}\right)^{15}=y^{990}$. Therefore, $y= \pm 1$ or $(x+y)^{2}=900$, i.e., $x+y= \pm 30$. Substituting $y= \pm 1$ into the second equation of the original system, we have
$$( \pm 1)^{x \pm 1}=x^{15},|x|=1$$
But $x=-1$ is not a solution, while $x=1, y= \pm 1$ are two ... | null | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,573 |
4. 159 Find the integer solutions of the equation $x+y=x^{2}-x y+y^{2}$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly. | [Solution] Transform the original equation into
$$y^{2}-(x+1) y+x^{2}-x=0$$
Since the equation has real roots, then
that is $\square$
$$\Delta=-3(x-1)^{2}+4 \geqslant 0$$
we have $\quad 1-\frac{2}{\sqrt{3}} \leqslant x \leqslant 1+\frac{2}{\sqrt{3}}$.
$$3(x-1)^{2} \leqslant 4$$
Since $x$ is an integer, we have $x=0... | null | Logic and Puzzles | other | Yes | Yes | inequalities | false | 735,574 |
4. 161 Solve the equation $2 \sqrt{x}+\sqrt{2 y}=\sqrt{32}, x, y$ are integers. | [Solution] It is easy to know that $x \geqslant 0, y \geqslant 0$.
From the original equation, we get
$$\sqrt{2 y}=\sqrt{32}-2 \sqrt{x}$$
Squaring both sides and rearranging, we obtain
$$y=16+2 x-8 \sqrt{2 x},$$
Thus, $8 \sqrt{2 x}$ should be an integer. Let it be $a$, then we have
$$2 x=\left(\frac{a}{8}\right)^{2}$... | x=0,2,8; y=16,4,0 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,575 |
4. 162 Find all integer pairs $(x, y)$ that satisfy the equation $x y=20-3 x+y$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | [Solution] The original equation can be transformed into
$$(x-1)(y+3)=17$$
Therefore,
$$\begin{array}{c}
\left\{\begin{array} { l }
{ x - 1 = 1 , } \\
{ y + 3 = 1 7 ; }
\end{array} \quad \text { or } \left\{\begin{array} { l }
{ x - 1 = 1 7 , } \\
{ y + 3 = 1 ; }
\end{array} , \text { or } \left\{\begin{array}{l}
x-... | null | Logic and Puzzles | other | Yes | Yes | inequalities | false | 735,576 |
4. 163 Find any set of positive integers $x$ and $y$ that satisfy the equation $x^{2}-51 y^{2}=1$.
Translate the text above into English, please keep the original text's line breaks and format, and output the translation result directly. | [Solution] The original equation can be transformed into
$$x^{2}=(7 y+1)^{2}+2 y(y-7)$$
Let $y=7$, we can get $x=50$.
Therefore, $x=50, y=7$ is a positive integer solution to the original equation. | null | Logic and Puzzles | other | Yes | Yes | inequalities | false | 735,577 |
4. 164 Solve the integer equation $x^{2} y+x y^{2}=30$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | ［Solution］ The original equation is transformed into
$$x y(x+y)=2 \times 3 \times 5$$
By symmetry, assume $|x| \leqslant|y|$, so we have $|x| \leqslant 5$. Let $x=1$, then equation (1) becomes $y(1+y)=30$. Solving this, we get $y=5$ or -6.
Let $x=2$, then equation (1) becomes $2 y(2+y)=30$, solving this, we get $y=3$... | (1,5) ;(5,1) ;(1,-6) ;(-6,1) ;(2,3) ;(3,2) ;(2,-5) ;(-5,2) ;(3,-5) ;(-5,3) ;(5,-6) ;(-6,5) | Logic and Puzzles | math-word-problem | Yes | Yes | inequalities | false | 735,578 |
4. 165 Find the integer solution of the equation $\underbrace{\sqrt{x+\sqrt{x+\sqrt{x+\cdots+\sqrt{x}}}}}_{1964 \text { terms }}=y$.
