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3.69 The geometric mean $\mathrm{G} \cdot \mathrm{M}$ of $k$ positive numbers $a_{1}, a_{2}, \cdots, a_{k}$ is defined as the $k$-th root of their product. For example, the G.M of $3,4,18$ is 6. Prove: The $G \cdot M$ of a set $S$ of $n$ positive numbers is equal to the $G \cdot M$ of the $G \cdot M$s of all non-empty ... | [Proof] Consider any fixed element $a_{i}$ of the set $S=\left\{a_{1}, a_{2}, \cdots, a_{n}\right\}$, let $k=$ $1,2, \cdots, n, S$ has $C_{n}^{k}$ $k$-element subsets, of which only $C_{n-1}^{k-1}$ contain $a_{i}$. The $\mathrm{G} \cdot \mathrm{M}$ of each such subset contains the factor $a_{i} \frac{1}{k}$, so the exp... | proof | Algebra | proof | Yes | Yes | inequalities | false | 735,399 |
1.39 Find all natural numbers $n, k$ such that $n^{n}$ has $k$ digits, and $k^{k}$ has $n$ digits. | 〔Solution〕 $n=k$, and $k=1,8$ or 9.
If $n^{n}$ has $k$ digits, and $k^{k}$ has $n$ digits, then
$$\begin{array}{l}
10^{k-1}<n^{n}<10^{k} \\
10^{n-1}<k^{k}<10^{n}
\end{array}$$
Assume without loss of generality that $n \geqslant k$, then we have
$$n<10 \text { and } k<10 \text {. }$$
If $n=1$, then since $n^{n}$ has o... | n=k=1, n=k=8, n=k=9 | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 735,400 |
$3 \cdot 70$ For every positive integer $n$, prove:
$$\begin{aligned}
& {[\sqrt{n}+\sqrt{n+1}]=[\sqrt{4 n+1}]=[\sqrt{4 n+2}] } \\
= & {[\sqrt{4 n+3}] . }
\end{aligned}$$
Here $[x]$ denotes the greatest integer not exceeding $x$. | [Proof] Let $x$ be a positive integer, and $x^{2}>4 n+1$.
If $x$ is even, then
$$x^{2}=4 m>4 n+1$$
Therefore,
$$\begin{array}{l}
m \geqslant n+1 \\
x^{2}=4 m \geqslant 4 n+4>4 n+3
\end{array}$$
If $x$ is odd, then
$$x^{2}=4 m+1>4 n+1$$
Therefore,
$$\begin{array}{c}
m \geqslant n+1 \\
x^{2} \geqslant 4 n+5
\end{array... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 735,401 |
$$\begin{array}{l}
\text { 3. } 72 \text { For all positive integers } n \text {, prove that } \\
\text { } (2 n-1)!-2!(2 n-2)!+\cdots-(2 n-2)!2!+(2 n-1)! \\
=\frac{(2 n)!}{n+1} .
\end{array}$$ | [Proof] This problem is equivalent to proving
$$\frac{1}{C_{2 n}^{1}}-\frac{1}{C_{2 n}^{2}}+\cdots-\frac{1}{C_{2 n}^{2 n-2}}+\frac{1}{C_{2 n}^{2 n-1}}=\frac{1}{n+1}$$
Since the equation $\frac{1}{C_{2 n+1}^{k}}+\frac{1}{C_{2 n+1}^{k+1}}=\frac{2 n-k+1 /}{(2 n+1) C_{2 n}^{k}}+\frac{k+1}{(2 n+1) C_{2 n}^{k}}$
$$=\frac{2 ... | proof | Combinatorics | proof | Yes | Yes | inequalities | false | 735,403 |
3.73 Observation
$$\frac{1}{1}=\frac{1}{2}+\frac{1}{2} ;-\frac{1}{2}=\frac{1}{3}+\frac{1}{6} ; \frac{1}{3}=\frac{1}{4}+\frac{1}{12} ; \frac{1}{4}=\frac{1}{5}+\frac{1}{20} .$$
Inspired by these examples, state the general rule and prove it. For any integer \( n \) greater than 1, prove that there exist integers \( i \)... | [Solution] Conjecture the general rule from the given examples
$$\frac{1}{n}=\frac{1}{n+1}+\frac{1}{n(n+1)}, n=1,2,3, \cdots$$
Since $\frac{1}{n+1}+\frac{1}{n(n+1)}=\frac{n+1}{n(n+1)}=\frac{1}{n}$
Therefore, equation (1) is valid.
Rewrite equation (1) in an equivalent form
$$\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1},... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 735,404 |
3・74 Let $1 \leqslant r \leqslant n$, consider all $r$-element subsets of the set $\{1,2, \cdots, n\}$ and the smallest element in each such subset. Let $F(n, r)$ denote the arithmetic mean of all such smallest numbers. Prove that:
$$F(n, r)=(n+1) /(r+1)$$ | [Proof] The number of $r$-element subsets of the set $\{1,2, \cdots, n\}$ is $C_{n}^{r}$. If the smallest element of an $r$-element subset is $j$, then the remaining $r-1$ elements must be chosen from $\{j+1, \cdots, n\}$. Therefore, the number of $r$-element subsets with the smallest element $j$ is $C_{n-j}^{r-1}$. Th... | proof | Combinatorics | proof | Yes | Yes | inequalities | false | 735,405 |
3.75 Let the sequence $u_{0}, u_{1}, u_{2}, \cdots$ be defined as follows:
$$\begin{array}{l}
u_{0}=2, u_{1}=\frac{5}{2} \\
u_{n+1}=u_{n}\left(u_{n-1}^{2}-2\right)-u_{1}, n=1,2, \cdots
\end{array}$$
Prove that $\left[u_{n}\right]=2^{\frac{2^{n}-(-1)^{n}}{3}}, n=1,2, \cdots$
Here $[x]$ denotes the greatest integer less... | ［Proof］ First, we use mathematical induction to prove: when $n>0$,
$$u_{n}=2^{\frac{2^{n}-(-1)^{n}}{3}}+2^{-\frac{2^{n}-(-1)^{n}}{3}}$$
In fact, by
$$\begin{array}{l}
u_{1}=\frac{5}{2}=2^{1}+2^{-1}=2^{\frac{2^{1}-(-1)^{1}}{3}}+2^{-\frac{2^{1}-(-1)^{1}}{3}} \\
u_{2}=\frac{5}{2}=2^{\frac{2^{2}-(-1)^{2}}{3}}+2^{-\frac{2^... | proof | Algebra | proof | Yes | Yes | inequalities | false | 735,406 |
3・76 Let $n$ be a natural number, try to prove the identity
$$\sum_{k=0}^{n} C_{n}^{k} 2^{k} C_{n-k}^{\left[\frac{n-k}{2}\right]}=C_{2 n+1}^{n}$$ | ［Solution］ Consider the generating function $(x+1)^{2 n+1}$, the coefficient of $x^{n}$ is $C_{2 n+1}^{n}$.
On the other hand, $(x+1)^{2 n+1}=\left(x^{2}+2 x+1\right)^{n}(x+1)$
$$=\left(x^{2}+2 x+1\right)^{n} x+\left(x^{2}+2 x+1\right)^{n}$$
and $\square$
$$\begin{aligned}
& \left(x^{2}+2 x+1\right)^{n} \\
= & \sum_{i... | proof | Combinatorics | proof | Yes | Yes | inequalities | false | 735,407 |
3. 77 Given a sequence of 0s and 1s $x_{1}, x_{2}, \cdots, x_{n}, A$ is the number of triples $\left(x_{i}, x_{j}, x_{k}\right)$ equal to $(0, 1, 0)$ or $(1, 0, 1)$, and $x_{j} \neq x_{i}$.
(1) Prove that: $A=C_{n}^{3}-C_{d_{1}}^{2}-C_{d_{2}}^{2}-\cdots-C_{d_{n}}^{2}$;
(2) For a given odd $n$, what is the maximum value... | [Solution] (1) The number of ternary tuples formed by $x_{1}, x_{2}, \cdots, x_{n}$ is $C_{n}^{3}$.
For each number $x_{i}$, by the problem's condition, the number of numbers equal to it before it and the number of numbers not equal to it after it is $d_{i}$.
Taking any two numbers from these $d_{i}$ numbers to form a... | C_{2 k+1}^{3}-n C_{k}^{2} | Combinatorics | proof | Yes | Yes | inequalities | false | 735,408 |
3.79 Let $A+B+C=\pi, \sin A-\sin B=\sin B-\sin C$, prove: $\operatorname{ctg} \frac{A}{2} \operatorname{ctg} \frac{C}{2}=3$ | [Proof] From $A+B+C=\pi$ we get $2 \sin \frac{A+C}{2} \cos \frac{A+C}{2}=2 \sin \frac{B}{2} \cos \frac{B}{2}$ $=\sin B$,
and from the given, $\sin B=\frac{1}{2}(\sin A+\sin C)$
$$=\sin \frac{A+C}{2} \cos \frac{A-C}{2}$$
Thus, $2 \cos \frac{A+C}{2}=\cos \frac{A-C}{2}$,
which means $2 \cos \frac{A}{2} \cos \frac{C}{2}-2... | proof | Algebra | proof | Yes | Yes | inequalities | false | 735,409 |
$1 \cdot 40$ Let $n$ be a five-digit number (the first digit is not zero), and $m$ be the four-digit number formed by removing the middle digit of $n$. Determine all $n$ such that $\frac{n}{m}$ is an integer. | [Solution] Let $n=\overline{x y z u v}=x \cdot 10^{4}+y \cdot 10^{3}+z \cdot 10^{2}+u \cdot 10+v$. Where $x, y, z, u, v$ are digits $0,1,2, \cdots, 8$ or $9, x \geqslant 1$.
$$m=\overline{x y u v}=x \cdot 10^{3}+y \cdot 10^{2}+u \cdot 10+v$$
We first prove that $9 m<n<11 m$.
