problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
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192*. The sequence $\left(x_{n}\right)$ is defined by the conditions:
$$
x_{1}=\frac{1}{2}, \quad x_{n+1}=\sqrt{\frac{1-\sqrt{1-x_{n}^{2}}}{2}}
$$
Find a formula for $x_{n}$. | $\triangle$ From the condition, it is clear that $0 \leq x_{n} \leq 1$. Let $x_{n}=\sin \alpha$, where $\alpha \in\left[0 ; \frac{\pi}{2}\right]$. We get:
$$
x_{n+1}=\sqrt{\frac{1-\sqrt{1-\sin ^{2} \alpha}}{2}}=\sqrt{\frac{1-\cos \alpha}{2}}=\sin \frac{\alpha}{2}
$$
The argument of $x_{n+1}$ under the sine sign turne... | x_{n}=\sin\frac{\pi}{3\cdot2^{n}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,614 |
193*. The sequence $\left(x_{n}\right)$ is defined by the conditions:
$$
x_{1}=2, \quad x_{n+1}=\frac{3 x_{n}-1}{3-x_{n}}
$$
Find a formula for $x_{n}$.
## $5.3^{*}$.
Let's tackle problems involving the calculation of the sum of the first $n$ terms of sequences. Such problems have already been encountered in § 3 (p... | $\triangle$ Let's write the identity
$$
(k+1)^{2}-k^{2}=2 k+1
$$
and set $k=0,1,2, \ldots, n$ in it. We will get a series of equalities, which we then add term by term:
$$
\begin{array}{r}
1^{2}=\quad 1, \\
+\quad 2^{2}-1^{2}=2 \cdot 1+1, \\
3^{2}-2^{2}=2 \cdot 2+1, \\
4^{2}-3^{2}=2 \cdot 3+1, \\
\cdots \cdots \cdot... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,615 |
198. Calculate the sum:
$$
S=(1+1^2)+(2+2^2)+(3+3^2)+\cdots+(n+n^2)
$$ | $\triangle$ Let's break the sum $S$ into two sums, and then apply formulas (1) and (2):
$$
\begin{aligned}
& S=(1+2+3+\cdots+n)+\left(1^{2}+2^{2}+3^{2}+\cdots+n^{2}\right)= \\
& =\frac{n(n+1)}{2}+\frac{n(n+1)(2 n+1)}{6}=\frac{1}{6}(3 n(n+1)+(n(n+1)(2 n+1))= \\
& =\frac{1}{6} n(n+1)(2 n+4)=\frac{1}{3} n(n+1)(n+2)
\end{... | \frac{1}{3}n(n+1)(n+2) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,616 |
206. Prove that the sum of the first $n$ terms of the Fibonacci sequence satisfies the condition:
$$
a_{1}+a_{2}+\cdots+a_{n}=a_{n+2}-1
$$ | $\triangle$ Let's represent the recurrence relation $a_{k+2}=a_{k+1}+a_{k}$ as
$$
a_{k}=a_{k+2}-a_{k+1}
$$
Set $k=1,2,3, \ldots, n$ in the last equation and add the resulting equations term by term:
$$
\begin{aligned}
& a_{1}=a_{3}-a_{2} \\
&+\quad a_{2}=a_{4}-a_{3} \\
& a_{3}=a_{5}-a_{4} \\
& \cdots \cdots \cdots \... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 34,617 |
214. One of the angles of a triangle is $120^{\circ}$, and the lengths of the sides form an arithmetic progression. Find the ratio of the lengths of the sides of the triangle. | $\triangle$ Let the lengths of the sides of the triangle in ascending order be $a$, $a+d$, and $a+2d$. Then the largest side is opposite the largest angle of the triangle, which is $120^{\circ}$.
By the Law of Cosines, we have:
$$
\begin{aligned}
& (a+2 d)^{2}=a^{2}+(a+d)^{2}-2 a(a+d) \cos 120^{\circ} \\
& a^{2}+4 a ... | 3:5:7 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 34,618 |
217. The sum of the first $n$ terms of a certain infinite numerical sequence ($a_{n}$) for any $n$ is expressed by the formula
$$
S_{n}=a n^{2}+b n
$$
where $a$ and $b$ are given numbers. Prove that this sequence is an arithmetic progression. | $\triangle$ By the condition for any $n>1$
$$
S_{n-1}=a(n-1)^{2}+b(n-1)
$$
Then we will have:
$a_{n}=S_{n}-S_{n-1}=a n^{2}+b n-a(n-1)^{2}-b(n-1)=$
$=a n^{2}+b n-a n^{2}+2 a n-a-b n+b=b-a+2 a n=$ $=(b-a)+2 a((n-1)+1)=a+b+2 a(n-1)$.
The obtained expression for $a_{n}$ represents the formula for the $n$-th term of an... | proof | Algebra | proof | Yes | Yes | olympiads | false | 34,619 |
219. An infinite arithmetic progression ( $a_{n}$ ) with a positive difference contains terms equal to 7, 15, and 27. Is it true that it must also contain a term equal to 1999? | $\triangle$ Let the first term of the progression be the number 7. Then
$$
15=7+d(k-1), \quad 27=7+d(m-1)
$$
where $d$ is the common difference of the progression, $k$ and $m$ are natural numbers greater than 1, and $m>k$. From this,
$$
8=d(k-1), \quad 20=d(m-1)
$$
61
Let $k-1=a$; then $d=\frac{8}{a}$.
The number... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 34,620 |
222. In the arithmetic progression $\left(a_{n}\right)$
$$
a_{k}=l, \quad a_{l}=k(k \neq l)
$$
Calculate $a_{n}$ | $\triangle$ Subtract the equalities term by term
$$
l=a_{1}+d(k-1), \quad k=a_{1}+d(l-1)
$$
where $d$ is the common difference of the arithmetic progression. We get:
$$
l-k=d(k-l), \quad d=-1
$$
Then
$$
l=a_{1}-(k-1), \quad a_{1}=k+l-1
$$
It remains to find $a_{n}$ :
$$
a_{n}=a_{1}+d(n-1)=k+l-1-(n-1)=k+l-n
$$
A... | a_{n}=k+-n | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,621 |
226. Prove that if the numbers $a, b$ and $c$ form an arithmetic progression, then the numbers
$$
a^{2}+a b+b^{2}, \quad a^{2}+a c+c^{2}, \quad b^{2}+b c+c^{2}
$$
also form (in the given order) an arithmetic progression. | $\triangle$ Let $d-$ be the difference of the given progression. We will have:
$\left(a^{2}+a c+c^{2}\right)-\left(a^{2}+a b+b^{2}\right)=(a c-a b)+\left(c^{2}-b^{2}\right)=$
$=(c-b)(a+c+b)=d(a+b+c)$
$\left(b^{2}+b c+c^{2}\right)-\left(a^{2}+a c+c^{2}\right)=(b c-a c)+\left(b^{2}-a^{2}\right)=$
$=(b-a)(c+b+a)=d(a+b... | proof | Algebra | proof | Yes | Yes | olympiads | false | 34,622 |
230*. Prove that the sequence of squares of natural numbers $1 ; 4$; $9 ; 16 ; 25 ; 36 \ldots$ does not contain any infinite arithmetic progression. | $\triangle$ Let's form a sequence of differences between consecutive terms (subsequent and preceding) of the given sequence:
$$
3 ; 5 ; 7 ; 9 ; 11 ; \ldots
$$
This last sequence increases indefinitely as $n$ increases indefinitely. Therefore, if we assume that such an arithmetic progression exists and denote its comm... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 34,623 |
243. Prove that if $a, b, c, d$ are consecutive terms of a geometric progression, then
$$
(a-d)^{2}=(a-c)^{2}+(b-c)^{2}+(b-d)^{2}
$$ | $\triangle$ Simplify this equality:
$a^{2}-2 a d+d^{2}=a^{2}-2 a c+c^{2}+b^{2}-2 b c+c^{2}+b^{2}-2 b d+d^{2}$, $b^{2}+c^{2}+a d=a c+b c+b d$.
Let's denote the common ratio of the geometric progression by $q$. We get:
$$
a^{2} q^{2}+a^{2} q^{4}+a^{2} q^{3}=a^{2} q^{2}+a^{2} q^{3}+a^{2} q^{4}
$$
The identity is prove... | proof | Algebra | proof | Yes | Yes | olympiads | false | 34,624 |
248. Can a geometric progression with $k$ terms be selected from the infinite sequence $1 ; \frac{1}{2} ; \frac{1}{3} ; \ldots ; \frac{1}{n} ; \ldots$ for any natural number $k>2$?