untranslated part:
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。
translated part:
Find the integer solution of the equation $\underbrace{\sqrt{x+\sqrt{x+\sqrt{x+\cdots+\sqrt{x}}}}}_{1964 \text... | [Solution] Let $x_{0}, y_{0}$ be integers that satisfy the equation. Then
$$\underbrace{\sqrt{x_{0}+\sqrt{x_{0}+\sqrt{x_{0}+\cdots+\sqrt{x_{0}}}}}}_{1964 \text { times }}=y_{0}$$
Squaring and rearranging terms, we get
$$\underbrace{\sqrt{x_{0}+\sqrt{x_{0}+\cdots+\sqrt{x_{0}}}}}_{1963 \text { times }}=y_{0}^{2}-x_{0}.$... | x=0, y=0 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,579 |
4- 166 Find all integer solutions $x, y$ that satisfy $x^{2}+x=y^{4}+y^{3}+y^{2}+y$. | ［Solution］Multiply both sides of the equation by 4, then add 1 to get
i.e. $\square$
$$\begin{array}{c}
(2 x+1)^{2}=\left(2 y^{2}+y\right)^{2}+3 y^{2}+4 y+1 \\
(2 x+1)^{2}=\left(2 y^{2}+y+1\right)^{2}-\left(y^{2}-2 y\right)
\end{array}$$
If $y$ is an integer different from $-1, 0, 1$, and 2, then
$$3 y^{2}+4 y+1>0 \t... | null | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,580 |
4. 167 Find the positive integer solutions of the equation $x^{3}-y^{3}=x y+61$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | ［Solution］ Clearly $x>y$. Let $x=y+d(d \geqslant 1)$.
Substitute into the original equation to get
$$\begin{array}{l}
(y+d)^{3}-y^{3}=(y+d) y+61 \\
(3 d-1) y^{2}+\left(3 d^{2}-d\right) y+d^{3}=61
\end{array}$$
From (1), we get $d^{3}<61$,
thus $d \leqslant 3$.
If $d=1$, then (1) becomes
$$\begin{array}{l}
2 y^{2}+2 y-... | x=6, y=5 | Logic and Puzzles | math-word-problem | Yes | Yes | inequalities | false | 735,581 |
4. 168 Find the integer solutions of the equation
$$\left(1+\frac{1}{m}\right)^{m+1}=\left(1+\frac{1}{1988}\right)^{1988}$$ | ［Solution］ The original equation can be transformed into
$$\frac{(m+1)^{m+1}}{m^{m+1}}=\frac{1989^{1988}}{1988^{1988}}$$
If $m>0$, then both sides of (1) are irreducible fractions, so it must be that
$$m^{m+1}=1988^{1988}$$
Notice that when $m \geqslant 1988$, $m^{m+1}>1988^{1988}$;
$$01988$ when, $n^{n}>1988^{1988}$... | m=-1989 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,582 |
4・169 Solve the equation in the set of natural numbers
$$x^{y}=y^{x}(x \neq y) .$$ | [Solution] If $x=1$, then $y=1$, which contradicts $x \neq y$. Therefore, $x \geqslant 2$ and $y \geqslant 2$.
Let $x^{y}=y^{x}=a^{d}$, where $a, d$ are natural numbers, and make $d$ as large as possible. Then $a \neq 1$, i.e., $a \geqslant 2$, and there must exist $b, c$ such that
$$x=a^{b}, y=a^{c} .$$
Substituting... | x=2, y=4 \text{ or } x=4, y=2 | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 735,583 |
4. 171 Find all positive integer pairs $(a, b)$ that satisfy the equation $2 a^{2}=3 b^{3}$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | [Solution] Let $(x, y)$ be a pair of positive integers that satisfy the given equation. Then
$$2 x^{2}=3 y^{3}$$
It is clear that $x$ is a multiple of 3. We set $x=3 x_{1}$, substitute and simplify to get
$$2 \cdot 3 \cdot x_{1}^{2}=y^{3}$$
Similarly, we can set $y=2 \cdot 3 \cdot y_{1}$, substitute and simplify to g... | (18 d^{3}, 6 d^{2}) | Logic and Puzzles | other | Yes | Yes | inequalities | false | 735,584 |
4・172 Find all positive integer solutions $(x, y, z, n)$ of the equation
$$x^{2 n+1}-y^{2 n+1}=x y z+2^{2 n+1}$$
satisfying the conditions $n \geqslant 2, z \leqslant 5 \cdot 2^{2 n}$. | [Solution] From the given equation, it is easy to see that $x-y>0$ and $x$ and $y$ have the same parity, so $x-y \geqslant 2$.