In fact, the inequality $9 m<n$ is equivale... | u = x y 000 = N \cdot 10^{3}, 10 \leqslant N \leqslant 99 | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 735,410 |
3・80 Given $A+B+C=180^{\circ}$, prove:
$$\sin ^{2} A+\sin ^{2} B+\sin ^{2} C-2 \cos A \cos B \cos C=2$$ | [Proof] Since $\cos ^{2} A+\cos ^{2} B+\cos ^{2} C$
$$\begin{array}{l}
=\frac{1+\cos 2 A}{2}+\frac{1+\cos 2 B}{2}+\frac{1+\cos 2 C}{2} \\
=\frac{3}{2}+\frac{1}{2}(\cos 2 A+\cos 2 B+\cos 2 C) \\
=\frac{3}{2}+\frac{1}{2}(-4 \cos A \cos B \cos C-1) \\
=1-2 \cos A \cos B \cos C,
\end{array}$$
Therefore,
Rearranging gives... | proof | Algebra | proof | Yes | Yes | inequalities | false | 735,411 |
$3 \cdot 82$ Try to prove:
$$\left(\operatorname{cot} \frac{\theta}{2}-\operatorname{tan} \frac{\theta}{2}\right)\left(1+\operatorname{tan} \theta \operatorname{tan} \frac{\theta}{2}\right)=2 \csc \theta .$$ | [Proof] $\begin{aligned} & \left(\operatorname{ctg} \frac{\theta}{2}-\operatorname{tg} \frac{\theta}{2}\right)\left(1+\operatorname{tg} \theta \operatorname{tg} \frac{\theta}{2}\right) \\ = & \frac{1-\operatorname{tg}^{2} \frac{\theta}{2}}{\operatorname{tg} \frac{\theta}{2}}\left(1+\frac{2 \operatorname{tg}^{2} \frac{\... | proof | Algebra | proof | Yes | Yes | inequalities | false | 735,413 |
$\begin{array}{l}3 \cdot 83 \text { Prove: If } \alpha, \beta, \gamma \text { are the interior angles of a right triangle, then } \\ \sin \alpha \sin \beta \sin (\alpha-\beta)+\sin \beta \sin \gamma \sin (\beta-\gamma)+\sin \gamma \sin \alpha \sin (\gamma-\alpha)+ \\ \sin (\alpha-\beta) \sin (\beta-\gamma) \sin (\gamma... | [Proof] Without loss of generality, let $\alpha=90^{\circ}=\beta+\gamma$, then we have
$$\begin{array}{l}
\sin a=1, \sin (\alpha-\beta)=\sin \gamma \\
\sin (\gamma-\alpha)=-\sin (\alpha-\gamma)=-\sin \beta
\end{array}$$
Therefore,
$$\begin{array}{l}
\sin \alpha \sin \beta \sin (\alpha-\beta)=\sin \beta \sin \gamma \\
... | proof | Geometry | proof | Yes | Yes | inequalities | false | 735,414 |
3.84 Prove: If $n$ is a natural number greater than 1, then
$$\cos \frac{2 \pi}{n}+\cos \frac{4 \pi}{n}+\cos \frac{6 \pi}{n}+\cdots+\cos \frac{2 n \pi}{n}=0 .$$ | [Proof] Establish a Cartesian coordinate system $x o y$ on a plane, and consider $n$ unit vectors $\overrightarrow{O A}_{1}$, $\overrightarrow{O A}_{2}, \cdots, \overrightarrow{O A}_{n-1}, \overrightarrow{O A}_{n}$, which form angles of $2 \pi / n, 4 \pi / n, \cdots, (2 n-2) \pi / n, 2 n \pi / n$ with the $o x$ axis (a... | proof | Algebra | proof | Yes | Yes | inequalities | false | 735,415 |
$3 \cdot 85$ Let $n \geqslant 2, a_{1}, a_{2}, \cdots \cdots, a_{n}$ be positive integers, and $a_{k} \leqslant k(1 \leqslant k \leqslant$ $n$ ). Prove that, when and only when $a_{1}+a_{2}+\cdots+a_{n}$ is even, it is possible to appropriately choose " + " and " - " signs, such that $a_{1} \pm a_{2} \pm \cdots \pm a_{... | [Proof] Necessity: Since the choice of plus or minus signs does not change the parity of $a_{1} \pm a_{2} \pm \cdots \pm a_{n}$, it is only possible to make $a_{1} \pm a_{2} \pm \cdots \pm a_{n}=0$ when $a_{1}+a_{2}+\cdots+a_{n}$ is even.
Sufficiency: When $n=2$, $a_{1}=1$. Since $a_{1}+a_{2}$ is even and $a_{2} \leqs... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 735,416 |
3. 87 Proof: If
$$\begin{array}{l}
\quad\left(x+\sqrt{x^{2}+1}\right)\left(y+\sqrt{y^{2}+1}\right)=1 \text {, } \\
\text { then } x+y=0 .
\end{array}$$ | [Proof] Multiplying both sides of the given equation by $x-\sqrt{x^{2}+1}$, we get
$$-\left(y+\sqrt{y^{2}+1}\right)=x-\sqrt{x^{2}+1}$$
Multiplying both sides of the given equation by $y-\sqrt{y^{2}+1}$, we get
$$-\left(x+\sqrt{x^{2}+1}\right)=y-\sqrt{y^{2}+1}$$
Adding (1) and (2) gives
$$\begin{aligned}
-y-x & =x+y \\... | proof | Algebra | proof | Yes | Yes | inequalities | false | 735,417 |
3-88 Prove the identity:
$$\begin{aligned}
& \frac{a_{1}}{a_{2}\left(a_{1}+a_{2}\right)}+\frac{a_{2}}{a_{3}\left(a_{2}+a_{3}\right)}+\cdots+\frac{a_{n}}{a_{1}\left(a_{n}+a_{1}\right)} \\
= & \frac{a_{2}}{a_{1}\left(a_{1}+a_{2}\right)}+\frac{a_{3}}{a_{2}\left(a_{2}+a_{3}\right)}+\cdots+\frac{a_{1}}{a_{n}\left(a_{n}+a_{1... | [Proof] We have
$$\begin{aligned}
& \frac{a_{1}}{a_{2}\left(a_{1}+a_{2}\right)}+\frac{a_{2}}{a_{3}\left(a_{2}+a_{3}\right)}+\cdots+\frac{a_{n}}{a_{1}\left(a_{n}+a_{1}\right)} \\
= & \left(\frac{1}{a_{2}}-\frac{1}{a_{1}+a_{2}}\right)+\left(\frac{1}{a_{3}}-\frac{1}{a_{2}+a_{3}}\right)+\cdots+\left(\frac{1}{a_{1}}-\frac{1... | proof | Algebra | proof | Yes | Yes | inequalities | false | 735,418 |
$3 \cdot 89$ integers $a, b, c$ such that
$$\frac{a}{b}+\frac{b}{c}+\frac{c}{a} \text{ and } \frac{a}{c}+\frac{c}{b}+\frac{b}{a}$$
are both integers, prove: $|a|=|b|=|c|$. | [Proof] If $(a, b, c)=d \neq 1$, then we can instead discuss $\frac{a}{d}, \frac{b}{d}, \frac{c}{d}$. Therefore, without loss of generality, assume $(a, b, c)=1$.
If the conclusion is not true, then at least one of $|a|,|b|,|c|$ is not equal to 1. Without loss of generality, assume $|a| \neq 1$.
Let $P$ be a prime fa... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 735,419 |
$1 \cdot 5 n$ is a fixed positive integer, find the sum of all positive integers that have the following property: in binary, this number has exactly $2 n$ digits, with $n$ ones and $n$ zeros (the leading digit cannot be $0$). | [Solution] The first digit of such a positive integer must be 1, and among the other $2n-1$ digits, exactly $n$ digits are 0, so there are $C_{2 n-1}^{n}$ such numbers. The 1 in the $k$-th position from the right $(1 \leqslant k \leqslant 2 n-1)$ represents $2^{k-1}$, and appears $C_{2 n-2}^{n}$ times (excluding the $k... | 2^{2 n-1} \times C_{2 n-1}^{n}+\left(2^{2 n-1}-1\right) \times C_{2 n-2}^{n} | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 735,420 |
1.41 Let $a, b$ be positive integers. When $a^{2}+b^{2}$ is divided by $a+b$, the quotient is $q$ and the remainder is $r$. Find all pairs $(a, b)$ such that $q^{2}+r=1977$.
---
The translation maintains the original format and line breaks as requested. | [Solution] Since $a^{2}+b^{2}=q(a+b)+r$, where $0 \leqslant r < q$, and if $r > 0$ it leads to a contradiction, so $q \leqslant 44$, and $\frac{a^{2}+b^{2}}{a+b} > 128$. Without loss of generality, assume $a \geqslant b$, then $a \geqslant \frac{1}{2}(a+b)$.
When $a \leqslant \frac{2}{3}(a+b)$, $b \geqslant \frac{1}{3... | (50,37),(50,7),(37,50),(7,50) | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 735,421 |
$3 \cdot 91 x, y$ are positive integers, $y>3$, and
$$x^{2}+y^{4}=2\left[(x-6)^{2}+(y+1)^{2}\right],$$
Prove: $x^{2}+y^{4}=1994$. | [Proof] Expanding the known equation, rearranging terms, we get
$$x^{2}+y^{4}-2\left(x^{2}-12 x+36\right)-2\left(y^{2}+2 y+1\right)=0$$
Simplifying the above equation, we have
which is
$$\left(y^{2}-1\right)^{2}-(x-12)^{2}=4 y-69$$
$$\left(y^{2}-x+11\right)\left(y^{2}+x-13\right)=4 y-69$$
If $4 y>69$, then since $y$... | 1994 | Algebra | proof | Yes | Yes | inequalities | false | 735,422 |
3.92 $\left\{x_{n} \mid n \in N\right\}$ is a sequence of real numbers, and for each positive integer $n \geqslant 2$, we have
$$x_{1}-C_{n}^{1} x_{2}+C_{n}^{2} x_{3}-\cdots+(-1)^{n} C_{n}^{n} x_{n+1}=0$$
Prove that for each positive integer $k, n, 1 \leqslant k \leqslant n-1$, we have
$$\sum_{p=0}^{n}(-1)^{p} C_{n}^{... | [Proof] Take the arithmetic sequence $\left\{a_{n} \mid n \in N\right\}$, where
$a_{1}=x_{1}, a_{2}=x_{2}$, and the common difference $d=x_{2}-x_{1}$.