$3^{*}$
67 | $\triangle$ This can be done, and in more than one way. For example, a geometric progression $1 ; \frac{1}{2} ; \frac{1}{4} ; \ldots ; \frac{1}{2^{k-1}}$ would work.
Answer: it is possible. | itispossible | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 34,625 |
250*. Can the numbers 2, 3, 5 be members of the same geometric progression with positive terms? | $\triangle$ Suppose it is possible. Let's consider 2 as the first term of a geometric progression, and the common ratio $q$ greater than 1. Then
$$
3=2 q^{k-1}, \quad 5=2 q^{n-1} \quad (k \in \mathbb{N}, n \in \mathbb{N}, 1<k<n)
$$
From each of the last two equations, express $q$ and equate the two obtained expressio... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 34,626 |
253. The digits of a three-digit number form a geometric progression with different terms. If this number is decreased by 200, the result is a three-digit number whose digits form an arithmetic progression. Find the original three-digit number. | $\triangle$ The number of three-digit numbers with distinct digits forming a geometric progression is small, and all of them can be easily found by enumeration:
$124,421,139,931,248,842,469,964$.
We will discard the numbers 124 and 139, as they are less than 200. Subtract 200 from the remaining numbers and find the n... | 842 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 34,627 |
256. Find the numbers $a$ and $b$, if the numbers $a, 1, b$ form an arithmetic progression, and the numbers $a^{2}, 1, b^{2}$ form a geometric progression. List all solutions. | $\triangle$ Let's use the characteristic properties of arithmetic and geometric progressions:
$$
\left\{\begin{array} { l }
{ 1 = \frac { a + b } { 2 } , } \\
{ 1 = \sqrt { a ^ { 2 } b ^ { 2 } } , }
\end{array} \quad \left\{\begin{array}{l}
a+b=2 \\
a^{2} b^{2}=1
\end{array}\right.\right.
$$
It remains to solve the ... | =b=1;\quad=1+\sqrt{2},b=1-\sqrt{2};\quad=1-\sqrt{2},b=1+\sqrt{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,628 |
262*. Given two geometric progressions $a_{1}, a_{2}, a_{3}$ and $b_{1}, b_{2}, b_{3}$ with positive terms, such that
$$
a_{1}+a_{2}+a_{3}=b_{1}+b_{2}+b_{3}
$$
The numbers $a_{1} b_{1}, a_{2} b_{2}, a_{3} b_{3}$ form an arithmetic progression. Prove that $a_{2}=b_{2}$. | $\triangle$ Let's denote the common ratio of the first geometric progression by $q$, and the second by $q_{1}$. We will have the system of equations:
$$
\left\{\begin{array}{l}
a_{1}+a_{1} q+a_{1} q^{2}=b_{1}+b_{1} q_{1}+b_{1} q_{1}^{2} \\
2 a_{1} b_{1} q q_{1}=a_{1} b_{1}+a_{1} b_{1} q^{2} q_{1}^{2}
\end{array}\right... | a_{2}=b_{2} | Algebra | proof | Yes | Yes | olympiads | false | 34,629 |
266*. Prove that any infinite arithmetic progression, all members of which are natural numbers, contains an infinite geometric progression. | $\triangle$ Let the first term of an arithmetic progression be denoted by $a$, and the common difference by $d$. Then the number
$$
a + da = a(1 + d)
$$
is a term of the arithmetic progression. Consequently, the numbers
$$
a(1+d)(1+d) = a(1+d)^{2}, a(1+d)^{3}, \ldots, a(1+d)^{n}(n \in \mathbb{N})
$$
are also terms ... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 34,630 |
274. Solve the equations:
a) $\left(2 x^{2}+3 x-1\right)^{2}-5\left(2 x^{2}+3 x+3\right)+24=0$;
b) $(x-1)(x+3)(x+4)(x+8)=-96$
c) $(x-1)(x-2)(x-4)(x-8)=4 x^{2}$. | $\triangle$ a) Let's introduce the substitution: $2 x^{2}+3 x-1=y$. Then
$$
y^{2}-5(y+4)+24=0, \quad y^{2}-5 y+4=0
$$
From here, $y_{1}=4, y_{2}=1$. Knowing $y$, we find $x$.
b) Multiply the first factor by the fourth and the second by the third in the left part of the equation:
$$
\left(x^{2}+7 x-8\right)\left(x^{... | )1,-2,\frac{1}{2},-\frac{5}{2};b)0,-7,\frac{-1\\sqrt{33}}{2};)5\\sqrt{17} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,631 |
276. Solve the equation:
$$
\left(x^{2}-16\right)(x-3)^{2}+9 x^{2}=0
$$ | $\triangle$ Transform the equation:
$x^{2}(x-3)^{2}-16(x-3)^{2}+9 x^{2}=0$, $x^{2}\left((x-3)^{2}+9\right)-16(x-3)^{2}=0$, $x^{2}\left(x^{2}-6 x+18\right)-16(x-3)^{2}=0$, $x^{4}-6 x^{2}(x-3)-16(x-3)^{2}=0$.
Divide the last equation by $(x-3)^{2}$ (check that no solutions are lost):
$$
\left(\frac{x^{2}}{x-3}\right)^... | -1\\sqrt{7} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,632 |
278. Solve the equation:
$$
(x+3)^{4}+(x+5)^{4}=16
$$ | $\triangle$ Let $x+4=y$. Why exactly $x+4$? Here's where it comes from:
$$
\frac{(x+3)+(x+5)}{2}=x+4
$$
We have:
$$
(y-1)^{4}+(y+1)^{4}=16
$$
Now we need to square $y-1$ and $y+1$ in the left part of the equation, and then square the result again. After simplifications, a biquadratic equation is formed:
$$
y^{4}+6... | -3,-5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,633 |
282. Solve the equation:
$$
x^{4}-5 x^{3}+8 x^{2}-5 x+1=0
$$ | $\triangle$ Let's divide this equation term by term by $x^{2}$:
$$
x^{2}-5 x+8-\frac{5}{x}+\frac{1}{x^{2}}=0, \quad\left(x^{2}+\frac{1}{x^{2}}\right)-5\left(x+\frac{1}{x}\right)+8=0
$$
Let $x+\frac{1}{x}=y$. Square this equality:
$$
x^{2}+2+\frac{1}{x^{2}}=y^{2}, \quad x^{2}+\frac{1}{x^{2}}=y^{2}-2
$$
The equation ... | 1,1,\frac{3\\sqrt{5}}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,634 |
285. Solve the equation:
$$
2 x^{5}+5 x^{4}-13 x^{3}-13 x^{2}+5 x+2=0
$$ | $\triangle$ This is a reciprocal equation of odd degree. Based on the statement of the reference problem 285, it has a root $x=-1$. Let's factorize the left-hand side of the equation:
$$
(x+1)\left(2 x^{4}+3 x^{3}-16 x^{2}+3 x+2\right)=0
$$
The equation
$$
2 x^{4}+3 x^{3}-16 x^{2}+3 x+2=0
$$
is also a reciprocal eq... | -1,2,1/2,-2\\sqrt{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,635 |
288. Solve the equation:
$$
x^{4}+2 x^{3}-11 x^{2}+4 x+4=0
$$ | $\triangle$ The given equation is not reciprocal. However, let's try to solve it using the same method as for a reciprocal equation, dividing the left part term by term by $x^{2}$:
$$
x^{2}+2 x-11+\frac{4}{x}+\frac{4}{x^{2}}=0, \quad\left(x^{2}+\frac{4}{x^{2}}\right)+2\left(x+\frac{2}{x}\right)-11=0
$$
Let $x+\frac{2... | 1,2,\frac{-5\\sqrt{17}}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,636 |
290. Prove that the equation
$$
\left(a^{2}+b^{2}+c^{2}\right) x^{2}+2(a+b+c) x+3=0
$$
has real roots only when $a=b=c \neq 0$. | $\triangle$ Let's write that the discriminant $D$ of this equation is non-negative:
$\frac{1}{4} D=(a+b+c)^{2}-3\left(a^{2}+b^{2}+c^{2}\right) \geq 0$,
$-2 a^{2}-2 b^{2}-2 c^{2}+2 a b+2 a c+2 b c \geq 0$
$2 a^{2}+2 b^{2}+2 c^{2}-2 a b-2 a c-2 b c \leq 0$,
$\left(a^{2}+b^{2}-2 a b\right)+\left(a^{2}+c^{2}-2 a c\righ... | proof | Algebra | proof | Yes | Yes | olympiads | false | 34,637 |
296*. Given three numbers $a, b, c$ such that $a<b<c$. Prove that the equation
$$
(x-a)(x-b)+(x-a)(x-c)+(x-b)(x-c)=0
$$
has two distinct roots $x_{1}$ and $x_{2}$, and that
$$
a<x_{1}<b<x_{2}<c
$$ | $\triangle$ Obviously, the discriminant of the quadratic equation is not enough here, so let's try another method of solving.