When $y=1, x=3$, from the equation we get
$$z=3^{2 n}-\frac{1}{3}\left(1+2^{2 n+1}\right) .$$
To make $z \leqslant 5 \cdot 2^{2 n}$, we should have
$$\begin{aligned}
3^{2 n} & \leqslant 5 \cd... | (3,1,70,2) | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 735,585 |
4. 173 Does the equation $x^{2}+y^{3}=z^{4}$ in $x, y, z$ have prime solutions? | [Solution] Transform the original equation into
$$y^{3}=\left(z^{2}-x\right)\left(z^{2}+x\right) .$$
Since $y$ is a prime number, we get
$$\left\{\begin{array}{l}
z^{2}-x=1 \\
z^{2}+x=y^{3}
\end{array}\right.$$
or $\left\{\begin{array}{l}z^{2}-x=y, \\ z^{2}+x=y^{2}\end{array}\right.$.
From (1), we get $x=(z-1)(z+1)$,... | proof | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 735,586 |
4. 174 Find the integer solutions of the following equation
$$\left[\frac{x}{1!}\right]+\left[\frac{x}{2!}\right]+\cdots+\left[\frac{x}{10!}\right]=1001$$ | [Solution] From the equation, we know that $x$ is a natural number not exceeding 1001, so $x<6$!, and thus for the terms $\left[\frac{x}{n!}\right]$ when $n \geqslant 6$, they can be omitted. Each $x<6!$ can be uniquely represented as
$$x=a \cdot 5!+b \cdot 4!+c \cdot 3!+d \cdot 2!+k \cdot 1!$$
where $a, b, c, d, k$ ar... | 584 | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 735,587 |
4-175 Find all possible integer values of $x, y$ that satisfy the condition $\frac{x+y}{x^{2}-x y+y^{2}}=\frac{3}{7}$
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | ［Solution］Let integers $x, y$ satisfy the given conditions, then we have
$$7(x+y)=3\left(x^{2}-x y+y^{2}\right)$$
Let $p=x+y, q=x-y$,
then $x=\frac{p+q}{2}, y=\frac{p-q}{2}$.
Substituting into (1) yields $28 p=3\left(p^{2}+3 q^{2}\right)$.
From this, we can deduce that $3 \mid p$, and $p$ is a non-negative integer. Le... | (5,4) \text{ or } (4,5) | Logic and Puzzles | math-word-problem | Yes | Yes | inequalities | false | 735,588 |
$1 \cdot 59$ Determine all positive rational triples $(x, y, z)$ such that $x+y+z, \frac{1}{x}+\frac{1}{y}+\frac{1}{z}, x y z$ are all integers. (Here $x \leqslant y \leqslant z$. ) | 【Solution】Since
$$x y+y z+z x=x y z\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)$$
Therefore, $x y+y z+z x$ is also an integer.
Let
$$\begin{aligned}
f(u) & =(u-x)(u-y)(u-z) \\
& =u^{3}-(x+y+z) u^{2}+(x y+y z+z x) u-x y z
\end{aligned}$$
Then $f(u)$ is a cubic polynomial with integer coefficients. Since $x, y, z$ ... | null | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 735,589 |
4・177 Prove that for each natural number $n$, the equation
$(x+\sqrt{3} y)^{n}=\sqrt{1+\sqrt{3}}$ has no rational solutions $x$ and $y$.
| [Proof] Suppose there exist rational numbers $x$ and $y$ satisfying the equation
$$x+\sqrt{3} y=\sqrt{1+\sqrt{3}}$$
Then $\quad(x+\sqrt{3} y)^{2}=1+\sqrt{3}$,
which simplifies to $x^{2}+3 y^{2}-1-(1-2 x y) \sqrt{3}$.
Since $x$ and $y$ are rational numbers and $\sqrt{3}$ is irrational, $x$ and $y$ must satisfy the syst... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 735,590 |
4. 178 (a) Let $n$ be a positive integer. Prove: there exist distinct positive integers $x, y$, $z$, such that
$$x^{n-1}+y^{n}=z^{n+1}$$
(b) Let $a, b, c$ be positive integers, and $a$ and $b$ be coprime, $c$ be coprime with either $a$ or $b$. Prove: there exist infinitely many distinct positive integer triples $(x, y,... | (a) Notice that $1+3=2^{2}$, so we can let
$$\left\{\begin{array}{l}
x=2^{n^{2}} \cdot 3^{n+1} \\
y=2^{n(n-1)} \cdot 3^{n} \\
z=2^{n^{2}-2 n+2} \cdot 3^{n-1}
\end{array}\right.$$
They satisfy the equation $x^{n}+y^{n}=z^{n+1}$.
