problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
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value | source stringclasses 8
values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
4.005. Find the number of terms of an arithmetic progression, the sum of all terms of which is 112, the product of the second term by the common difference is 30, and the sum of the third and fifth terms is 32. Write the first three terms of this progression. | Solution.
$$
\begin{aligned}
& \text { From the condition we have }\left\{\begin{array}{l}
\left(a_{1}+d\right) d=30, \\
\left(a_{1}+2 d\right)+\left(a_{1}+4 d\right)=32
\end{array} S_{n}=112 \Rightarrow\right. \\
& \Rightarrow a_{1}=16-3 d, \quad(16-2 d) d=30, \quad 2 d^{2}-16 d+30=0
\end{aligned}
$$
or
$$
\begin{a... | 7,10,13or1,6,11 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,647 |
4.007. When the ninth term of an arithmetic progression is divided by the second term, the quotient is 5, and when the thirteenth term is divided by the sixth term, the quotient is 2 and the remainder is 5. Find the first term and the common difference of the progression. | Solution.
From the condition we have $\left\{\begin{array}{l}a_{9}=5 a_{2}, \\ a_{13}=2 a_{6}+5 .\end{array}\right.$
Using formula (4.1), we get
$$
\left\{\begin{array} { l }
{ a _ { 1 } + 8 d = 5 ( a _ { 1 } + d ) , } \\
{ a _ { 1 } + 1 2 d = 2 ( a _ { 1 } + 5 d ) + 5 }
\end{array} \Leftrightarrow \left\{\begin{ar... | 3;4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,648 |
4.008. Find four numbers that form a geometric progression, where the sum of the extreme terms is -49, and the sum of the middle terms is 14. | ## Solution.
From the condition, we have $\left\{\begin{array}{l}b_{1}+b_{4}=-49, \\ b_{2}+b_{3}=14 .\end{array}\right.$
Using formula (4.6), we get
$$
\begin{aligned}
& \left\{\begin{array} { l }
{ b _ { 1 } + b _ { 1 } q ^ { 3 } = - 4 9 , } \\
{ b _ { 1 } q + b _ { 1 } q ^ { 2 } = 1 4 }
\end{array} \Leftrightarro... | 7,-14,28,-56;-56,28,-14,7 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,649 |
4.009. Find the third term of an infinite geometric progression with a common ratio $|q|<1$, the sum of which is equal to $8 / 5$, and the second term is $-1 / 2$. | Solution.
Using the formulas $S=\frac{b_{1}}{1-q}$ and $b_{n}=b_{1} q^{n-1}$, we get $\left\{\begin{array}{l}\frac{b_{1}}{1-q}=\frac{8}{5}, \\ b_{1} q=-\frac{1}{2}\end{array} \Rightarrow\right.$
$\Rightarrow b_{1}=\frac{8-5 q}{5}, 16 q^{2}-16 q-5=0$, from which we find $q_{1}=-\frac{1}{4}, q_{2}=\frac{5}{4}>1$ (does n... | \frac{1}{8} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,650 |
4.011. The sum of three numbers forming an arithmetic progression is 2, and the sum of the squares of these same numbers is 14/9. Find these numbers. | Solution.
From the condition we have $\left\{\begin{array}{l}a_{1}+a_{2}+a_{3}=2, \\ a_{1}^{2}+a_{2}^{2}+a_{3}^{2}=\frac{14}{9} .\end{array}\right.$
Using formula (4.1), we get
$$
\begin{aligned}
& \left\{\begin{array} { l }
{ a _ { 1 } + a _ { 1 } + d + a _ { 1 } + 2 d = 2 , } \\
{ a _ { 1 } ^ { 2 } + ( a _ { 1 } ... | 1,\frac{2}{3},\frac{1}{3};\frac{1}{3},\frac{2}{3},1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,651 |
4.012. The sum of the third and ninth terms of an arithmetic progression is 8. Find the sum of the first 11 terms of this progression. | Solution.
From the condition, we have $a_{3}+a_{9}=8$. Using formula (4.1), we get $a_{1}+2 d+a_{1}+8 d=8, 2 a_{1}+10 d=8$, and using formula (4.5), we find $S_{11}=\frac{2 a_{1}+10 d}{2} \cdot 11=44$.
Answer: 44. | 44 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,652 |
4.013. The sum of the first three terms of an increasing arithmetic progression is 15. If 1 is subtracted from the first two terms of this progression, and 1 is added to the third term, the resulting three numbers will form a geometric progression. Find the sum of the first 10 terms of the arithmetic progression. | ## Solution.
From the condition, we have: $a_{1}-1, a_{1}+d-1, a_{1}+2 d+1 \cdots$ - three consecutive terms of a geometric progression. Using formula (4.5), we find $S_{3}=\frac{2 a_{1}+2 d}{2} \cdot 3=15$ or $a_{1}+d=5$. Using formula (4.7), we get $\left(a_{1}+d-1\right)^{2}=\left(a_{1}-1\right)\left(a_{1}+2 d+1\ri... | 120 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,653 |
4.015. Calculate
$$
\left(1+3^{2}+5^{2}+\ldots+(2 n-1)^{2}+\ldots+199^{2}\right)-\left(2^{2}+4^{2}+6^{2}+(2 n)^{2}+\ldots+200^{2}\right)
$$ | ## Solution.