Let's denote the quadratic trinomial on the left side of the equation as $f(x)$. Let's determine the signs of $f(a)$, $f(b)$, and $f(c)$:
$$
\begin{gathered}
f(a)=(a-b)(a-c)>0, \quad f(b)=(b-a... | proof | Algebra | proof | Yes | Yes | olympiads | false | 34,638 |
298. Solve the equation:
$$
x^{3}-2 x^{2}-\left(a^{2}-a-1\right) x+a^{2}-a=0
$$ | $\triangle$ One root of the equation can be guessed here: $x_{1}=1$ (check it!). We get:
$$
(x-1)\left(x^{2}-x+a-a^{2}\right)=0
$$
The roots of the quadratic equation
$$
x^{2}-x+a-a^{2}=0
$$
can also be guessed: $x_{2}=a, x_{3}=1-a$.
Answer: $1, a, 1-a$. | 1,,1- | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,639 |
303. Solve the equation:
$$
\left(a-x^{2}\right)^{2}=a+x
$$ | $\triangle$ Let's expand the brackets, collect all terms on the left side of the equation:
$$
x^{4}-2 a x^{2}-x+a^{2}-a=0
$$
Let's try the following approach: consider this equation as a quadratic in terms of $a$:
$$
a^{2}-\left(2 x^{2}+1\right) a+\left(x^{4}-x\right)=0
$$
Find its discriminant:
$$
D=\left(2 x^{2}... | \begin{aligned}&if\geq\frac{3}{4},thenx_{1,2}=\frac{-1\\sqrt{4a-3}}{2},\quadx_{3,4}=\frac{1\\sqrt{4a+1}}{2}\\&if-\frac{1}{4}\leq | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,640 |
305. Solve the equation:
$$
10 x^{4}+3 x^{3}+5 x^{2}+5 x+8=0
$$ | $\triangle$ Let's make the equation reduced:
$$
x^{4}+\frac{3}{10} x^{3}+\frac{1}{2} x^{2}+\frac{1}{2} x+\frac{4}{5}=0
$$
Complete the sum $x^{4}+\frac{3}{10} x^{3}$ to the square of a sum:
$$
\left(x^{2}+\frac{3}{20} x\right)^{2}+\left(\frac{191}{400} x^{2}+\frac{1}{2} x+\frac{4}{5}\right)=0
$$
The discriminant of... | nosolutions | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,641 |
309*. Solve the equation:
$$
x^{3}+x^{2}+x=-\frac{1}{3}
$$ | $\triangle$ Transform the equation:
$$
\begin{gathered}
3 x^{3}+3 x^{2}+3 x+1=0 \\
\left(x^{3}+3 x^{2}+3 x+1\right)=-2 x^{3} \\
(x+1)^{3}=-2 x^{3}
\end{gathered}
$$
Then
$$
x+1=-x \sqrt[3]{2}, \quad x=-\frac{1}{1+\sqrt[3]{2}}
$$
Answer: $-\frac{1}{1+\sqrt[3]{2}}$. | -\frac{1}{1+\sqrt[3]{2}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,642 |
314*. Prove that all roots of the equation
$$
(x+1)(x+2)(x+3) \cdots(x+2001)=2001
$$
are less than the number $\frac{1}{2000 \text { ! }}$; | $\triangle$ If $x=a$ is a negative root of the equation, then it is certainly less than $\frac{1}{2000 \text { ! }}$.
If the root $x=a$ of the equation is positive, we have:
$$
a+1=\frac{2001}{(a+2)(a+3) \cdots(a+2001)}<\frac{2001}{2 \cdot 3 \cdot 4 \cdots \cdots \cdot 2001}=\frac{1}{2000!}
$$
Therefore, the number ... | proof | Algebra | proof | Yes | Yes | olympiads | false | 34,643 |
315*. Find all roots of the equation
$$
8 x\left(2 x^{2}-1\right)\left(8 x^{4}-8 x^{2}+1\right)=1
$$
satisfying the condition $0<x<1$. | $\triangle$ Let's introduce a trigonometric substitution $x=\cos \alpha$, where $\alpha \in\left(0 ; \frac{\pi}{2}\right)$. Then
$\cos 2 \alpha=2 \cos ^{2} \alpha-1=2 x^{2}-1$,
$\cos 4 \alpha=2 \cos ^{2} 2 \alpha-1=2\left(2 x^{2}-1\right)^{2}-1=8 x^{4}-8 x^{2}+1$.
The original equation transforms into a trigonometri... | \cos\pi/9,\cos\pi/3,\cos2\pi/7 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,644 |
324. Solve the system of equations:
$$
\left\{\begin{array}{l}
x^{2}+x y+y=1 \\
y^{2}+x y+x=1
\end{array}\right.
$$ | $\triangle$ Subtract the equations of the system:
$$
x^{2}+y-y^{2}-x=0, \quad(x-y)(x+y-1)=0
$$
Now we need to consider two cases: $x-y=0$ and $x+y-1=0$. Work through these cases on your own.
Answer: $(-1 ;-1),(x ; 1-x)$, where $x-$ is any number. | (-1,-1),(x,1-x) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,645 |
327. Solve the system of equations:
$$
\left\{\begin{array}{l}
y+z=x y z \\
z+x=x y z \\
x+y=x y z
\end{array}\right.
$$ | $\triangle$ Subtract the first and second, as well as the first and third equations of the system:
$$
y-x=0, \quad z-x=0
$$
From here, $x=y=z$. Then the first equation reduces to $2 x=x^{3}$. Therefore, $x_{1}=0, x_{2,3}= \pm \sqrt{2}$.
Answer: $(0 ; 0 ; 0),(\sqrt{2} ; \sqrt{2} ; \sqrt{2}),(-\sqrt{2} ;-\sqrt{2} ;-\s... | (0;0;0),(\sqrt{2};\sqrt{2};\sqrt{2}),(-\sqrt{2};-\sqrt{2};-\sqrt{2}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,646 |
329. Solve the system of equations:
$$
\left\{\begin{array}{l}
x y^{2}-2 y^{2}+3 x=18 \\
3 x y+5 x-6 y=24
\end{array}\right.
$$ | $\triangle$ Transform each equation of the system, factoring out the multiplier $x-2$ in the left part:
$y^{2}(x-2)+3 x=18, \quad 3 y(x-2)+5 x=24$
$y^{2}(x-2)+(3 x-6)=18-6, \quad 3 y(x-2)+(5 x-10)=24-10$
$(x-2)\left(y^{2}+3\right)=12, \quad(x-2)(3 y+5)=14$.
Divide the first equation of the last system by the second... | (3;3),(75/13;-3/7) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,647 |
330. Solve the system of equations:
$$
\left\{\begin{array}{l}
x^{3}+y^{3}=7(x+y) \\
x^{3}-y^{3}=13(x-y)
\end{array}\right.
$$ | $\triangle$ Let's gather all terms in each equation of the system to the left side and factor out $(x+y)$ in the first equation and $-x-y$ in the second:
$$(x+y)\left(x^{2}-x y+y^{2}-7\right)=0, \quad(x-y)\left(x^{2}+x y+y^{2}-13\right)=0.$$
When solving the last system of equations, we need to consider four cases.