(b) Let $P \in \mathbb{N}, P \geqslant 3, Q=P^{c}-1>1$. If the given equation
$$x^{a}+y^{b}... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 735,591 |
4. 180 Determine all quadruples $\left(p_{1}, p_{2}, p_{3}, p_{4}\right)$, where $p_{1}, p_{2}, p_{3}, p_{4}$ are all prime numbers, and satisfy:
(1) $p_{1}<p_{2}<p_{3}<p_{4}$;
(2) $p_{1} p_{2}+p_{2} p_{3}+p_{3} p_{4}+p_{4} p_{1}=882$. | ［Solution］ Rewrite condition (2) as
$$\left(p_{1}+p_{3}\right)\left(p_{2}+p_{4}\right)=2 \times 3^{2} \times 7^{2},$$
The right side of this equation cannot be divisible by 4, so one of the factors on the left side must be odd, thus
$$p_{1}=2, p_{1}+p_{3} \text { is an odd factor of } 882.$$
Since $p_{1}+p_{3}<p_{2}+... | (2,5,19,37),(2,11,19,31),(2,13,19,29) | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 735,593 |
4. 181 Find all integer pairs $(a, b)$, where $a \geqslant 1, b \geqslant 1$, and satisfy the equation $a^{b^{2}}=b^{a}$ | ［Solution］ If $a=1$, then from the equation we get $b=1$.
If $b=1$, then from the equation we get $a=1$.
If $a, b$ are both not less than 2, at this point, we let
$$t=\frac{b^{2}}{a}$$
Then from the equation in the problem, we get
$$\dot{a}^{a t}=(a t)^{\frac{a}{2}},$$
which means $\square$
$$\begin{array}{l}
a^{2 t}... | (a, b)=(1,1),(16, 2), (27,3) | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 735,594 |
4. 182 Find all positive integer solutions $(x, y)$ of the following equation: $x^{2}+y^{2}-5 x y+5=0$. | [Solution] Clearly, $x \neq y$. Let
$$\begin{array}{l}
S=\left\{(x, y) \mid x^{2}+y^{2}-5 x y+5=0, x, y \in \mathbb{N}, y>x\right\} \\
S^{\prime}=\left\{(x, y) \mid x^{2}+y^{2}-5 x y+5=0, x, y \in \mathbb{N}, x>y\right\}
\end{array}$$
Then $S \cup S^{\prime}$ is the set of all positive integer solutions to the origina... | null | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,595 |
4. 183 Find the number of real pairs $(a, b)$ that make the system of equations $\left\{\begin{array}{l}a x+b y=1, \\ x^{2}+y^{2}=50 .\end{array}\right.$ have at least one solution, and all solutions are integer solutions. | [Solution] The graph of the equation $x^{2}+y^{2}=50$ is a circle with center at $(0,0)$ and radius $5 \sqrt{2}$, and the graph of $a x+b y=1$ is a straight line. The problem is equivalent to finding the number of lines that intersect the circle and have each intersection point as an integer point.
There are only 12 i... | 72 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,596 |
4・184 For any three positive integers $x, y, z$, let
$$f(x, y, z)=[1+2+3+\cdots+(x+y-2)]-z$$
Find all positive integer quadruples $(a, b, c, d)$, such that
$$f(a, b, c)=f(c, d, a)=1993$$ | [Solution] Let $(a, b, c, d)$ be the required positive integer tuple. Then, by the problem statement, we have
$$\left\{\begin{array}{l}
f(a, b, c)=1993 \\
f(c, d, a)=1993
\end{array}\right.$$
Since
$$f(x, y, z)=\frac{1}{2}(x+y-2)(x+y-1)-z,$$
we have
$$\left\{\begin{array}{l}
\frac{1}{2}(a+b-2)(a+b-1)-c=1993 \\
\frac{... | (23,42,23,42) | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,597 |
4-186 $p$ is an integer, try to prove:
$$x^{2}-2 x-\left(10 p^{2}+10 p+2\right)=0$$
has no integer solutions. | [Solution] Transform the original equation into
$$x(x-2)=2[5 p(p+1)+1] .$$
It is easy to see that $x$ is even, so the left side is a multiple of 4. However, the right side is not a multiple of 4, hence the original equation has no integer solutions. | proof | Algebra | proof | Yes | Yes | inequalities | false | 735,599 |
4. 188 Find all three-digit numbers that satisfy the following condition: the quotient when divided by 11 equals the sum of the squares of its digits. | [Solution] Let the required three-digit number be
$$n=100 a+10 b+c$$
where $b, c \in\{0,1,2,3, \cdots, 9\}, a \in\{1,2, \cdots, 9\}$.