Notice that
$$\begin{aligned}
\sum_{p=0}^{n}(-1)^{p} C_{n}^{p} a_{p+1} & =\sum_{p=0}^{n}(-1)^{p} C_{n}^{p}\left(a_{1}+p d\right) \\
& =a_{1} \sum_{p=0}^{n}(-1)^{p} C_{n... | proof | Algebra | proof | Yes | Yes | inequalities | false | 735,423 |
3.93 For non-negative integers $n$ and $k$, define $Q(n, k)$ as the coefficient of $x^k$ in the expansion of $\left(1+x+x^{2}+x^{3}\right)^{n}$. Prove:
$$Q(n, k)=\sum_{j=0}^{n}\binom{n}{j} \cdot\binom{n}{k-2 j},$$
where $\binom{a}{b}=C_{a}^{b}$, and when $a \geqslant b \geqslant 0$, $\binom{a}{b}=\frac{a!}{b!(a-b)!}$,... | [Proof]
$$\begin{array}{l}
\sum_{k \geqslant 0} Q(n, k) \cdot x^{k}=\left(1+x+x^{2}+x^{3}\right)^{n} \\
=(1+x)^{n}\left(1+x^{2}\right)^{n} \\
=\sum_{i \geqslant 0}\binom{n}{i} x^{i} \cdot \sum_{j \geqslant 0}\binom{n}{j} x^{2 i} \\
=\sum_{i \geqslant 0} \sum_{j \geqslant 0} x^{i+2 j} \cdot\binom{n}{i} \cdot\binom{n}{j}... | proof | Combinatorics | proof | Yes | Yes | inequalities | false | 735,424 |
4.1 If $p, q_{1}$ and $q_{2}$ are real numbers, and $p=q_{1}+q_{2}+1$, then at least one of the two equations
$$x^{2}+x+q_{1}=0, x^{2}+p x+q_{2}=0$$
has two distinct real roots. | [Proof] If $x^{2}+x+q_{1}=0$ does not have two distinct real roots, then $\Delta_{1}=1-4 q_{1} \leqslant 0$, i.e., $q_{1} \geqslant \frac{1}{4}$.
In this case, the discriminant of the equation $x^{2}+p x+q_{2}=0$ is
$$\begin{aligned}
\Delta_{2} & =p^{2}-4 q_{2}=\left(q_{1}+q_{2}+1\right)^{2}-4 q_{2} \\
& =q_{2}^{2}+2\l... | proof | Algebra | proof | Yes | Yes | inequalities | false | 735,425 |
4-2 Find all positive values of $a$ such that the equation $a^{2} x^{2}+a x+1-7 a^{2}=0$ has two integer roots. | [Solution] If the original equation has two integer roots, then by Vieta's formulas, $-\frac{1}{a}$ is an integer, so $a$ is a number of the form $\frac{1}{n}(n \in \mathbb{N})$. Let $a=\frac{1}{n}$, and transform the original equation into
$$x^{2}+n x+n^{2}-7=0$$
Then $\quad \Delta=n^{2}-4 n^{2}+28=-3 n^{2}+28 \geqsl... | a=1, \frac{1}{2}, \frac{1}{3} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,426 |
$4 \cdot 3$ Given the equation $2 x^{2}-9 x+8=0$, find a quadratic equation whose one root is the reciprocal of the sum of the roots of the original equation, and the other root is the square of the difference of the roots of the original equation. | [Solution] Let $x_{1}, x_{2}$ be the roots of the equation $2 x^{2}-9 x+8=0$. By Vieta's formulas, we have
$$\left\{\begin{array}{l}
x_{1}+x_{2}=\frac{9}{2} \\
x_{1} x_{2}=4
\end{array}\right.$$
Let the required equation be $x^{2}+p x+q=0$, with its roots being $x^{\prime}{ }_{1}, x^{\prime}{ }_{2}$. According to the ... | 36 x^{2}-161 x+34=0 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,427 |
4.4 Let $x_{1}$ and $x_{2}$ be the roots of the equation
$$x^{2}-(a+d) x+a d-b c=0$$
Prove that $x_{1}^{3}$ and $x_{2}^{3}$ are the roots of the equation
$$y^{2}-\left(a^{3}+d^{3}+3 a b c+3 b c d\right) y+(a d-b c)^{3}=0$$ | [Proof] For the equation
$$x^{2}-(a+d) x+(a d-b c)=0$$
applying Vieta's formulas yields
$$a+d=x_{1}+x_{2}, a d-b c=x_{1} x_{2}$$
Therefore, \(a^{3}+d^{3}+3 a b c+3 b c d=a^{3}+d^{3}+3(a+d) b c\)
$$\begin{array}{l}
=(a+d)^{3}-3(a+d)(a d-b c) \\
=\left(x_{1}+x_{2}\right)^{3}-3\left(x_{1}+x_{2}\right) x_{1} x_{2}=x_{1}^... | proof | Algebra | proof | Yes | Yes | inequalities | false | 735,428 |
4.5 Let $p$ and $q$ be two odd integers. Prove that the equation
$$x^{2}+2 p x+2 q=0$$
cannot have rational roots. | [Proof] (1) The root of equation (1) cannot be an odd number. In fact, if $x$ is odd, then $x^{2}$ is odd, while $2 p x + 2 q$ is even, so $x^{2} + 2 p x + 2 q$ takes an odd value, which cannot be zero.
(2) The root of equation (1) cannot be an even number. In fact, if $x$ is even, then $x^{2} + 2 p x$ is divisible by ... | proof | Algebra | proof | Yes | Yes | inequalities | false | 735,429 |
4・6 Prove: If the quadratic equation with integer coefficients
$$a x^{2}+b x+c=0 \quad(a \neq 0)$$
has a rational root, then at least one of the numbers $a, b, c$ is even. | [Proof] In the equation
$$a x^{2}+b x+c=0$$
let $x=y / a$, and then multiply both sides by $a$, obtaining the equation with integer coefficients
$$y^{2}+b y+a c=0$$
If $x$ is a rational number, then the number $y=a x$ is also a rational number. Therefore, equation (2) has a rational root just like equation (1). We kn... | proof | Algebra | proof | Yes | Yes | inequalities | false | 735,430 |
1.42 Find all quadruples of real numbers $\left(x_{1}, x_{2}, x_{3}, x_{4}\right)$ such that the sum of any one of the numbers and the product of the other three is equal to 2. | [Solution] Let $x_{1} x_{2} x_{3} x_{4}=d$,
so $x_{i}+\frac{d}{x_{i}}=2,(i=1,2,3,4)$,
thus $x_{i}=1 \pm \sqrt{1-d},(i=1,2,3,4)$.
Clearly, $d \leqslant 1$. There are five cases:
(1) $d=(1+\sqrt{1-d})^{4} \geqslant 1 \geqslant d$,
then $d=1, x_{i}=1,(i=1,2,3,4)$.
$$\text { (2) } \begin{aligned}
d & =(1-\sqrt{1-d})(1+\sq... | x_{1}=x_{2}=x_{3}=x_{4}=1 \text{ or three of them are } -1 \text{ and one is } +3 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,431 |
4. 7 If the equations $x^{2}+a x+b=0$ and $x^{2}+p x+q=0$ have a non-zero common root, find the quadratic equation whose roots are the distinct roots of these equations. | [Solution] Let the common root be $\alpha$. By Vieta's formulas, the distinct roots are $\frac{b}{\alpha}$ and $\frac{q}{\alpha}$. Since $\alpha$ is the common root, we have $\left\{\begin{array}{l}\alpha^{2}+a \alpha+b=0, \\ a^{2}+p \alpha+q=0 .\end{array}\right.$ Subtracting the two equations, we get $\quad \alpha=\f... | x^{2}-\frac{(b+q)(a-p)}{q-b} x+\frac{b q(a-p)^{2}}{(q-b)^{2}}=0 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,432 |
4・8 For what real number $k$ do the equation $4 x^{2}-(2-k) x+k-5=0$ have two negative roots? | [Solution] Let the two roots of the equation be $\alpha, \beta$. According to the problem, the following conditions must be satisfied:
$$\left\{\begin{array}{l}
\alpha+\beta=\frac{2-k}{4}0 \\
(k-2)^{2}-16(k-5) \geqslant 0
\end{array}\right.$$
From (1) $2-k2$.
From (2) $k-5>0$, i.e., $k>5$.
From (3) $k^{2}-20 k+84=(k-1... | k \geqslant 14 \text{ or } 5 < k \leqslant 6 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,433 |
$4 \cdot 9$ Given that a quadratic trinomial achieves an extremum of 25 when $x=\frac{1}{2}$; the sum of the cubes of the roots of this quadratic trinomial is 19. Try to find this quadratic trinomial. | [Solution] From the first given condition, we know that the leading coefficient of this quadratic trinomial is not equal to 0, and it can be written as
$$a\left(x-\frac{1}{2}\right)^{2}+25$$
which is $a x^{2}-a x+\frac{1}{4} a+25$.
Thus, by Vieta's formulas, we have
$$x_{1}+x_{2}=1, x_{1} x_{2}=\frac{1}{4}+\frac{25}{a... | -4 x^{2}+4 x+24 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,434 |
4・10 Given the equation $x^{2}-2 x+\lg \left(2 a^{2}-a\right)=0$ has one positive root and one negative root, try to find the range of real values for $a$.
| [Solution] From the given, we know that $\lg \left(2 a^{2}-a\right)<0$. Therefore, $0<2 a^{2}-a<1$,
Solving this, we get that $a$ should be in the range
$$-\frac{1}{2}<a<0 \text { or } \frac{1}{2}<a<1 \text {. }$$ | -\frac{1}{2}<a<0 \text { or } \frac{1}{2}<a<1 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,435 |
4・ 11 When $a, b, c$ are real numbers, prove that the equation
$$x^{2}-(a+b) x+\left(a b-c^{2}\right)=0$$
has two real roots, and find the condition for these roots to be equal. | [Solution] The discriminant of the given equation is
$$\Delta=(a+b)^{2}-4\left(a b-c^{2}\right)=(a-b)^{2}+4 c^{2},$$
Since $a$, $b$, and $c$ are all real numbers, hence $\Delta \geqslant 0$, so the equation has two real roots. The condition for these two roots to be equal is
i.e., $\square$
$$(a-b)^{2}+4 c^{2}=0$$
$$... | a=b, c=0 | Algebra | proof | Yes | Yes | inequalities | false | 735,436 |
4. 15 Given that $\operatorname{tg} \alpha$ and $\operatorname{tg} \beta$ are the roots of $x^{2}+p x+q=0$, try to find
the value of
$$\sin ^{2}(\alpha+\beta)+p \sin (\alpha+\beta) \cos (\alpha+\beta)+q \cos ^{2}(\alpha+\beta) \text { }$$ | [Solution] By the relationship between roots and coefficients, we have
$$\operatorname{tg} \alpha + \operatorname{tg} \beta = -p \text{ and } \operatorname{tg} \alpha \operatorname{tg} \beta = q,$$
Therefore, $\operatorname{tg}(\alpha+\beta) = \frac{-p}{1-q}$.