... | (0;0),(3;1),(1;3),(-3;-1),(-1;-3),(\sqrt{13};-\sqrt{13}),(-\sqrt{13};\sqrt{13}),(\sqrt{7};\sqrt{7}),(-\sqrt{7};-\sqrt{7}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,648 |
332. Solve the system of equations:
$$
\left\{\begin{array}{l}
2 x^{2}-7 x y-4 y^{2}+9 x-18 y+10=0 \\
x^{2}+2 y^{2}=6
\end{array}\right.
$$ | $\triangle$ Let's consider the first equation of the system as a quadratic equation in $x$. Arrange its terms in the left part in descending order of $x$:
$$
2 x^{2}+(9-7 y) x-4 y^{2}-18 y+10=0
$$
Find the discriminant of the quadratic equation:
$D=(9-7 y)^{2}-8\left(-4 y^{2}-18 y+10\right)=$
$=81-126 y+49 y^{2}+32... | (2;1),(-2;-1),(-22/9;-1/9) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,649 |
334. Solve the system of equations:
$$
\left\{\begin{array}{l}
x^{2}+y z=y+z \\
y^{2}+x z=x+z \\
z^{2}+x y=x+y
\end{array}\right.
$$ | $\triangle$ Subtract the second equation from the first equation of the system, and the third equation from the second:
$$
\begin{array}{ll}
x^{2}-y^{2}-z(x-y)=y-x, & y^{2}-z^{2}-x(y-z)=z-y \\
(x-y)(x+y-z+1)=0, & (y-z)(y+z-x+1)=0
\end{array}
$$
Now we need to consider four cases, depending on which of the factors in ... | (0;0;0),(1;1;1),(-1;1;1),(1;-1;1),(1;1;-1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,650 |
338. Solve the system of equations:
$$
\left\{\begin{array}{l}
x+y+z=6 \\
x y+y z+z x=11 \\
x y z=6
\end{array}\right.
$$ | $\triangle$ Express $x+y$ from the first equation of the system, and $-xy$ from the third:
$$
x+y=6-z, \quad xy=\frac{6}{z}
$$
Now transform the second equation, expressing its left side through $z$:
$$
\begin{aligned}
& xy+z(x+y)=11, \quad \frac{6}{z}+z(6-z)=11 \\
& 6+6z^2-z^3=11z, \quad z^3-6z^2+11z-6=0
\end{align... | (1;2;3),(1;3;2),(2;1;3),(2;3;1),(3;1;2),(3;2;1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,651 |
340. Solve the systems of equations:
a) $\left\{\begin{array}{l}x+y+x y=5, \\ x y(x+y)=6 ;\end{array}\right.$
b) $\left\{\begin{array}{l}x^{3}+y^{3}+2 x y=4, \\ x^{2}-x y+y^{2}=1 .\end{array}\right.$ | $\triangle$ a) Let's introduce the main symmetric polynomials $u$ and $v$. We obtain:
$$
u+v=5, \quad u v=6
$$
We find the solutions to the last system of equations:
$$
u_{1}=2, \quad v_{1}=3 ; \quad u_{2}=3, \quad v_{2}=2
$$
Returning to the variables $x$ and $y$, we will have two systems of equations:
$$
\left\{... | (2;1),(1;2);(1;1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,652 |
342. Solve the system of equations:
$$
\left\{\begin{array}{l}
x^{2}+y=13 \\
x^{6}+y^{3}=793
\end{array}\right.
$$ | $\triangle$ The given system is not symmetric. Let $x^{2}=z$. Then we obtain a symmetric system:
$$
\left\{\begin{array}{l}
z+y=13 \\
z^{3}+y^{3}=793
\end{array}\right.
$$
Introduce the elementary symmetric polynomials $u=z+y, v=zy$. We will have:
$$
u=13, \quad u^{3}-3uv=793
$$
From this, $u=13, v=36$, i.e.,
$$
z... | (3;4),(-3;4),(2;9),(-2;9) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,653 |
343. Solve the equation:
$$
x \frac{19-x}{x+1}\left(x+\frac{19-x}{x+1}\right)=84
$$ | $\triangle$ Let $\frac{19-x}{x+1}=y$. Then
$$
x y(x+y)=84
$$
In addition, transform the equality with the substitution:
$$
19-x=x y+y, \quad x+y+x y=19
$$
This results in a symmetric system of equations:
$$
x y(x+y)=84, \quad x+y+x y=19
$$
Complete the solution on your own.
Answer: $3,4,6 \pm \sqrt{29}$. | 3,4,6\\sqrt{29} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,654 |
346. Solve the system of equations:
$$
\left\{\begin{array}{l}
(x+y)^{2}-8(x+y)=33 \\
(x-y)^{2}+2(x-y)=80
\end{array}\right.
$$ | $\triangle$ Let's introduce the substitutions:
$$
x+y=z, \quad x-y=t .
$$
We have the system of equations:
$$
z^{2}-8 z-33=0, \quad t^{2}+2 t-80=0
$$
Let's solve each of the quadratic equations in this system:
$$
z_{1}=11, \quad z_{2}=-3 ; \quad t_{1}=8, \quad t_{2}=-10
$$
Now, to find $x$ and $y$, we need to com... | (11/2;3/2),(21/2;1/2),(5/2;-11/2),(-13/2;7/2) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,655 |
348. Solve the systems of equations:
a) $\left\{\begin{array}{l}x^{2}-3 x y-4 y^{2}=0, \\ x^{3}+y^{3}=65 ;\end{array}\right.$
b) $\left\{\begin{array}{l}x^{2}+2 y^{2}=17, \\ 2 x y-x^{2}=3\end{array}\right.$ | $\triangle$ a) The first equation of the system is homogeneous. We divide it term by term by $y^{2}$. In this process, no solutions are lost: although the pair $(0 ; 0)$ satisfies the first equation, it does not satisfy the second. We get:
$$
\left(\frac{x}{y}\right)^{2}-3 \frac{x}{y}-4=0
$$
Let $\frac{x}{y}=t$. Then... | (4;1);(3;2),(-3;-2),(\sqrt{3}/3;5\sqrt{3}/3),(-\sqrt{3}/3;-5\sqrt{3}/3) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,656 |
351*. Solve the system of equations:
$$
\left\{\begin{array}{l}
x^{2}+y^{2}=1 \\
4 x y\left(2 y^{2}-1\right)=1
\end{array}\right.
$$ | $\triangle$ Given the first equation of the system, we introduce trigonometric substitutions
$$
x=\cos \alpha, \quad y=\sin \alpha
$$
where $\alpha \in[0 ; 2 \pi]$. We transform the second equation:
$4 \cos \alpha \sin \alpha \cos 2 \alpha=1, \quad \sin 4 \alpha=1$.
Then
$$
4 \alpha=\frac{\pi}{2}+2 \pi k, \quad \a... | (\cos\pi/8;\sin\pi/8),(\cos5\pi/8;\sin5\pi/8),(\cos9\pi/8;\sin9\pi/8),(\cos13\pi/8;\sin13\pi/8) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,657 |
353. Solve the systems of equations:
a) $\left\{\begin{array}{l}2 x=(y+z)^{2}, \\ 2 y=(z+x)^{2}, \\ 2 z=(x+y)^{2}\end{array}\right.$,
b) $\left\{\begin{array}{l}x^{2}-x y-x z+z^{2}=0, \\ x^{2}-x z-y z+3 y^{2}=2, \\ y^{2}+x y+y z-z^{2}=2 .\end{array}\right.$ | $\triangle$ a) From the equations of the system, it is clear that the unknowns $x, y$ and $z$ must be non-negative.
Let's try using inequalities. Could it be that $x > y$?
Assume that $x > y$. Then from the first two equations, it follows that
$$
(y+z)^{2} > (z+x)^{2}, \quad y+z > z+x, \quad y > x
$$
But the inequa... | )(0;0;0),(1/2;1/2;1/2),b)(1;1;1),(-1;-1;-1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,658 |
357*. Solve the systems of equations:
a) $\left\{\begin{array}{l}x^{2}+x y+y^{2}=7, \\ y^{2}+y z+z^{2}=13, \\ z^{2}+z x+x^{2}=19 ;\end{array}\right.$
b) $\left\{\begin{array}{l}x y+x z=x^{2}+2, \\ x y+y z=y^{2}+3, \\ x z+y z=z^{2}+4 .\end{array}\right.$ | $\triangle$ a) Multiply the first equation of the system by $x-y$, the second by $y-z$, and the third by $z-x$. Why? To obtain a linear equation with $x, y$, and $z$. We will have:
$$
x^{3}-y^{3}=7(x-y), \quad y^{3}-z^{3}=13(y-z), \quad z^{3}-x^{3}=19(z-x)
$$
Add all these equations:
$$
\begin{aligned}
0 & =7 x-7 y+... | )(2;1;3),(-2;-1;-3);b)(\frac{2\sqrt{15}}{3};\frac{3\sqrt{15}}{5};\frac{4\sqrt{15}}{15}),(-\frac{2\sqrt{15}}{3};-\frac{3\sqrt{15}}{5} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,659 |
359*. Solve the system of equations:
$$
\left\{\begin{array}{l}
x^{2}-y z=3 \\
y^{2}-x z=4 \\
z^{2}-x y=5
\end{array}\right.