The necessary and sufficient condition for $n$ to be divisible by 11 is that $a-b+c$ is divisible by 11. Since
$$-8 \leqslant a-b+c \leqslant 18$$
Thus, only $\square$
$$\begin{array}{... | 550 \text{ and } 803 | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 735,601 |
4・189 There is an inaccurate double-pan balance with unequal arm lengths and unequal pan weights. Three objects of different weights, $A$, $B$, and $C$, are weighed one by one. When they are placed on the left pan, their weights are measured as $A_{1}, B_{1}, C_{1}$; when $A$ and $B$ are placed on the right pan, their ... | [Solution] Let the left arm length of the balance be $l_{1}$, and the right arm length be $l_{2}$; the weight of the left pan be $m_{1}$, and the right pan be $m_{2}$. According to the problem, we have
$$\left\{\begin{array}{l}
\left(A+m_{1}\right) l_{1}=\left(A_{1}+m_{2}\right) l_{2}, \\
\left(B+m_{1}\right) l_{1}=\le... | C=\left(C_{1}-A_{1}\right) \sqrt{\frac{A_{2}-B_{2}}{A_{1}-B_{1}}}+\frac{A_{1} \sqrt{A_{2}-B_{2}}+A_{2} \sqrt{A_{1}-B_{1}}}{\sqrt{A_{1}-B_{1}}+\sqrt{A_{2}-B_{2}}} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,602 |
4. 190 Prove: There does not exist such an integer, that by moving its first digit to the last position, the resulting number is twice the original number. | [Proof] Suppose there exists such an integer, whose first digit is $a$, and the number formed by the remaining digits is $x$, where $x$ has $k$ digits, and
$$10 x + a = 2\left(a \times 10^{k} + x\right)$$
Then
$$8 x = a\left(2 \times 10^{k} - 1\right),$$
which means
$$8 x = \underbrace{199 \cdots 9}_{k} \cdot a$$
It... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 735,603 |
4. 192 Find such a three-digit number, which equals the sum of the factorials of its digits. | [Solution] Let $M=\overline{abc}=100a+10b+c$ be the desired three-digit number. According to the problem, we have $\overline{abc}=a!+b!+c!$
Analyzing both sides of (1), considering
$$4!=24, 5!=120, 6!=720, 7!=5040$$
we can conclude that none of the three digits of $M$ exceed 6. Therefore, $M$ does not exceed 666. This... | 145 | Logic and Puzzles | math-word-problem | Yes | Yes | inequalities | false | 735,604 |
4- 195 Two drops of water fall successively from a steep cliff that is 300 meters high. When the first drop has fallen 0.001 millimeters, the second drop begins to fall. What is the distance between the two drops when the first drop reaches the bottom of the cliff (the answer should be accurate to 0.1 millimeters; air ... | [Solution] Let $d$ be the distance between the two water droplets when the second droplet starts to fall, and $s_{1}$ and $s_{2}$ be the distances traveled by the first and second droplets, respectively, when the first droplet reaches the foot of the mountain. If the distances $d, s_{1}$, and $s_{2}$ are traveled in ti... | 34.6 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,605 |
4. 196 On a horizontal plane, three points are 100 meters, 200 meters, and 300 meters away from the base of an antenna. The sum of the angles of elevation from these three points to the antenna is $90^{\circ}$. What is the height of the antenna? | [Solution] Let the height of the antenna be $x$ meters. The angles of elevation from points 100 meters, 200 meters, and 300 meters away from the base of the antenna are $\alpha, \beta, \gamma$ respectively. Then
$$\operatorname{tg} \alpha=\frac{x}{100}, \operatorname{tg} \beta=\frac{x}{200}, \operatorname{tg} \gamma=\f... | 100 | Geometry | math-word-problem | Yes | Yes | inequalities | false | 735,606 |
4-197 Prove: If $a, b, c$ are the side lengths of a triangle, and $\alpha, \beta, \gamma$ are the angles opposite to them, and
$a(1-2 \cos \alpha)+b(1-2 \cos \beta)+c(1-2 \cos \gamma)=0$, then such a triangle is an equilateral triangle. | [Proof] Let $R$ be the radius of the circumcircle of a triangle. According to the Law of Sines,
$$a=2 R \sin \alpha, b=2 R \sin \beta, c=2 R \sin \gamma$$
The equation given in the problem can be transformed into
That is, $\square$
$$\sin \alpha(1-2 \cos \alpha)+\sin \beta(1-2 \cos \beta)+\sin \gamma(1-2 \cos \gamma)... | proof | Geometry | proof | Yes | Yes | inequalities | false | 735,607 |
4. 198 Prove that the number $u=\operatorname{ctg} 22.5^{\circ}$ is a root of a quadratic equation, and the number $v=\frac{1}{\sin 22.5^{\circ}}$ is a root of a quartic equation, and that the coefficients of both equations are integers, with the leading coefficient equal to 1. | [Proof] The angle $22.5^{\circ}$ is a quarter of a right angle, and can be constructed as follows: on the extension of the right-angle side $CA$ of the isosceles right triangle $ABC$, from vertex $A$ outward, construct a line segment $AD = AB$, and connect points $D$ and $B$. In the isosceles triangle $DAB$, $\angle D ... | proof | Algebra | proof | Yes | Yes | inequalities | false | 735,608 |
4. 199 A car departs from point $O$ along a straight highway, maintaining a constant speed $v$. At the same time the car starts, a person on a bicycle departs from a point that is a distance $a$ from $O$ and a distance $b$ from the highway, aiming to deliver a letter to the car's driver. What is the minimum speed the c... | [Solution] Let $b>0$. If $b=0$, it means the cyclist is on the road, and the solution to the problem is obvious.