Thus, the value we seek is
$$\begin{aligned}
& \sin ^{2}(\... | q | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,438 |
4・ For what value of $16 b$ does the equation $1988 x^{2}+b x+8891=0$ and $8891 x^{2}+b x+1988=0$ have a common root. | ［Solution］Suppose the given two equations have a common root $x$, then
$$1988 x^{2}+b x+8891=8891 x^{2}+b x+1988$$
Simplifying, we get $x= \pm 1$. Substituting $x$ back into the given equations, we get
$$b=\mp 10879 .$$ | b=\mp 10879 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,439 |
1.43 Find all integers $a, b, c$, where $1<a<b<c$, such that $(a-1)(b-1)(c-1)$ is a divisor of $a b c-1$. | [Solution] Given $a \geqslant 2, b \geqslant 3, c \geqslant 4$, hence
$$\frac{a-1}{a}=1-\frac{1}{a} \geqslant \frac{1}{2}, \frac{b-1}{b} \geqslant \frac{2}{3}, \frac{c-1}{c} \geqslant \frac{3}{4} .$$
Therefore, $a b c \leqslant 4(a-1)(b-1)(c-1)$,
$$S=\frac{a b c-1}{(a-1)(b-1)(c-1)}b>a>1$$, which implies $a \geqslant 3... | null | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 735,440 |
4. 17 If \(a, b, c\) are the three sides of a triangle, then the roots of the equation
\[
b^{2} x^{2}+\left(b^{2}+c^{2}-a^{2}\right) x+c^{2}=0
\]
are complex conjugates. Prove it. | [Proof] $\Delta=\left(b^{2}+c^{2}-a^{2}\right)^{2}-4 b^{2} c^{2}$
$$=(a+b+c)(b+c-a)(b-c-a)(b-c+a) .$$
Since $a, b, c$ are the three sides of a triangle. Therefore,
$$\begin{array}{l}
a+b+c>0, b+c-a>0, b-c-a<0, b-c+a>0
\end{array}$$
Thus,
the equation has no real roots. Hence, the original proposition is proved. | proof | Algebra | proof | Yes | Yes | inequalities | false | 735,441 |
4. For the equation $x^{2} + a x + b + 1 = 0$, if the roots are natural numbers, prove that $a^{2} + b^{2}$ is a composite number. | [Proof] Let $x_{1}$ and $x_{2}$ be the two roots of the original equation. By Vieta's formulas, we have
$$\left\{\begin{array}{l}
x_{1}+x_{2}=-a \\
x_{1} x_{2}=b+1
\end{array}\right.$$
Therefore,
$$\begin{aligned}
a^{2}+b^{2} & =\left(x_{1}+x_{2}\right)^{2}+\left(x_{1} x_{2}-1\right)^{2} \\
& =x_{1}^{2}+x_{2}^{2}+x_{1... | proof | Algebra | proof | Yes | Yes | inequalities | false | 735,442 |
4- 19 Given the quadratic trinomial $f(x)=a x^{2}+b x+c$ such that the equation $f(x)=x$ has no real roots. Prove: The equation $f(f(x))=x$ also has no real roots. | [Proof] Given that $f(x)=x$ has no real roots.
Therefore, when $a>0$, for all real numbers $x$ we have $f(x)>x$, thus we have
$$f(f(x))>f(x)>x$$
which means $f(f(x))>x$,
so when $a>0$, the equation $f(f(x))=x$ has no real roots.
Similarly, when $a<0$, we get $f(x)<x$, thus we have
$$f(f(x))<f(x)<x$$
so when $a<0$, th... | proof | Algebra | proof | Yes | Yes | inequalities | false | 735,443 |
4.20 There is a $\triangle A B C, a, b, c$ are the sides opposite to $\angle A, \angle B, \angle C$ respectively. If $b$ is the arithmetic mean of $a$ and $c$, and $\operatorname{tg} \frac{B}{2}$ is the geometric mean of $\operatorname{tg} \frac{A}{2}$ and $\operatorname{tg} \frac{C}{2}$, try to form a quadratic equati... | [Solution] From the problem, we have $A+B+C=180^{\circ}$,
$$\begin{aligned}
2 b= & a+c \\
& \left(\operatorname{tg} \frac{B}{2}\right)^{2}=\operatorname{tg} \frac{A}{2} \cdot \operatorname{tg} \frac{C}{2} .
\end{aligned}$$
From (2) and the Law of Sines, $2 \sin B=\sin A+\sin C$,
From (1) $2 \sin (A+C)=\sin A+\sin C$,
... | 3 x^{2}-2 \sqrt{3} x+1=0 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,444 |
$4 \cdot 21$ is written on the blackboard with the equation
$x^{3}+\cdots x^{2}+\cdots x+\cdots=0$ , two people are playing a game: the first person fills in one of the blanks with a non-zero integer (positive or negative integer), then the second person fills in another blank with an integer, and finally the first per... | [Proof] If the first player places -1 before $x$, and in his second turn, he fills the last blank with a number of the opposite sign to the one filled by the second player, then a polynomial of the form $x^{3}-a x^{2}-x+a=\left(x^{2}-1\right)(x-a)$ is obtained. The roots of this polynomial are $-1, 1, a$. They are all ... | proof | Algebra | proof | Yes | Yes | inequalities | false | 735,445 |
4・22 On the blackboard, there is the equation
$$x^{3}+\ldots x^{2}+$$
$\qquad$ $\qquad$ $=0$,
Two people play a game according to the following rules: Player A first says any number, and Player B then fills that number into any of the three blank spaces; then Player A says another number, and Player B fills it into any... | [Solution] The answer is affirmative.
In fact, Player A can adopt the following strategy: He first says 0. If Player B places 0 in the last position, the left side of the equation becomes the polynomial $x^{3}+a x^{2}+b x$ (coefficients $a$ and $b$ are yet to be determined). Player A then says 2 and -3, and based on th... | proof | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,446 |
4. 24 Given that the difference of the squares of the two roots of $x^{2} \cos A-2 x+\cos A=0$ is $\frac{3}{8}$, find the value of $A$ $(0<A<\pi)$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | [Solution] Let $\cos A \neq 0$ (if $\cos A=0$, it does not meet the problem's requirements).
Rewrite the original equation as: $x^{2}-\frac{2}{\cos A} x+1=0$. Let the two roots be $\alpha$ and $\beta$. By Vieta's formulas and the problem's conditions, we have
$$\left\{\begin{array}{l}
\alpha+\beta=\frac{2}{\cos A} \\
\... | A \approx 5^{\circ} 20^{\prime} \text { or } A \approx 174^{\circ} 40^{\prime} | Logic and Puzzles | math-word-problem | Yes | Yes | inequalities | false | 735,448 |
$4 \cdot 25$ (1) Find the necessary and sufficient condition for the quadratic equations
$$\left\{\begin{array}{l}
x^{2}+p_{1} x+q_{1}=0, \\
x^{2}+p_{2} x+q_{2}=0 .
\end{array}\right.$$
to have a common root.
(2) Prove that if the quadratic equations $\circledast$ have a common root, but the equations are not identica... | [Solution] (1) Let a number $x$ satisfy the two equations:
$$\left\{\begin{array}{l}
x^{2}+p_{1} x+q_{1}=0 \\
x^{2}+p_{2} x+q_{2}=0
\end{array}\right.$$
By eliminating $x$ from these two equations, we can find the necessary and sufficient condition mentioned in the problem. For example, we can proceed as follows.
If ... | proof | Algebra | proof | Yes | Yes | inequalities | false | 735,449 |
4. 26 Prove: If real numbers $a, b, c$ satisfy the condition
$$\frac{a}{m+2}+\frac{b}{m+1}+\frac{c}{m}=0$$
where $m$ is a positive number, then the equation
$$a x^{2}+b x+c=0$$
has a root between 0 and 1. | [Proof] It is sufficient to prove this problem for $a \geqslant 0$. Because the case for $a<0$ can be derived from the case for $a>0$. We now prove in two scenarios.
(1) If $a=0$.
If $b \neq 0$, then equation (2) has a root
$$x_{0}=-\frac{c}{b}$$
From (1), we know $\frac{c}{b}=-\frac{m}{m+1}$, thus
$$x_{0}=\frac{m}{m... | proof | Algebra | proof | Yes | Yes | inequalities | false | 735,450 |
$1 \cdot 44$ increasing sequence $2,3,5,6,7,10,11, \cdots$, includes all positive integers that are neither perfect squares nor perfect cubes. Try to find the 500th term of this sequence. | [Solution] It is easy to know that $[\sqrt{500}]=22, [\sqrt[3]{500}]=7, [\sqrt[6]{500}]=2$ (where $[x]$ represents the greatest integer not exceeding $x$).
Therefore, there are $22+7-2=27$ square numbers or cube numbers within the natural number 500. Now consider how many square numbers or cube numbers there are from ... | 528 | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 735,451 |
4.28 $a, b, c, A, B, C$ are 6 positive real numbers, such that the equations
$a x^{2}-b x+c=0$ and $A x^{2}-B x+C=0$ have real roots. Prove that for any real number $u$ between the two real roots of the equation $a x^{2}-b x+c=0$ and any real number $v$ between the two real roots of the equation $A x^{2}-B x+C=0$, the ... | [Proof] From the given conditions,
$$\begin{array}{l}
a u^{2}-b u+c \leqslant 0 \\
A v^{2}-B v+C \leqslant 0
\end{array}$$
Since \(a, b, c, A, B, C\) are all positive real numbers, therefore
$$u>0, v>0$$
From (1) and (2), we get
$$0<a u+\frac{c}{u} \leqslant b$$
and
$$0<A v+\frac{C}{v} \leqslant B$$
Adding the abov... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,453 |
4・30 Find all solutions to the equation $x^{2}-8[x]+7=0$. Here, $[x]$ denotes the greatest integer less than or equal to $x$.