$$ | $\triangle$ Multiply the first equation by $y$, the second by $z$, and the third by $x$ to form a linear equation with $x, y$, and $z$:
$$
x^{2} y-y^{2} z=3 y, \quad y^{2} z-x z^{2}=4 z, \quad x z^{2}-x^{2} y=5 x
$$
After adding these equations, we get:
$$
5 x+3 y+4 z=0
$$
Now multiply the first equation of the ori... | (11/6;-1/6;-13/6),(-11/6;1/6;13/6) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,660 |
362. Solve the equation:
$$
\frac{x^{2}+x+2}{3 x^{2}+5 x-14}=\frac{x^{2}+x+6}{3 x^{2}+5 x-10}
$$ | $\triangle$ Let's get rid of the denominators in the equation. It is reasonable to consider the sums $x^{2}+x$ and $3 x^{2}+5 x$ in the left and right parts of the equation as single terms:
$$
\frac{\left(x^{2}+x\right)+2}{\left(3 x^{2}+5 x\right)-14}=\frac{\left(x^{2}+x\right)+6}{\left(3 x^{2}+5 x\right)-10}
$$
Now ... | 2,-4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,661 |
366. Solve the equation:
$$
\left(\frac{x+6}{x-1}\right)^{2}+3 \frac{x-1}{x+6}=\frac{515}{8}
$$ | $\triangle$ The substitution $\frac{x+6}{x-1}=y$ suggests itself. We obtain:
$$
y^{2}+\frac{3}{y}=\frac{515}{8}, \quad 8 y^{3}-515 y+24=0
$$
To find the integer roots of the last equation, we need to check the divisors of the constant term 24. We can notice that if the equation has an integer root, then it is even. A... | 2,\frac{-211\14\sqrt{262}}{69} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,662 |
368*. Solve the equation:
$$
x^{2}+\frac{x^{2}}{(x+1)^{2}}=1
$$ | $\triangle$ Subtract $\frac{2 x^{2}}{x^{2}+1}$ from both sides of the equation to obtain a square of a difference on the left side:
$$
\left(x-\frac{x}{x+1}\right)^{2}=1-\frac{2 x^{2}}{x+1}, \quad \frac{x^{4}}{(x+1)^{2}}=1-\frac{2 x^{2}}{x+1}
$$
Now, the obvious substitution $-\frac{x^{2}}{x+1}=y$. Complete the solut... | \frac{(\sqrt{2}-1\\sqrt{2\sqrt{2}-1})}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,663 |
370. Solve the equation:
$$
\frac{2}{x+8}+\frac{5}{x+9}=\frac{3}{x+15}+\frac{4}{x+6}
$$ | $\triangle$ Let's represent the equation in the following form:
$$
\frac{5}{x+9}-\frac{4}{x+6}=\frac{3}{x+15}-\frac{2}{x+8}
$$
We will bring the differences in the left and right parts of this equation to common denominators:
$\frac{x-6}{(x+9)(x+6)}=\frac{x-6}{(x+15)(x+8)}$,
$(x-6)\left(\frac{1}{(x+9)(x+6)}-\frac{1... | 6,-\frac{33}{4} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,664 |
372. Solve the system of equations:
$$
\left\{\begin{array}{l}
\frac{1}{x}+\frac{1}{y}=1 \\
\frac{1}{3-x}+\frac{1}{3-y}=2
\end{array}\right.
$$ | $\triangle$ Transform each equation of the system into an algebraic form:
$y+x=x y, \quad 6-x-y=18-6 x-6 y+2 x y$;
$x y=x+y, \quad 2 x y=5 x+5 y-12$.
We obtain a symmetric system (see § 8, point 8.3). Therefore, introduce new variables
$$
u=x+y, \quad v=x y
$$
This results in a system of linear equations
$$
v=u, ... | (2;2) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,665 |
374. Solve the system of equations:
$$
\left\{\begin{array}{l}
\frac{1}{x}+\frac{1}{y+z}=\frac{1}{2} \\
\frac{1}{y}+\frac{1}{z+x}=\frac{1}{3} \\
\frac{1}{z}+\frac{1}{x+y}=\frac{1}{4}
\end{array}\right.
$$ | $\triangle$ In each of the equations of the system, we will eliminate the denominators of the fractions:
$$
\begin{aligned}
& 2(x+y+z)=x y+x z \\
& 3(x+y+z)=x y+y z \\
& 4(x+y+z)=x z+y z
\end{aligned}
$$
Let $x+y+z=t$. Then
$$
x y+x z=2 t, \quad x y+y z=3 t, \quad x z+y z=4 t .
$$
From this system, we will express ... | (\frac{23}{10};\frac{23}{6};\frac{23}{2}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,666 |
376. Solve the systems of equations:
a) $\left\{\begin{array}{l}\frac{1}{x}+\frac{2}{y}-\frac{3}{z}=3 \\ \frac{4}{x}-\frac{1}{y}-\frac{2}{z}=5 \\ \frac{3}{x}+\frac{4}{y}+\frac{1}{z}=23\end{array}\right.$
b) $\left\{\begin{array}{l}\frac{1}{x-y}+x^{2}=1, \\ \frac{x^{2}}{x-y}=-2 .\end{array}\right.$ | $\triangle$ a) Let's introduce the substitutions:
$$
\frac{1}{x}=t, \quad \frac{1}{y}=u, \quad \frac{1}{z}=v
$$
We will have:
$$
t+2 u-3 v=3, \quad 4 t-u-2 v=5, \quad 3 t+4 u+v=23
$$
This system of linear equations can be solved in various ways. For example, add the first and third equations and subtract the second... | )(1/3;1/3;1/2);\quadb)(\sqrt{2};1+\sqrt{2}),(-\sqrt{2};1-\sqrt{2}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,667 |
378*. Solve the system of equations:
$$
\left\{\begin{array}{l}
x+y=\frac{5 x y}{1+x y} \\
y+z=\frac{6 y z}{1+y z} \\
z+x=\frac{7 z x}{1+z x}
\end{array}\right.
$$ | $\triangle$ Let's get rid of the denominator in the first equation:
$$
(x+y)(1+x y)=5 x y, \quad x+y+x y(x+y)=5 x y
$$
Now, it seems reasonable to divide all terms of the last equation by $x y$. But then we have to consider two cases.
1) Let $x y=0$.
If here $x=0$, then, as can be seen from the original system, $y=... | (0;0;0),(\frac{3\\sqrt{5}}{2};1;2\\sqrt{3}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,668 |
380. Solve the system of equations:
$$
\left\{\begin{array}{l}
x=\frac{y+1}{3 y-5} \\
y=\frac{3 z-2}{2 z-3} \\
z=\frac{3 x-1}{x-1}
\end{array}\right.
$$ | $\triangle$ Substitute the expression for $z$ from the third equation of the system into the second:
$$
\begin{aligned}
& y=\left(3 \frac{3 x-1}{x-1}-2\right) /\left(2 \frac{3 x-1}{x-1}-3\right) \\
& y=\frac{9 x-3-2 x+2}{6 x-2-3 x+3}, \quad y=\frac{7 x-1}{3 x+1}
\end{aligned}
$$
Substitute the last expression for $y$... | (0,-1,1),(3,2,4) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,669 |
382. Solve the system of equations:
$$
\left\{\begin{array}{l}
\frac{x y z}{x+y}=\frac{6}{5} \\
\frac{x y z}{y+z}=2 \\
\frac{x y z}{z+x}=\frac{3}{2}
\end{array}\right.
$$ | $\triangle$ In each of the equations of the system, we will transition to the inverse quantities and in the left parts of the resulting equations, we will divide the numerators of the fractions term by term by the denominators:
$\frac{x+y}{x y z}=\frac{5}{6}, \quad \frac{y+z}{x y z}=\frac{1}{2}, \quad \frac{z+x}{x y z}... | (3;2;1),(-3;-2;-1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,670 |
385*. Solve the system of equations:
$$
\left\{\begin{array}{l}
x=\frac{2 y^{2}}{1+z^{2}} \\
y=\frac{2 z^{2}}{1+x^{2}} \\
z=\frac{2 x^{2}}{1+y^{2}}
\end{array}\right.
$$ | $\triangle$ From the system, it is clear that all unknowns $x, y$, and $z$ are non-negative.