Let $M$ be the point where the cyclist is located, and $S$ be the point where the cyclist meets the car. Let $\angle M O S=\alpha, t$ be the time the cyclist spends from departure to meeting, and $x$ repres... | x_{\min }=v \sin \alpha=\frac{v b}{a} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,609 |
$4 \cdot 200$ A dance party has 42 participants. Lady $A_{1}$ danced with 7 male partners, Lady $A_{2}$ danced with 8 male partners $\cdots \cdots$ Lady $A_{n}$ danced with all male partners. How many ladies and male partners are there at the dance party? | [Solution] Let there be $n$ ladies at the ball, so the number of male partners is $42-n$. The $k$-th lady $(1 \leqslant k \leqslant n)$ has danced with $k+6$ male partners. Therefore, the $n$-th lady has danced with $n+6$ male partners. According to the given condition, the number of male partners is $n+6$. Thus, $n+6=... | 18 \text{ ladies and } 24 \text{ male partners} | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 735,610 |
$4 \cdot 201$ (1) How much boiling water should be added to $a$ liters of water at $t_{1} \mathrm{C}$ to obtain water at $t_{2}{ }^{\circ} \mathrm{C}$ $\left(t_{1}<100^{\circ} \mathrm{C}\right)$?
(2) If the approximate measured values are:
$a=3.641$ liters, $t_{1}=36.7^{\circ} \mathrm{C}, t_{2}=57.4^{\circ} \mathrm{C}$... | [Solution] (1) Let the added boiling water be $x$ liters,
then we have $c \cdot x\left(100-t_{2}\right)=c \cdot a\left(t_{2}-t_{1}\right)$ (where $c$ is the specific heat), so $x=\frac{a\left(t_{2}-t_{1}\right)}{100-t_{2}}$.
(2) When $a=3.641$ liters, $t_{1}=36.7^{\circ} \mathrm{C}, t_{2}=57.4^{\circ} \mathrm{C}$, then... | 1.769 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,611 |
4. 202 City A has 157 tons of goods to be transported to City B. The load capacity of a large truck is 5 tons, and that of a small truck is 2 tons. The fuel consumption of a large truck is 20 liters, and that of a small truck is 10 liters. How many large trucks and small trucks should be used to minimize fuel consumpti... | [Solution] The fuel consumption for transporting goods with a large truck is 4 liters/ton, and for a small truck, it is 5 liters/ton. Therefore, to minimize fuel consumption, we must use the large truck as much as possible.
157 divided by 5 gives a quotient of 31 and a remainder of 2, so we use 31 large trucks and 1 sm... | 31 \text{ large trucks and } 1 \text{ small truck} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,612 |
4・203 There are two coal mines, A and B. Coal from mine A releases 4 calories when burned per gram, and coal from mine B releases 6 calories when burned per gram. The price of coal at the origin is: 20 yuan per ton for mine A, and 24 yuan per ton for mine B. It is known that: the transportation cost of coal from mine A... | [Solution] Let the transportation cost per ton from Coal Mine B to City N be $x$ yuan when the cost is the same as Coal Mine A. According to the problem, we have:
$$(20+8):(24+x)=4: 6$$
That is
$$\begin{array}{l}
24+x=\frac{28 \times 6}{4} \\
x=42-24=18 \text { (yuan) }
\end{array}$$
When the transportation cost per ... | 18 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,613 |
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