保持源文本的换行和格式如下：
4・30 Find all solutions to the equation $x^{2}-8[x]+7=0$. Here, $[x]$ denotes the greatest integer less than or equal to $x$. | [Solution] Let $x$ be the root of the given equation, and $[x]=n$. From the original equation, we have
$$x^{2}+7=8 n$$
This indicates that $n>0$. Using $n \leqslant x < n+1$, we get
$$\left\{\begin{array}{l}
n^{2}+7<8(n+1) \\
n^{2}-8 n+7 \leqslant 0
\end{array}\right.$$
Solving this, we get $\quad 1 \leqslant n<2$ or... | x=1, \sqrt{33}, \sqrt{41}, 7 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,455 |
4.33 Find all real numbers that satisfy the equation $|x+3|-|x-1|=x+1$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | [Solution] When $x \leqslant -3$, the equation becomes
$$-(x+3)-(1-x)=x+1$$
which yields $\square$
$$x=-5 .$$
When $-3 < x \leqslant 1$, the equation becomes
$$-(x+3)+(x-1)=x+1$$
which yields $\square$
$$x=-1 .$$
When $x > 1$, the equation becomes
$$(x+3)-(x-1)=x+1$$
which yields $\square$
$$x=3 .$$
Therefore, th... | -5, -1, 3 | Logic and Puzzles | math-word-problem | Yes | Yes | inequalities | false | 735,456 |
4. 34 Find the range of real values of $x$ that satisfy the following relation:
$$|3 x-2|+|3 x+1|=3$$ | [Solution] $\because|3 x-2|+|3 x+1|=|2-3 x|+|3 x+1|=3$,
and $|2-3 x+3 x+1|=3$,
$\therefore \quad|2-3 x|+|3 x+1|=|(2-3 x)+(3 x+1)|$,
$\therefore \quad 2-3 x$ and $3 x+1$ must have the same sign, or one of them is 0,
If $\left\{\begin{array}{l}2-3 x \geqslant 0 \\ 3 x+1 \geqslant 0,\end{array}\right.$, then $-\frac{1}{3}... | -\frac{1}{3} \leqslant x \leqslant \frac{2}{3} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,457 |
$4 \cdot 36$ Try to find the $n$ $n$-th roots of 1, and find the sum of their $n$-th powers.
untranslated part:
试求 1 的 $n$ 个 $n$ 次方根,并求它们 $n$ 次幕的和.
translated part:
Try to find the $n$ $n$-th roots of 1, and find the sum of their $n$-th powers. | [Solution] $x^{n}=1$, that is, $x^{n}=\cos 2 k \pi+i \sin 2 k \pi \cdot(k=0, \pm 1, \pm 2, \cdots)$, so by De Moivre's theorem, the $n$ roots are:
$$x_{k}=\cos \frac{2 k \pi}{n}+i \sin \frac{2 k \pi}{n} \cdot(k=0,1,2, \cdots, n-1)$$
Since $\quad x_{k}^{n}=1 .(k=0,1,2, \cdots, n-1)$
Therefore, $x_{0}^{n}+x_{1}^{n}+\cdo... | n | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,458 |
$4 \cdot 37$ Try to point out, if the equation
$$x^{3}+a x^{2}+b x+c=0$$
has three real roots in arithmetic progression, what necessary and sufficient conditions should the real numbers $a, b, c$ satisfy? | ［Solution］ Let the real-coefficient cubic equation be
$$x^{3}+a x^{2}+b x+c=0$$
with three real roots $x_{0}-y, x_{0}, x_{0}+y$ forming an arithmetic sequence. By the relationship between roots and coefficients, we have
$$\left(x_{0}-y\right)+x_{0}+\left(x_{0}+y\right)=-a,$$
so $x_{0}=-\frac{1}{3} a$.
Substituting $x... | 2 a^{3}-9 a b+27 c=0 \text{ and } a^{2}-3 b \geqslant 0 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,459 |
$4 \cdot 38 \quad$ For the equation
$$x^{3}+a x^{2}+b x+c=0$$
to have three different real roots that form a geometric progression, the real numbers $a, b, c$ should satisfy what conditions? | [Solution] Let $x_{1}, x_{1} q, x_{1} q^{2}$ be the three roots of equation (1), where $x_{1}, q$ are real numbers, $x_{1} \neq 0, q \neq 0, \pm 1$. These three roots satisfy the requirements of the problem. According to the relationship between roots and coefficients, we have
$$\left\{\begin{array}{l}
x_{1}+x_{1} q+x_... | b^{3}=a^{3} c, \\ c \neq 0, \\ c \text { lies between }-a^{3} \text { and } a^{3} / 27. | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,460 |
4. 39 Prove: If two cubic equations with integer coefficients have a common irrational root, then they have another common root.
| [Proof] Let the polynomials be
$$\begin{array}{l}
P(x)=a_{0} x^{3}+a_{1} x^{2}+a_{2} x+a_{3} \\
Q(x)=b_{0} x^{3}+b_{1} x^{2}+b_{2} x+b_{3}
\end{array}$$
have a common irrational root $\alpha$, where the coefficients $a_{i}, b_{i} (i=0,1,2,3)$ are integers, $a_{0} \neq 0, b_{0} \neq 0$.
The number $\alpha$ is also a ro... | proof | Algebra | proof | Yes | Yes | inequalities | false | 735,461 |
4. 41 Solve the equation $5 x^{2}+x-x \cdot \sqrt{5 x^{2}-1}-2=0$. | ［Solution］ Transform the original equation into
$$\left(5 x^{2}-1\right)-x \cdot \sqrt{5 x^{2}-1}+(x-1)=0,$$
Factorizing gives
$$\left(\sqrt{5 x^{2}-1}-x+1\right)\left(\sqrt{5 x^{2}-1}-1\right)=0$$
From $\sqrt{5 x^{2}-1}-x+1=0$, we get $x_{1}=\frac{1}{2}, x_{2}=-1$.
From $\sqrt{5 x^{2}-1}-1=0$, we get $x_{3,4}= \pm \... | \pm \frac{\sqrt{10}}{5} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,462 |
4.43 For any real number $t$, let $[t]$ denote the greatest integer less than or equal to $t$. For example, $[8]=8$, $[\pi]=3$, $\left[-\frac{5}{2}\right]=-3$. Prove that the equation $[x]+[2 x]+[4 x]+[8 x]+[16 x]+[32 x]=12345$ has no real solution.
Translate the above text into English, please retain the original tex... | [Proof] Let $f(x)=[x]+[2 x]+[4 x]+[8 x]+[16 x]+[32 x]$
Assume the equation has a real solution $x$, then
$$\begin{array}{l}
f(x)=12345 \\
f(195)=1228512345
\end{array}$$
and $f(x)$ is increasing, so we have
$$195<x<196$$
Let $y=x-195$, then $0<y<1$.
$$\begin{aligned}
f(y) & =[x-195]+[2 x-2 \times 195]+\cdots+[32 x-32... | proof | Number Theory | proof | Yes | Yes | inequalities | false | 735,463 |
4. 44 Let $m$ be a real number, solve the equation for $x$: $\left|x^{2}-1\right|+\left|x^{2}-4\right|=m x$.
---
The provided text has been translated into English while preserving the original formatting and line breaks. | [Solution] Since the two terms on the left side of the equation $\left|x^{2}-1\right|+\left|x^{2}-4\right|=m x$ cannot be zero simultaneously, the left side is always positive, so $m \neq 0$.
First, consider the case where $m>0$, in which case $x>0$.
If $0<x<1$, then by (1) we get $5-m x=0$.
If $1 \leq x \leq 2$, then ... | When\ m \geqslant 3,\ x=\frac{-m \pm \sqrt{m^{2}+40}}{4};\ When\ \frac{3}{2} \leqslant m<3,\ x=\frac{m-\sqrt{m^{2}+40}}{4},\ \frac{3}{m};\ When\ m\ takes\ other\ values,\ equation\ (1)\ has\ no | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,464 |
4. $45 f(x)=x^{3}-3 x+\left(x^{2}-1\right) \sqrt{x^{2}-4}$, Express $\frac{f(x)-2}{f(x)+2}$ as a fraction where both the numerator and the denominator are in the form $u \sqrt{v}$, where $u, v$ are linear polynomials in $x$, and state for which values of $x$ your equation holds. | [Solution] If $x \geqslant 2$, then
$$\begin{aligned}
\frac{f(x)-2}{f(x)+2} & =\frac{x^{3}-3 x-2+\left(x^{2}-1\right) \sqrt{x^{2}-4}}{x^{3}-3 x+2+\left(x^{2}-1\right) \sqrt{x^{2}-4}} \\
& =\frac{(x+1)^{2}(x-2)+(x+1)(x-1) \sqrt{x^{2}-4}}{(x-1)^{2}(x+2)+(x+1)(x-1) \sqrt{x^{2}-4}} \\
& =\frac{(x+1) \sqrt{x-2}[(x+1) \sqrt{... | \frac{(x+1) \sqrt{x-2}}{(x-1) \sqrt{x+2}} \text{ for } x \geqslant 2 \text{ and } \frac{-(x+1) \sqrt{-x+2}}{(x-1) \sqrt{-x-2}} \text{ for } x \leqslant -2 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,465 |
4.46 Let the equation
$$x^{n}+a_{n-1} x^{n-1}+\cdots+a_{1} x+a_{0}=0$$
have real coefficients that satisfy the condition $0<a_{0} \leqslant a_{1} \leqslant \cdots \leqslant a_{n-1} \leqslant 1$. Given that $\lambda$ is a complex root of this equation and $|\lambda| \geqslant 1$. Prove: $\lambda^{n+1}=1$. | [Solution] Since $\lambda$ is a root of the equation, we have
$$\lambda^{n}+a_{n-1} \lambda^{n-1}+\cdots+a_{1} \lambda+a_{0}=0 .$$
Multiplying both sides of the above equation by $\lambda-1$, we get
$$\begin{aligned}
0= & (\lambda-1)\left(\lambda^{n}+a_{n-1} \lambda^{n-1}+\cdots+a_{1} \lambda+a_{0}\right) \\
= & \lamb... | proof | Algebra | proof | Yes | Yes | inequalities | false | 735,466 |
4. 47. Let $a, b$ be real numbers, and $x^{4}+a x^{3}+b x^{2}+a x+1=0$ has at least one real root. Try to find the minimum value of $a^{2}+b^{2}$. | ［Solution］First, consider the equation
$$x+\frac{1}{x}=y$$
Here $y$ is a real number. This equation can be written as a quadratic equation in $x$: $x^{2}-y x+1=0$. It has real roots if and only if its discriminant is not less than 0, i.e.,
$$y^{2}-4 \geqslant 0 \text { or }|y| \geqslant 2 \text {. }$$
The original eq... | \frac{4}{5} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,467 |
$$\begin{array}{l}
\text { 4・48 } \quad \text { If } p_{1}(x)=x^{2}-2 . \\
p_{i}(x)=p_{1}\left[p_{i-1}(x)\right], i=2,3,4, \cdots
\end{array}$$
Prove that for any natural number $n$, the solutions of the equation $p_{n}(x)=x$ are distinct real numbers. | [Proof] It is not difficult to prove by mathematical induction that when $|x|>2$, there must be $p_{n}(x)>x$. Therefore, we discuss $p_{n}(x)$ on the domain $[-2,2]$, and make the substitution
$$x=2 \cos t, t \in[0, \pi]$$
First, we use mathematical induction to prove that for any natural number $n$,
$$p_{n}(2 \cos t)... | proof | Algebra | proof | Yes | Yes | inequalities | false | 735,468 |
4. 49 Solve the equation $x^{2}+\left(\frac{x}{x+1}\right)^{2}=3$.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly.