Multiply all the equations of the system and reduce the resulting equation to an algebraic one:
$$
\begin{aligned}
& x y z=\frac{8 x^{2} y^{2} z^{2}}{\left(1+x^{2}\right)\left(1+y^{2}\right)\left(1+z^{2}\right)}, \quad x y z\... | (0;0;0),(1;1;1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,671 |
388. Solve the equations:
a) $\frac{x^{3}}{\sqrt{4-x^{2}}}+x^{2}-4=0$
b) $\sqrt{17 x^{2}+7 x+0.5}=13 x^{2}+5 x+0.5$. | $\triangle$ a) Transform the equation:
$$
\frac{x^{3}}{\sqrt{4-x^{2}}}=4-x^{2}, \quad x^{3}=\left(4-x^{2}\right)^{\frac{3}{2}}
$$
Extract the cubic root from both sides of the last equation (thus obtaining an equivalent equation):
$$
x=\left(4-x^{2}\right)^{\frac{1}{2}}, \quad x=\sqrt{4-x^{2}}
$$
Square both sides ... | )\sqrt{2};b)\frac{-7\\sqrt{23}}{26},\frac{-3\\sqrt{35}}{26} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,672 |
390. Solve the equation:
$$
\sqrt[3]{x-1}+\sqrt[3]{2 x-1}=1
$$ | $\triangle$ Let's raise both sides of the equation to the third power, using the formula
$$
(a+b)^{3}=a^{3}+b^{3}+3 a b(a+b)
$$
We will have:
$x-1+2 x-1+3 \sqrt[3]{(x-1)(2 x-1)} \cdot(\sqrt[3]{x-1}+\sqrt[3]{2 x-1})=1$,
$\sqrt[3]{(x-1)(2 x-1)} \cdot(\sqrt[3]{x-1}+\sqrt[3]{2 x-1})=1-x$.
But what now? Now let's use t... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,673 |
394. Solve the equation:
$$
\sqrt[3]{x}+\sqrt[3]{x+19}=5
$$ | $\triangle$ Equations of this type have already been encountered in section 11.1, with a different method of solution (see the solution to problem 390). Let's introduce two new variables:
$$
\sqrt[3]{x}=y, \quad \sqrt[3]{x+19}=z
$$
We obtain a system of rational equations:
$$
\left\{\begin{array}{l}
y+z=5 \\
y^{3}=x... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,674 |
397. Solve the equation:
$$
\sqrt{9 x^{2}-12 x+11}+\sqrt{5 x^{2}-8 x+10}=2 x-1
$$ | $\triangle$ Multiply both sides of the equation by the conjugate of its left part
$$
\sqrt{9 x^{2}-12 x+11}+\sqrt{5 x^{2}-8 x+10}
$$
Since it does not turn into zero, we will obtain an equation equivalent to the given one. We will have:
$$
\begin{aligned}
& \left(9 x^{2}-12 x+11\right)-\left(5 x^{2}-8 x+10\right)= \... | \frac{1}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,675 |
403*. Solve the equation:
$$
\sqrt[4]{1-x^{2}}+\sqrt[4]{1-x}+\sqrt[4]{1+x}=3
$$ | $\triangle$ The domain of the equation is the interval $[-1 ; 1]$. In this domain, we can apply the inequality between the geometric mean and the arithmetic mean of two non-negative numbers to each of the radicals in the left-hand side:
$\sqrt[4]{1-x^{2}}=\sqrt{\sqrt{1+x} \cdot \sqrt{1-x}} \leq \frac{\sqrt{1+x}+\sqrt{... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,677 |
405. Solve the equation:
$$
\sqrt{4 x^{2}+5 x-1}-2 \sqrt{x^{2}-3 x+3}=\frac{17 x-13}{7}
$$ | $\triangle$ Let's introduce three new variables:
$$
\sqrt{4 x^{2}+5 x-1}=y, \quad 2 \sqrt{x^{2}-3 x+3}=z, \quad \frac{17 x-13}{7}=t
$$
Then
$$
y-z=t, \quad y^{2}-z^{2}=17 x-13=7 t
$$
It seems natural to divide the second of the obtained equations by the first. But first, let's consider the case $t=0$:
$$
t=0, \qua... | \frac{13}{17},2,-\frac{746}{495} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,678 |
407. Solve the equation:
$$
\sqrt{1-x^{2}}=4 x^{3}-3 x
$$ | $\triangle$ The domain of the equation is the interval $[-1 ; 1]$.
Let's introduce a trigonometric substitution $x=\cos \alpha$, where $\alpha \in[0 ; \pi]$. Then
$$
\sqrt{1-x^{2}}=\sqrt{1-\cos ^{2} \alpha}=\sin \alpha, \quad 4 x^{3}-3 x=4 \cos ^{3} \alpha-3 \cos \alpha=\cos 3 \alpha
$$
(here, in the first equality,... | -\frac{\sqrt{2}}{2},\frac{\sqrt{2+\sqrt{2}}}{2},-\frac{\sqrt{2-\sqrt{2}}}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,679 |
412. Solve the system of equations:
$$
\left\{\begin{array}{l}
x+y+\sqrt{x^{2}-y^{2}}=12 \\
y \sqrt{x^{2}-y^{2}}=12
\end{array}\right.
$$ | $\triangle$ Let $\sqrt{x^{2}-y^{2}}=z$, where $z \geq 0$. We obtain the system of rational equations
$$
\left\{\begin{array}{l}
x+y+z=12 \\
y z=12 \\
x^{2}-y^{2}=z^{2}
\end{array}\right.
$$
Since
$$
x^{2}-y^{2}=z^{2}, \quad x+y=12-z
$$
then $x-y=\frac{z^{2}}{12-z}$. From the system of equations
$$
x+y=12-z, \quad ... | (5;3),(5;4) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,680 |
414. Solve the system of equations:
$$
\left\{\begin{array}{l}
x \sqrt{y}+y \sqrt{x}=6 \\
x^{2} y+y^{2} x=20
\end{array}\right.
$$ | $\triangle$ Let's introduce two new variables:
$$
x=\sqrt{y}, \quad y \sqrt{x}=t
$$
where $z \geq 0, t \geq 0$. Then
$$
z+t=6, \quad z^{2}+t^{2}=20
$$
The solutions of the last system are $(4 ; 2)$ and $(2 ; 4)$. Now let's consider two cases.
1) For the first solution of this system, we get:
$$
x \sqrt{y}=4, \qua... | (4;1),(1;4) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,681 |
416*. Solve the system of equations:
$$
\left\{\begin{array}{l}
2 \sqrt{2 x+3 y}+\sqrt{5-x-y}=7 \\
3 \sqrt{5-x-y}-\sqrt{2 x+y-3}=1
\end{array}\right.
$$ | $\triangle$ Let's introduce three new variables:
$$
\sqrt{2 x+3 y}=z, \quad \sqrt{5-x-y}=t, \quad \sqrt{2 x+y-3}=u
$$
where $z, t, u$ are non-negative. This results in a system of five rational equations with five unknowns:
$$
\left\{\begin{array}{l}
2 x+3 y=z^{2} \\
5-x-y=t^{2} \\
2 x+y-3=u^{2} \\
2 x+t=7 \\
3 t-u=... | (3;1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,682 |
418. Solve the system of equations:
$$
\left\{\begin{array}{l}
x^{2}+x \sqrt[3]{x y^{2}}=80 \\
y^{2}+y \sqrt[3]{x^{2} y}=5
\end{array}\right.
$$ | $\triangle$ Let $\frac{x}{y}=t$, i.e., $y=t x$, considering that from the first equation $x \neq 0$. We will have:
$$
\begin{aligned}
& x^{2}+x^{2 \sqrt[3]{t^{2}}}=80, \quad t^{2} x^{2}+t x^{23} \sqrt{t^{2}}=5 \\
& x^{2}\left(1+\sqrt[3]{t^{2}}\right)=80, \quad x^{2} t^{\frac{4}{3}}\left(1+\sqrt[3]{t^{2}}\right)=5
\end... | (\8;\1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,683 |
419. Solve the system of equations:
$$
\left\{\begin{array}{l}
\frac{x^{2}}{y^{2}}+2 \sqrt{x^{2}+1}+y^{2}=3 \\
x+\frac{y}{\sqrt{x^{2}+1}+x}+y^{2}=0
\end{array}\right.