4. 49 Solve the equation $x^{2}+\left(\frac{x}{x+1}\right)^{2}=3$. | [Solution] Since $x^{2}-2 \cdot \frac{x^{2}}{x+1}+\left(\frac{x}{x+1}\right)^{2}=\left(x-\frac{x}{x+1}\right)^{2}$
$$=\left(\frac{x^{2}}{x+1}\right)^{2}$$
The original equation can be transformed into
$$\begin{array}{l}
x^{2}+\left(\frac{x}{x+1}\right)^{2}-2 \cdot \frac{x^{2}}{x+1}+2 \cdot \frac{x^{2}}{x+1}=3 \\
\left... | x_{1}=\frac{-3+\sqrt{3} i}{2}, x_{2}=\frac{-3-\sqrt{3} i}{2}, x_{3}=\frac{1+\sqrt{5}}{2}, x_{4}=\frac{1-\sqrt{5}}{2} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,469 |
$4 \cdot 50$ Solve the equation
$$\frac{1}{x^{2}-10 x-29}+\frac{1}{x^{2}-10 x-45}-\frac{2}{x^{2}-10 x-69}=0$$ | ［Solution］ Let $x^{2}-10 x-49=t$. Then the original equation becomes
$$\frac{1}{t+20}+\frac{1}{t+4}-\frac{2}{t-20}=0$$
Eliminating the denominators, we get
$$(t+4)(t-20)+(t+20)(t-20)-2(t+20)(t+4)=0$$
Expanding and simplifying, we get
That is,
$$-64 t=640$$
Substituting, we get
$$t=-10$$
That is
$$\begin{array}{l}
... | x^2 - 10x - 39 = 0 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,470 |
4.51 Find the sum of all roots of the following equation
$$\sqrt[4]{x}=\frac{12}{7-\sqrt[4]{x}}$$ | [Solution] Let $y=\sqrt[4]{x}$, then $y=\frac{12}{7-y}$, which gives $y^{2}-7 y+12=0$,
solving for
$$y_{1}=3, y_{2}=4$$
$$\text { hence } \quad x_{1}=3^{4}, x_{2}=4^{4} \text {. }$$
Therefore, the sum of all roots $x_{1}+x_{2}=3^{4}+4^{4}=337$. | 337 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,471 |
4. 55 Try to find real numbers \(x, y, z\) greater than 1, satisfying the equation
$$x+y+z+\frac{3}{x-1}+\frac{3}{y-1}+\frac{3}{z-1}=2(\sqrt{x+2}+\sqrt{y+2}+\sqrt{z+2}) .$$ | [Solution] Let $f(t)=t+\frac{3}{t-1}-2 \sqrt{t+2}$, then
$$\begin{aligned}
f(t) & =\frac{1}{t-1}\left(t^{2}-t+3-2(t-1) \sqrt{t+2}\right) \\
& =\frac{1}{t-1}\left[t^{2}-2 t+1+(\sqrt{t+2})^{2}-2(t-1) \sqrt{t+2}\right] \\
& =\frac{1}{t-1}[t-1-\sqrt{t+2}]^{2}
\end{aligned}$$
Obviously, when $t>1$, $f(t) \geqslant 0$, and ... | x=y=z=\frac{3+\sqrt{13}}{2} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,473 |
4.56 Let $a$ and $b$ be two roots of $x^{4}+x^{3}-1=0$. Prove that $ab$ is a root of $x^{6}+x^{4}+x^{3}-x^{2}-1=0$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly. | [Proof] Let $a, b, c, d$ be the four roots of the equation $x^{4}+x^{3}-1=0$. Then, by Vieta's formulas, we have
$$\begin{array}{l}
a+b+c+d=-1 \\
a b+a c+a d+b c+b d+c d=0 \\
b c d+a c d+a d b+a b c=0 \\
a b c d=-1
\end{array}$$
Let $p=a+b, q=a b, r=c+d, s=c d$, then the equations (1) and (4) can be transformed into
$... | proof | Algebra | proof | Yes | Yes | inequalities | false | 735,474 |
1. 47 If a natural number $n$ can be uniquely represented as the sum $a_{1}+a_{2}+\cdots+a_{k}$ and the product $a_{1} a_{2} \cdots a_{k}$ of $k(k \geqslant 2)$ natural numbers, then this natural number is called a "good" number. For example, 10 is a "good" number, because $10=5+2+1+1+1=5 \times 2 \times 1 \times 1$ $\... | [Solution] If $n$ is a prime number, then $n$ is not a "good" number. Because $n=n \cdot 1<n+1$, such $n$ cannot be expressed as the sum and product of the same integers.
If $n=p \cdot q$, where $p$ and $q$ are both primes, then
$n=p q=p+q+1 \times(p q-p-q)$, and this representation is unique.
If $n=a b c$, where $a, ... | n \text{ is a "good" number if and only if } n \text{ is the product of two prime numbers.} | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 735,475 |
4.57 If $2 a^{2}<5 b$, prove that the equation
$$x^{5}+a x^{4}+b x^{3}+c x^{2}+d x+e=0$$
does not have all real roots. | [Proof] By contradiction. If all the roots of the equation are real, let these roots be \(x_{i}(1 \leqslant i \leqslant 5)\), then by Vieta's formulas we have
$$\begin{array}{l}
-a=\sum_{i=1}^{5} x_{i}, b=\sum_{i<j} x_{i} x_{j} . \\
2 a^{2}-5 b<0 \text { is equivalent to } \\
2\left(\sum_{i=1}^{5} x_{i}\right)^{2}-5 \s... | proof | Algebra | proof | Yes | Yes | inequalities | false | 735,476 |
4. 59 The equation $z^{6}+z^{3}+1=0$ has a complex root, and on the complex plane, the argument $\theta$ of this root lies between $90^{\circ}$ and $180^{\circ}$. Find the degree measure of $\theta$.
The equation $z^{6}+z^{3}+1=0$ has a complex root, and on the complex plane, the argument $\theta$ of this root lies be... | [Solution] Let $\omega=z^{3}$. Then the original equation can be transformed into
$$\begin{array}{l}
\omega^{2}+\omega+1=0 \\
\omega=\cos \frac{2 \pi}{3} \pm i \sin \frac{2 \pi}{3}
\end{array}$$
Since $z$ is the cube root of $w$, we have
$$\begin{array}{l}
z_{1}=\cos \frac{2 \pi}{9}+i \sin \frac{2 \pi}{9} \\
z_{2}=\co... | 160^{\circ} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,477 |
4・60 Quartic Equation
$$x^{4}-18 x^{3}+k x^{2}+200 x-1984=0$$
The product of two of the four roots of the equation is -32. Determine the value of $k$. | [Solution] Let the four roots of the equation be $x_{1}, x_{2}, x_{3}, x_{4}$. Then, by Vieta's formulas, we have
$$\begin{array}{l}
x_{1}+x_{2}+x_{3}+x_{4}=18 \\
x_{1} x_{2}+x_{1} x_{3}+x_{1} x_{4}+x_{2} x_{3}+x_{2} x_{4}+x_{3} x_{4}=k \\
x_{1} x_{2} x_{3}+x_{1} x_{2} x_{4}+x_{1} x_{3} x_{4}+x_{2} x_{3} x_{4}=-200 \\
... | 86 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,478 |
4.61 For what real values of $a, b, c$ does the equation
$$\begin{aligned}
& |a x+b y+c z|+|b x+c y+a z|+|c x+a y+b z| \\
= & |x|+|y|+|z|
\end{aligned}$$
hold for all real numbers $x, y, z$? | [Solution] Let $x=y=z=1$, we get
$$|a+b+c|=1$$
Let $x=y=0, z=1$, we get
$$|a|+|b|+|c|=1$$
Let $x=1, y=-1, z=0$, we get
$$|a-b|+|b-c|+|c-a|=2 .$$
From (1) and (2), we know that, except for 0, the signs of $a, b, c$ are the same, i.e.,
$$a b \geqslant 0, b c \geqslant 0, c a \geqslant 0$$
Since
$$\begin{aligned}
& |a... | not found | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,479 |
4.63 Find all values of \( x \) that satisfy the equation
$$\sqrt{x^{2}-p}+2 \sqrt{x^{2}-1}=x$$
where \( p \) is a real parameter. | [Solve] $\sqrt{x^{2}-p}+2 \sqrt{x^{2}-1}=x$
Rearrange and square both sides to get
$$x^{2}-p=x^{2}+4\left(x^{2}-1\right)-4 x \sqrt{x^{2}-1}$$
which simplifies to $4\left(x^{2}-1\right)+p=4 x \sqrt{x^{2}-1}$.