$$ | $\triangle$ Let's eliminate the irrationality $\sqrt{x^{2}+1}$ from the system. For this, we express $\sqrt{x^{2}+1}$ from the first equation:
$$
\sqrt{x^{2}+1}=\frac{3 y^{2}-x^{2}-y^{4}}{2 y^{2}}
$$
Now we need to substitute this expression into the second equation, first moving the irrationality from the denominato... | (0,-1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,684 |
421. Solve the equations with two unknowns:
a) $x^{2}+y^{2}-4 x+6 y+13=0$
b) $x y-1=x-y$. | $\triangle$ a) Transform the equation:
$$
\left(x^{2}-4 x+4\right)+\left(y^{2}+6 y+9\right)=0, \quad(x-2)^{2}+(y+3)^{2}=0
$$
Then
$$
x-2=0, \quad y+3=0 \Rightarrow x=2, \quad y=-3
$$
b) Gather all terms on the left side and factor the left side of the new equation:
$$
x y-1-x+y=0, \quad(x+1)(y-1)=0
$$
In addition... | (2,-3);(-1,y),(x,1)wherex\neq-1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,685 |
426. Solve the equation:
$$
\sqrt{4 x-y^{2}}=\sqrt{y+2}+\sqrt{4 x^{2}+y}
$$ | $\triangle$ Let's square this equation; we will obtain an equation equivalent to the given one. We will have:
$$
\begin{aligned}
& 4 x-y^{2}=y+2+4 x^{2}+y+2 \sqrt{(y+2)\left(4 x^{2}+y\right)} \\
& \left(4 x^{2}-4 x+1\right)+\left(y^{2}+2 y+1\right)+2 \sqrt{(y+2)\left(4 x^{2}+y\right)}=0 \\
& (2 x-1)^{2}+(y+1)^{2}+2 \s... | (\frac{1}{2},-1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,686 |
428. Solve the equation:
$$
x^{2}-2 x \sin y+1=0
$$ | $\triangle$ Let's represent 1 on the left side of the equation as $\sin ^{2} y+\cos ^{2} y$. Then
$$
\left(x^{2}-2 x \sin y+\sin ^{2} y\right)+\cos ^{2} y=0, \quad(x-\sin y)^{2}+\cos ^{2} y=0
$$
Therefore,
$$
x-\sin y=0, \quad \cos y=0
$$
If $\cos y=0$, then
$$
\sin y= \pm 1, \quad x=\sin y= \pm 1
$$
Let's consid... | (1;\pi/2+2\pik)(k\inZ);(-1;-\pi/2+2\pin)(\mathrm{n}\inZ) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,687 |
431. Solve the equation:
$$
x^{3}+\sqrt{x}=y^{3}+\sqrt{y}
$$ | $\triangle$ Given that $x$ and $y$ are non-negative. Let's consider two cases.
1) Suppose $y=x$. We get the set of solutions $(x ; x)$, where $x$ is any non-negative number.
2) Suppose $y \neq x$, for example, $y>x$.
The functions $z=x^{3}$ and $z=\sqrt{x}$ are increasing on the interval $[0 ;+\infty)$. Since the sum... | (x;x),x\geq0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,688 |
436. Solve the equation:
$$
3 x-4 y=5 \sqrt{x^{2}+y^{2}}
$$ | $\triangle$ Let's introduce vectors $\bar{u}$ and $\bar{v}$, and choose their coordinates in such a way that the left side of the equation expresses the dot product of the vectors in coordinates, while the right side expresses the product of the lengths of the vectors. Suitable vectors are $\bar{u}(x ; y)$ and $\bar{v}... | (x;-\frac{4}{3}x), | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,689 |
439. Solve the system of equations:
$$
\left\{\begin{array}{l}
x+y+z=3 \\
x^{2}+y^{2}+z^{2}=3
\end{array}\right.
$$ | $\triangle$ Using the left part of the first equation of the system, let's introduce vectors in space $\bar{u}(x ; y ; z)$ and $\bar{v}(1 ; 1 ; 1)$. We get:
$$
\bar{u} \cdot \bar{v}=x+y+z \leq|\bar{u}| \cdot|\bar{v}|=\sqrt{\left(x^{2}+y^{2}+z^{2}\right) 3}=\sqrt{3 \cdot 3}=3
$$
Since the left part of this inequality ... | (1;1;1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,690 |
442*. Solve the system of equations:
$$
\left\{\begin{array}{l}
x+y^{2}+z^{3}=\frac{13}{6} \\
9 x^{2}+16 y^{4}+144 z^{6}=26
\end{array}\right.
$$ | $\triangle$ Vectors $\bar{u}$ and $\bar{v}$ are chosen differently than before. Using the left part of the second equation, we introduce the vector $\bar{u}\left(3 x ; 4 y^{2} ; 12 z^{3}\right)$. Now, we introduce the vector $\bar{v}$ such that $\bar{u} \cdot \bar{v}=x+y^{2}+z^{3}$. The vector $\bar{v}\left(\frac{1}{3}... | (4/3;\\sqrt{3}/2;\sqrt[3]{18}/12) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,691 |
447. A motorcyclist left point A at a speed of 45 km/h. After 40 minutes, a car left A in the same direction at a speed of 60 km/h. How much time after the car's departure will the distance between it and the motorcyclist be 36 km? | $\triangle$ Important question: at the moment when the car is 36 km away from the motorcycle, will it be ahead or behind the motorcycle?
In 40 minutes, the motorcycle will travel a distance of $45 \cdot \frac{2}{3}$ km $=30$ km, which is less than 36 km. Therefore, at the moment the car departs, it is 30 km behind the... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,692 |
448. A cyclist set off from point A to point B, and 15 minutes later, a car set off after him. Halfway from A to B, the car caught up with the cyclist. When the car arrived at B, the cyclist still had to cover another third of the entire distance. How long will it take the cyclist to travel the distance from A to B? | $\triangle$ Let's take the path AB as a unit. (This can be done in cases where there is no distance given in linear units in the problem data.) Let's denote the speed of the cyclist by $x$ (in fractions of the path per hour).
During the time it took the car to travel the second half of the path, the cyclist traveled $... | 45 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,693 |
452. Two points, moving along a circle in the same direction, meet every 12 minutes, with the first point completing a circle 10 seconds faster than the second. What part of the circle does each point cover in 1 second? | $\triangle$ Let's accept the circumference length as a unit. Denote the speeds of the first and second points as $v_{1}$ and $v_{2}$ (in fractions of this unit per second). Then, according to the first condition, $v_{1}-v_{2}=\frac{1}{12 \cdot 60}$, and according to the second condition, $\frac{1}{v_{2}}-\frac{1}{v_{1}... | \frac{1}{80} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,694 |
455*. From point A to point B, which is 40 km away from A, two tourists set off simultaneously: the first on foot at a speed of 6 km/h, and the second on a bicycle. When the second tourist overtook the first by 5 km, the first tourist got into a passing car traveling at a speed of 24 km/h. Two hours after leaving A, th... | $\triangle$ Let's make a drawing (Fig. 2). $\qquad$
Fig. 2
Let $\mathrm{A}_{1}$ and $\mathrm{B}_{1}$ be the points where the first and second tourists are, respectively, at the moment when the second tourist overtakes the first by $5 \mathrm{km}$, and let $\mathrm{K}$ be the point where the first tourist catches up wi... | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,695 |
457. A motorcyclist and a cyclist set off towards each other from points A and B simultaneously and met 4 km from B. At the moment when the motorcyclist arrived at B, the cyclist was 15 km from A. Find the distance AB. | $\triangle$ Let the distance AB be denoted as $x$ km.
By the time they meet, the motorcyclist and the cyclist have traveled $(x-4)$ km and $x$ km, respectively, and by the time the motorcyclist arrives at $\mathrm{B}$, the distances are $x$ km and $(x-4)$ km, respectively.
Thus, the ratio of the speeds of the motorcy... | 20 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,696 |
463. From two ports A and B, two steamships set off simultaneously towards each other across the sea. The speed of each is constant. The first steamship arrived at B 16 hours after the meeting, while the second arrived at A 25 hours after the meeting. How long does it take for each steamship to travel the entire distan... | $\triangle$ Let the path AB be a unit. Denote the speeds of the steamships as $v_{1}$ and $v_{2}$ (in fractions of a unit per hour).