Square both sides again to get
$$16\left(x^{2}-1\right)^{2}+8 p\left(x^{2}-1\right)+p^{2}=16 x^{2}\left(x^{2}-1... | x=\frac{4-p}{2 \sqrt{2(2-p)}} \text{ for } 0 \leqslant p \leqslant \frac{4}{3} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,481 |
$4 \cdot 64$ Find the product of the real roots of the equation $x^{2}+18 x+30=2 \sqrt{x^{2}+18 x+45}$ | [Solution] Let $y=\sqrt{x^{2}+18 x+45}$. From the original equation, we get
$$y^{2}-15=2 y,$$
Rearranging gives $y^{2}-2 y-15=0$,
Solving yields $y_{1}=5, y_{2}=-3$ (discard $y<0$).
Thus, we have
$$\sqrt{x^{2}+18 x+45}=5$$
which simplifies to
$$x^{2}+18 x+20=0$$
By the discriminant $\Delta=18^{2}-4 \times 20>0$, we ... | 20 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,482 |
4. 65 Given the cubic equation $x^{3}+a x^{2}+b x+c=0$ has three real roots. Prove: $a^{2}-3 b \geqslant 0$, and $\sqrt{a^{2}-3 b}$ is not greater than the difference between the largest and smallest roots. | [Proof] Without loss of generality, let the three real roots of the equation $p, q, r$ satisfy $p \leqslant q \leqslant r$. By Vieta's formulas, we have
$$\begin{array}{l}
a=-(p+q+r) \\
b=p q+q r+r p
\end{array}$$
Thus,
$$\begin{aligned}
a^{2}-3 b & =(p+q+r)^{2}-3(p q+q r+r p) \\
& =p^{2}+q^{2}+r^{2}-p q-q r-r p \\
& ... | proof | Algebra | proof | Yes | Yes | inequalities | false | 735,483 |
4.66 Find all real roots of the equation
$$x^{4}-\left(2 \cdot 10^{10}+1\right) x^{2}-x+10^{20}+10^{10}-1=0$$
accurate to four decimal places. | [Solution] By Descartes' Rule of Signs:
If the coefficient sequence of the real-coefficient $n$-degree algebraic equation
$$a_{0} x^{n}+a_{1} x^{n-1}+a_{2} x^{n-2}+\cdots+a_{n-1} x+a_{n}=0\left(a_{0} \neq 0, a_{n} \neq 0\right)$$
is
$$\left\{a_{0}, a_{1}, a_{2}, \cdots, a_{n-1}, a_{n}\right\}$$
and the number of sign... | x \approx 10^{5} \pm 0.0016 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,484 |
1.48 In the expression $x_{1}: x_{2}: \cdots: x_{n}$, use parentheses to indicate the order of operations, and the result can be written in fractional form:
$$\frac{x_{i_{1}} x_{i_{2}} \cdots x_{i_{k}}}{x_{j_{j_{1}}}^{x_{j_{2}}} \cdots x_{j_{n-k}}}$$
(At the same time, each letter in $x_{1}, x_{2}, \cdots, x_{n}$ may a... | [Solution] Clearly, $x_{1}$ will be in the numerator of the resulting fraction.
When parentheses are added arbitrarily, the division sign before $x_{2}$ or the expression involving $x_{2}$ will be in the denominator, so $x_{2}$ will always be in the denominator.
We will use induction to prove that each of the remainin... | 2^{n-2} | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 735,485 |
4. 67 Euler's conjecture was disproved by American mathematicians in 1960, who confirmed the existence of a positive real number $n$, such that
$$133^{5}+110^{5}+84^{5}+27^{5}=n^{5},$$
Find the value of $n$. | [Solution] Clearly, $n \geqslant 134$. On the other hand,
$$\begin{array}{l}
n^{5}= 133^{5}+110^{5}+84^{5}+27^{5} \\
<133^{5}+110^{5}+(84+27)^{5} \\
<3 \cdot 133^{5} \\
<\frac{3125}{1024} \cdot 133^{5} \\
=\left(\frac{5 \times 133}{4}\right)^{5} \\
=\left(166 \frac{1}{4}\right)^{5}
\end{array}$$
Therefore, $n \leqsla... | 144 | Number Theory | math-word-problem | Yes | Yes | inequalities | false | 735,486 |
4-68 In a competition with $n(n \geqslant 2)$ participants lasting $k$ days, each participant's score each day is exactly $1,2, \cdots, n$ (no two participants have the same score), and after $k$ days, each participant's total score is 26. Try to find all possible values of $(n, k)$.
In a competition with $n(n \geqsla... | [Sol] Calculate the total score of all players in two different ways, we get
$$k(1+2+\cdots+n)=26 n$$
So $k(n+1)=52$.
And $52=2^{2} \times 13$, which has 6 factors: $1,2,4,13,26,52$. Since $n+1 \geqslant 3, k \geqslant$ 2 (if there is only one day, it is impossible for all players to have the same total score), there ... | (n, k)=(25,2),(12,4),(3,13) | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 735,487 |
4-69 Let $a$ be a positive real number, $b=\left(a+\sqrt{a^{2}+1}\right)^{\frac{1}{3}}+\left(a-\sqrt{a^{2}+1}\right)^{\frac{1}{3}}$, prove: $b$ is a positive integer if and only if $a$ is a positive integer of the form $\frac{1}{2} n\left(n^{2}+3\right)$ $(n \in$ $N)$ | [Proof] Let $\alpha=\left(a+\sqrt{a^{2}+1}\right)^{\frac{1}{3}}, \beta=\left(a-\sqrt{a^{2}+1}\right)^{\frac{1}{3}}$, then
$$b^{3}=(\alpha+\beta)^{3}=\alpha^{3}+\beta^{3}+3 \alpha \beta(\alpha+\beta)=2 a-3 b$$
That is, $b^{3}=2 a-3 b$.
This means that $b$ is a root of the cubic equation
$$x^{3}+3 x-2 a=0$$
Necessity: ... | proof | Algebra | proof | Yes | Yes | inequalities | false | 735,488 |
4-70 The sequence of functions $\left\{f_{n}(x)\right\}$ is recursively defined by:
$$\left\{\begin{array}{l}
f_{1}(x)=\sqrt{x^{2}+48} \\
f_{n+1}(x)=\sqrt{x^{2}+6 f_{n}(x)}
\end{array} \text { for each } n \geqslant 1 .\right.$$
Find all real solutions to the equation $f_{n}(x)=2 x$. | [Solution] From the given information, we have
$$f_{n}(x)>0 \quad(n=1,2, \cdots)$$
Therefore, the equation $f_{n}(x)=2 x$ has only positive solutions.
First, we use mathematical induction to prove that $x=4$ is a solution to the equation
$$f_{n}(x)=2 x$$
In fact, when $n=1$, substituting $x=4$ into the left side of (... | 4 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,489 |
4. 71 If the roots of the equation $x^{3}-x^{2}-x-1=0$ are $a, b, c$. Prove:
(1) $a, b, c$ are distinct;
(2) $\frac{a^{1982}-b^{1982}}{a-b}+\frac{b^{1982}-c^{1982}}{b-c}+\frac{c^{1982}-a^{1982}}{c-a}$ is an integer. | [Proof] (1) By Vieta's formulas, we have
$$\left\{\begin{array}{l}
a+b+c=1 \\
a b+b c+c a=-1 \\
a b c=1
\end{array}\right.$$
If two of $a, b, c$ are equal, without loss of generality, let $a=b$. Then substituting into the above equations, we get
$$\left\{\begin{array}{l}
2 a+c=1 \\
a^{2}+2 a c=-1 \\
a^{2} c=1
\end{arr... | proof | Algebra | proof | Yes | Yes | inequalities | false | 735,490 |
4・72 Let $z_{1}, z_{2}, \cdots, z_{n}$ be complex numbers, satisfying
$$\left|z_{1}\right|+\left|z_{2}\right|+\cdots+\left|z_{n}\right|=1$$
Prove that among the above $n$ complex numbers, there must exist some complex numbers, the modulus of whose sum is not less than $\frac{1}{6}$. | [Proof] Let $z_{k}=a_{k}+b_{k} i\left(i^{2}=-1, k=1,2, \cdots, n\right)$. Then
$$\left|z_{k}\right| \leqslant\left|a_{k}\right|+\left|b_{k}\right|$$
Thus, we have
$$\begin{aligned}
1 & =\sum_{k=1}^{n}\left|z_{k}\right| \\
& \leqslant \sum_{k=1}^{n}\left|a_{k}\right|+\sum_{k=1}^{n}\left|b_{k}\right| \\
& =\sum_{a_{k} \... | proof | Inequalities | proof | Yes | Yes | inequalities | false | 735,491 |
$4 \cdot 73$ Let $n$ be a natural number, prove that the equation
$$z^{n+1}-z^{n}-1=0$$
has a complex root of modulus 1 if and only if $n+2$ is divisible by 6. | $\cos \frac{\pi}{3} \pm i \sin \frac{\pi}{3}$ (i.e., the points $w_{1}$ and $w_{2}$ in the figure), and
$$w-1=\cos \frac{2 \pi}{3} \pm i \sin \frac{2 \pi}{3}$$
Thus,
$$\begin{aligned}
1 & =w^{n}(w-1) \\
& =\left(\cos \frac{\pi}{3} \pm i \sin \frac{\pi}{3}\right)^{n} \cdot\left(\cos \frac{2 \pi}{3} \pm i \sin \frac{2 \... | proof | Algebra | proof | Yes | Yes | inequalities | false | 735,492 |
$4 \cdot 74$ If $\alpha, \beta, \gamma$ are the roots of the equation $x^{3}-x-1=0$, find
$$\frac{1+\alpha}{1-\alpha}+\frac{1+\beta}{1-\beta}+\frac{1+\gamma}{1-\gamma}$$
the value. | ［Solution］From the given, we have
$$x^{3}-x-1=(x-\alpha)(x-\beta)(x-\gamma),$$
Therefore,
$$\left\{\begin{array}{l}
\alpha+\beta+\gamma=0 \\
\alpha \beta+\beta \gamma+\gamma \alpha=-1 \\
\alpha \beta \gamma=1
\end{array}\right.$$
Thus, we have
$$\begin{array}{l}
\frac{1+\alpha}{1-\alpha}+\frac{1+\beta}{1-\beta}+\fra... | -7 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,493 |
4・76 Let $n \geqslant 4, \alpha_{1}, \alpha_{2}, \cdots, \alpha_{n} ; \beta_{1}, \beta_{2}, \cdots, \beta_{n}$ be two sets of real numbers, satisfying $\sum_{j=1}^{n} \alpha_{j}^{2}<1, \sum_{j=1}^{n} \beta_{j}^{2}<1$.