Suppose the steamships met at point C. Then $\frac{A C}{C B}=\frac{v_{1}}{v_{2}}$. From this,
$$
\frac{A C}{C B}+1=\frac{v_{1}}{v_{2}}+1, \quad \frac{1}{C B}=\frac{v_{1}+v_{2}}{v_{2}}, ... | 36 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,697 |
468. Two schoolchildren left home for school at the same time and with the same speed. After 3 minutes, one of them remembered that he had forgotten his notebook at home, and ran back at a speed greater than the initial one by 60 m/min. After picking up the notebook, he ran back at the same speed and caught up with his... | $\triangle$ Let's denote the initial speed of the schoolchildren as $v$ m/min. Then in 3 minutes, they covered a distance of $3 v \, \text{m}$. Therefore, the schoolchild who ran for the notebook traveled a distance of $(400 + 3 v)$ m at a speed of $(v + 60)$ m/min. The second schoolchild, in the same time, covered a d... | 3\, | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,698 |
471. A motorcyclist left A for B and at the same time a pedestrian set off from B to A. Upon meeting the pedestrian, the motorcyclist gave him a ride, brought him to A, and immediately set off again for B. As a result, the pedestrian reached A 4 times faster than he would have if he had walked the entire way. How many ... | $\triangle$ Let's take the path AB as a unit. Denote the speeds of the pedestrian and the motorcyclist as $v_{1}$ and $v_{2}$ (in fractions of this unit per hour).
Suppose the pedestrian and the motorcyclist met at point C. Since they moved for $\frac{1}{v_{1}+v_{2}}$ hours before meeting, we have
$$
A C=v_{2} \cdot ... | 2.75 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 34,699 |
478. Two pedestrians set out simultaneously from point A in the same direction. The first one met a tourist walking towards A, 20 minutes after leaving A, while the second one met the tourist 5 minutes later than the first. Ten minutes after the second meeting, the tourist arrived at A. Find the ratio of the speeds of ... | Let $B$ and $M$ be the points of meeting of the first and second pedestrian with the tourist (Fig. 3).
Fig. 3
Let the path $AB$ be taken as a unit. Denote the speeds of the first and second... | \frac{15}{8} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,700 |
489*. Three cyclists set off simultaneously: the first and second from point A, and the third towards them from point B. After 1.5 hours, the first cyclist was at an equal distance from the other two, and after 2 hours from the start, the third cyclist was at an equal distance from the first and second. How many hours ... | $\triangle$ Let's take the path AB as a unit. Denote the speeds of the first, second, and third cyclists as $v_{1}, v_{2}$, and $v_{3}$ (in fractions of this path per hour).
We will denote the positions of the first, second, and third cyclists in each of the three specified situations by the letters $\mathbf{M}_{1}, \... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,701 |
493. Four pumps of the same capacity, working together, filled the first tanker and a third of the second tanker (of a different volume) in 11 hours. If three pumps had filled the first tanker and then one of them filled a quarter of the second tanker, the work would have taken 18 hours. How many hours would it take fo... | $\triangle$ Let one pump fill the first tanker in $x$ hours, and the second tanker in $y$ hours. Then four pumps, working together, will fill the first tanker in $\frac{x}{4}$ hours, and the second in $\frac{y}{4}$ hours.
We have the system of equations:
$$
\left\{\begin{array}{l}
\frac{x}{4}+\frac{y}{4 \cdot 3}=11 \... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,702 |
498. Three automatic lines produce the same product, but have different productivity. The combined productivity of all three lines working simultaneously is 1.5 times the productivity of the first and second lines working simultaneously. A shift assignment for the first line can be completed by the second and third lin... | $\triangle$ Let's accept the shift task for the first line as a unit.
Let the first line complete its shift task in $x$ hours, and the third line the same task in $y$ hours. Then the second line completes the task of the first in $(x-2)$ hours.
Therefore, the productivity of the first, second, and third lines are res... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,703 |
503. Three workers need to make 80 identical parts. Together, they make 20 parts per hour. The first worker started the job alone. He made 20 parts, spending more than 3 hours on their production. The remaining work was done by the second and third workers together. The entire job took 8 hours. How many hours would it ... | Let's denote the productivity of the first, second, and third workers as \( x, y, \) and \( z \) parts per hour, respectively. Based on the conditions of the problem, we obtain a system of two equations with three unknowns:
\[
\left\{\begin{array}{l}
x+y+z=20 \\
\frac{20}{x}+\frac{60}{y+z}=8
\end{array}\right.
\]
To ... | 16 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,704 |
507. Fresh mushrooms contain $90\%$ water, while dried ones contain $12\%$ water. How many kilograms of dried mushrooms can be obtained from 44 kg of fresh mushrooms? | $\triangle$ According to the condition, 44 kg of fresh mushrooms contain $44 \cdot 0.9 = 39.6$ kg of water, which means there is $44 - 39.6 = 4.4$ kg of dry matter.
Let's denote the mass of dried mushrooms that can be obtained from 44 kg of fresh mushrooms by $x$ kg. These $x$ kg consist of $0.12 x$ kg of water and $0... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,705 |
1. A father and son measured the length of the courtyard with their steps in winter, starting from the same place and walking in the same direction. In some places, the father's and son's footprints coincided exactly. In total, there were 61 footprints along the measurement line on the snow. What is the length of the c... | 1. Let's calculate the ratio of the father's step length to the son's step length, which is: $0.72: 0.54 = 4: 3$. Therefore, four steps of the son equal three steps of the father, and on such a segment, we get 6 footprints. In total, $(61-1): 6=10$ segments (Fig. 9). The length of the courtyard is $10 \cdot 3$ steps of... | 21.6() | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 34,706 |
2. Several people were collecting mushrooms. One of them found 6 mushrooms, and each of the others found 13. Another time, the number of collectors was different. This time, one found 5 mushrooms, and each of the others found 10. In both cases, the same number of mushrooms was collected, more than 100 but less than 200... | 2. The first time there were
Fig. 9 is more than 8 but less than 15. The second time it was more than 10 but less than 20. In the second case, one collected 5 mushrooms, the others collecte... | 1418 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 34,707 |
4. A dog is chasing a hare at a speed of 17 m/s, the hare is running at a speed of $14 \boldsymbol{m} /$ s. The distance between them before the chase was 150 m. Will the dog catch the hare if there are bushes $520 \mathcal{M}$ from the hare where he can hide? | 4. It will not catch up, as the dog can only catch the hare after 50 seconds. In this time, the hare can run 700 m, while the bushes are located 520 m away. | Itwillnotcatchup | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,709 |
7. A mowing team had to mow two meadows, one twice as large as the other. For half a day, the team mowed the larger meadow. After that, they split in half: one half stayed on the large meadow and finished it by evening, while the other half mowed the smaller meadow but did not finish it. How many mowers were in the tea... | 7. If the whole brigade mowed a large meadow for half a day and half of the brigade mowed for another half a day, it is clear that the brigade mowed $\frac{2}{3}$ of the meadow in half a day, and half of the brigade mowed $\frac{1}{3}$ of the meadow in half a day. Since the second meadow is half the size of the first, ... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,712 |
11. Three boys collectively bought a volleyball. How much money did each contribute to the purchase, if it is known that the third boy contributed 6 rubles and 40 kopecks more than the first, and $\frac{1}{2}$ of the first boy's contribution is equal to $\frac{1}{3}$ of the second boy's contribution or $\frac{1}{4}$ of... | 11. 6 rub. 40 kop., 9 rub. 60 kop., 12 rub. 80 kop. | 6rub.40kop.,9rub.60kop.,12rub.8 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,716 |
14. A box with a square base is required to be made for laying boxes that are 9 cm wide and 21 cm long. What should be the minimum length of the side of the square base so that the boxes fit snugly in the box? | 14. The side of the base of the box is 63 cm (you need to find the LCM of the numbers 9 and 21). | 63 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 34,719 |
15. When adding several numbers, the following errors were made: in one of the numbers, the digit 3 in the tens place was taken for 8, and the digit 7 in the hundreds place was taken for 4; in another, the digit 2 in the thousands place was taken for 9. The sum obtained was 52000. Find the correct sum. | 15. The obtained sum is greater than the true one by 7 thousand and 5 tens, but less than it by three hundred. To get the true sum, you need to subtract 7050 from 52000 and add 300 to the result. Answer: 42250. | 42250 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 34,720 |
16. Columbus discovered America in the 15th century. In which year did this discovery take place, if the following is known:
a) the sum of the digits representing this year is 16,
b) if the tens digit is divided by the units digit, the quotient is 4 and the remainder is 1. | 16. Let $x$ be the number of tens, $y$ be the number of units of the year of discovery. Then: $\left\{\begin{array}{l}x+y=11 \\ x=4 y+1 .\end{array}\right.$ Answer: 1492. | 1492 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 34,721 |
23. A pond is overgrown with duckweed in 64 days. In how many days will a quarter of the pond be overgrown if the amount of duckweed doubles every day? | 23. 62 days (the problem should be solved from the end). | 62 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 34,728 |
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