Let
$$\begin{array}{l}
A^{2}=1-\sum_{j=1}^{n} \alpha_{j}^{2}, B^{2}=1-\sum_{j=1}^{n} \beta_{j}^{2} \\
... | [Solution] Clearly, when $\lambda=0$, the equation in the problem has only real roots.
Now we prove that when $\lambda \neq 0$, the roots of the equation in the problem cannot all be real.
In fact, if $\lambda \neq 0$, and the $n$ roots of the equation in the problem are all real. At this time, we denote these $n$ root... | \lambda=0 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,494 |
$1 \cdot 49$ integers $1,2, \cdots, n$ are arranged in a permutation such that each number is either greater than all the numbers before it, or less than all the numbers before it. How many such permutations are there? | ［Solution］ Let the number of permutations we are looking for be $a_{n}$.
When $n=1$, there is only the number 1, so obviously $a_{1}=1$.
For $n \geqslant 2$, if the number $n$ is placed in the $i$-th position, then the $n-i$ numbers after it are completely determined, i.e., they can only be $n-i, n-i-1, \cdots, 1$. The... | 2^{n-1} | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 735,495 |
$4 \cdot 77$ Given three quadratic trinomials:
$$\begin{array}{l}
p_{1}(x)=x^{2}+p_{1} x+q_{1} \\
p_{2}(x)=x^{2}+p_{2} x+q_{2} \\
p_{3}(x)=x^{2}+p_{3} x+q_{3}
\end{array}$$
Prove that the equation $\left|p_{1}(x)\right|+\left|p_{2}(x)\right|=\left|p_{3}(x)\right|$ has at most 8 roots. | [Proof] Each root of the original equation should be a root of
$$\pm p_{1}(x) \pm p_{2}(x)= \pm p_{3}(x)$$
and thus a root of
$$\pm p_{1}(x) \pm p_{2}(x)=p_{3}(x).$$
That is, a root of one of the following four quadratic equations:
$$\begin{array}{l}
p_{1}(x)+p_{2}(x)=p_{3}(x), \\
p_{1}(x)-p_{2}(x)=p_{3}(x), \\
-p_{1... | proof | Algebra | proof | Yes | Yes | inequalities | false | 735,496 |
4. 78 Solve the equation $3^{x^{2}+1}+3^{x^{2}-1}-270=0$.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | ［Solution］ The original equation is
$$\begin{aligned}
3 \cdot 3^{x^{2}}+\frac{1}{3} \cdot 3^{x^{2}}-270 & =0 \\
10 \cdot 3^{x^{2}} & =810 \\
3^{x^{2}} & =3^{4} \\
x^{2} & =4 \\
x & = \pm 2
\end{aligned}$$
Upon verification, $x= \pm 2$ are solutions to the original equation. | null | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,497 |
$4 \cdot 79$ Solve the equation $\sqrt[r]{9}+\sqrt[x]{6}=\sqrt[x]{4}$.
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 is
$$\left(\sqrt[x]{\frac{3}{2}}\right)^{2}+\sqrt[x]{\frac{3}{2}}-1=0$$
Let $y=\sqrt[x]{\frac{3}{2}}$, then we get $y^{2}+y-1=0$.
Solving it, we get $y=\frac{-1 \pm \sqrt{5}}{2}$.
$\because y>0, \therefore$ take $y=\frac{\sqrt{5}-1}{2}$.
That is $\square$
$$\sqrt[x]{\frac{3}{2}}=\frac{... | null | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,498 |
4. 81 Solve the equation in the range of positive rational numbers:
$$x^{y}=y^{x} \quad(x \neq y)$$ | [Solution] Without loss of generality, let $y > x$, and set $y = kx$ ($k$ is a rational number greater than 1). Thus, from the original equation, we have
$$x^{kx} = (kx)^x,$$
which implies
$$\begin{array}{l}
x^k = kx \\
x^{k-1} = k \\
x = k^{\frac{1}{k-1}} \\
y = kx = k^{\frac{k}{k-1}}
\end{array}$$
Let $\frac{1}{k-1... | null | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,499 |
4・82 Solve the equation $\log _{3} x+\log _{x} 3-2 \log _{3} x \log _{x} 3=\frac{1}{2}$. | [Solution] The original equation is
$$\log _{3} x+\log _{x} 3-2=\frac{1}{2} .$$
Substituting $\log _{x} 3=\frac{1}{\log _{3} x}$ into the above equation and simplifying, we get
$$2 \log _{3}^{2} x-5 \log _{3} x+2=0$$
Thus, $\log _{3} x=\frac{1}{2}, \log _{3} x=2$.
This means $x=\sqrt{3}, x=9$.
Upon verification, both... | x=\sqrt{3}, x=9 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,500 |
4.83 Solve the equation $a^{\log _{\sqrt{b}} x}-5 a^{\log _{b} a}+6=0$.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly. | [Solution] The original equation is
$$a^{2 \log _{b} x}-5 a^{\log _{b} a}+6=0 .$$
Let $\log _{b} x=y$, then $x=b^{y}$. Substituting into the equation and rearranging, we get
$$\begin{array}{l}
a^{2 y}-5 a^{y}+6=0 \\
a^{y}=3, a^{y}=2
\end{array}$$
Therefore, $y=\log _{a} 3, y=\log _{a} 2$.
That is, $\log _{b} x=\log _... | null | Logic and Puzzles | math-word-problem | Yes | Yes | inequalities | false | 735,501 |
4. 84 In the equation $\sin ^{2} A+\sin ^{2} B+\sin ^{2} C=1$, let $A, B, C$ all be acute angles. Prove: $\frac{\pi}{2} \leqslant A+B+C \leqslant \pi$.
Given the equation $\sin ^{2} A+\sin ^{2} B+\sin ^{2} C=1$, where $A, B, C$ are all acute angles, we need to prove that $\frac{\pi}{2} \leqslant A+B+C \leqslant \pi$. | [Proof] From the given,
$$\begin{aligned}
\sin ^{2} A & =1-\sin ^{2} B-\sin ^{2} C \\
& =\cos ^{2} B-\sin ^{2} C \\
& =\cos ^{2} B-\sin ^{2} C \cos ^{2} B+\sin ^{2} C \cos ^{2} B-\sin ^{2} C \\
& =\cos ^{2} B \cos ^{2} C-\sin ^{2} C \sin ^{2} B \\
& =(\cos B \cos C-\sin C \sin B)(\cos B \cos C+\sin C \sin B) \\
& =\cos... | proof | Algebra | proof | Yes | Yes | inequalities | false | 735,502 |
1. 50 An ordered pair of non-negative integers $(m, n)$ is called "simple" if no carrying is required when adding $m+n$, and $m+n$ is called the sum of the ordered pair $(m, n)$. Find the number of "simple" ordered pairs of non-negative integers whose sum is 1492. | [Solution] The unit digits of $m, n$ can be 2, 0; 1, 1; 0, 2, three possibilities.
Similarly, the tens digit has ten possibilities, the hundreds digit has five, and the thousands digit has two. Therefore, the total number is $3 \times 10 \times 5 \times 2=300$. | 300 | Combinatorics | math-word-problem | Yes | Yes | inequalities | false | 735,504 |
4・87 Solve the equation $\log _{x} 3 \cdot \log _{3 x} 3=\log _{9} 3$.
| [Solution] By the definition of logarithms, we know that $x>0$, and $x \neq 1, x \neq \frac{1}{3}, x \neq \frac{1}{9}$. Using the change of base formula, the original equation can be rewritten as:
i.e., $\square$
$$\frac{1}{\log _{3} x} \cdot \frac{1}{\log _{3} 3 x}=\frac{1}{\log _{3} 9 x}$$
$$\begin{array}{l}
\log _{... | x=3^{\pm \sqrt{2}} | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,505 |
4.90 Given a quadratic equation in terms of $\cos x$
$a \cos ^{2} x+b \cos x+c=0$, where $a, b, c$ are known real numbers. Find a quadratic equation whose roots are $\cos 2 x$. In the case of $a=4, b=2, c=-1$, compare the given equation with the newly derived equation. | [Solution] Transform the given equation into
$$a \cos ^{2} x+c=-b \cos x .$$
Square both sides to get
$$\begin{array}{l}
a^{2} \cos ^{4} x+2 a \cos ^{2} x \cdot c+c^{2}=b^{2} \cos ^{2} x \\
a^{2} \cos ^{4} x+\left(2 a c-b^{2}\right) \cos ^{2} x+c^{2}=0
\end{array}$$
Substitute $\cos ^{2} x=\frac{1+\cos 2 x}{2}$ into ... | 4 \cos ^{2} 2 x+2 \cos 2 x-1=0 | Algebra | math-word-problem | Yes | Yes | inequalities | false | 735,507 |
4.93 Let \( F_{r}=x^{r} \sin (r A)+y^{r} \sin (r B)+z^{r} \sin (r C) \), where \( x, y, z \), and \( A, B, C \) are real numbers, and \( A+B+C \) is an integer multiple of \( \pi \). Prove that if \( F_{1}=F_{2}=0 \), then for all positive integers \( r \), \( F_{r}=0 \). | [Proof] Let the complex numbers $\alpha=x(\cos A+i \sin A), \beta=y(\cos B+i \sin B)$, $\gamma=z(\cos C+i \sin C)$. Suppose $\alpha, \beta, \gamma$ are the roots of the cubic equation
$$u^{3}-a u^{2}+b u-c=0$$
By Vieta's formulas, we have:
$$\begin{aligned}
a & =\alpha+\beta+\gamma=x \cos A+y \cos B+z \cos C+i F_{1}=\... | proof | Algebra | proof | Yes | Yes | inequalities | false | 735,508 |
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