From the condition we have
$$
\begin{aligned}
& 1+3^{2}+5^{2}+\ldots+(2 n-1)^{2}+\ldots+199^{2}-2^{2}-4^{2}-6^{2}-(2 n)^{2}-\ldots-200^{2}= \\
& =(1-2)^{2}+\left(3^{2}-4^{2}\right)+\left(5^{2}-6^{2}\right)+\ldots+\left((2 n-1)^{2}-(2 n)^{2}\right)+\ldots+\left(199^{2}-200^{2}\right)= \\
& =(1-2)(1+2)+(3-... | -20100 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,654 |
4.016. Find four numbers that form a geometric progression, where the second term is 35 less than the first, and the third term is 560 more than the fourth. | ## Solution.
From the condition, we have $\left\{\begin{array}{l}b_{1}-b_{2}=35, \\ b_{3}-b_{4}=560 .\end{array}\right.$
Using formula (4.6), we get
$$
\begin{aligned}
& \left\{\begin{array} { l }
{ b _ { 1 } - b _ { 1 } q = 3 5 , } \\
{ b _ { 1 } q ^ { 2 } - b _ { 1 } q ^ { 3 } = 5 6 0 }
\end{array} \Leftrightarro... | 7,-28,112,-448;-\frac{35}{3},-\frac{140}{3},-\frac{560}{3},-\frac{2240}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,655 |
4.017. Find four numbers that form a geometric progression, where the third term is 9 more than the first, and the second term is 18 more than the fourth. | ## Solution.
From the condition, we have $\left\{\begin{array}{l}b_{3}-b_{1}=9, \\ b_{2}-b_{4}=18 .\end{array}\right.$
Using formula (4.6), we get
$$
\left\{\begin{array} { l }
{ b _ { 1 } q ^ { 2 } - b _ { 1 } = 9 , } \\
{ b _ { 1 } q - b _ { 1 } q ^ { 3 } = 1 8 }
\end{array} \Leftrightarrow \left\{\begin{array}{l... | 3,-6,12,-24 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,656 |
4.018. The denominator of the geometric progression is $1 / 3$, the fourth term of this progression is $1 / 54$, and the sum of all its terms is 121/162. Find the number of terms in the progression. | Solution.
From the condition we have $\left\{\begin{array}{l}b_{4}=\frac{1}{54}, \\ S_{n}=\frac{121}{162}\end{array}\right.$.
Using formulas (4.6) and (4.11), we get
$$
\begin{aligned}
& b_{4}=b_{1} q^{3}=b_{1}\left(\frac{1}{3}\right)^{3} ; \frac{b_{1}}{27}=\frac{1}{54}, b_{1}=\frac{1}{2} ; \\
& S_{n}=\frac{b_{1}\le... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,657 |
4.019. Find the first term and the common ratio of the geometric progression, given that $b_{4}-b_{2}=-45 / 32$ and $b_{6}-b_{4}=-45 / 512$. | ## Solution.
From the condition, we have $\left\{\begin{array}{l}b_{4}-b_{2}=-\frac{45}{32}, \\ b_{6}-b_{4}=-\frac{45}{512} .\end{array}\right.$
Using formula (4.6), we get
$$
\left\{\begin{array} { l }
{ b _ { 1 } q ^ { 3 } - b _ { 1 } q = - \frac { 4 5 } { 3 2 } , } \\
{ b _ { 1 } q ^ { 5 } - b _ { 1 } q ^ { 3 } ... | 1)-6,-\frac{1}{4};2)6,\frac{1}{4} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,658 |
4.020. Find the first and fifth terms of a geometric progression if it is known that its common ratio is 3, and the sum of the first six terms is 1820. | ## Solution.
From the condition, we have $q=3, S_{6}=1820$.
Using formula (4.11), we get $\frac{b_{1}\left(1-3^{6}\right)}{1-3}=1820, b_{1}=5$.
Using formula (4.6), we find $b_{5}=b_{1} q^{4}=5 \cdot(3)^{4}=405$.
Answer: $5,405$. | 5,405 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,659 |
4.021. An arithmetic progression has the following property: for any $n$ the sum of its first $n$ terms is equal to $5 n^{2}$. Find the common difference of this progression and its first three terms. | ## Solution.
Let $n=2$ and $n=3$. By formula (4.5) we find
$$
\left\{\begin{array} { l }
{ \frac { 2 a _ { 1 } + d } { 2 } \cdot 2 = 5 \cdot 2 ^ { 2 } , } \\
{ \frac { 2 a _ { 1 } + 2 d } { 2 } \cdot 3 = 5 \cdot 3 ^ { 2 } }
\end{array} \Leftrightarrow \left\{\begin{array} { l }
{ 2 a _ { 1 } + d = 2 0 , } \\
{ a _ ... | 10;5,15,25 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,660 |
4.022. The product of the first three terms of a geometric progression is 1728, and their sum is 63. Find the first term and the common ratio of this progression. | ## Solution.
From the condition, we have $\left\{\begin{array}{l}b_{1} \cdot b_{2} \cdot b_{3}=1728, \\ b_{1}+b_{2}+b_{3}=63 .\end{array}\right.$
Using formula (4.6), we get
$\left\{\begin{array}{l}b_{1} \cdot b_{1} q \cdot b_{1} q^{2}=1728, \\ b_{1}+b_{1} q+b_{1} q^{2}=63\end{array} \Leftrightarrow\left\{\begin{arra... | 1)3,4;2)48,\frac{1}{4} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,661 |
4.024. The first term of an arithmetic progression is 429, and its difference is -22. How many terms of this progression need to be taken so that their sum is equal to 3069? | ## Solution.
From the condition, we have $a_{1}=429, d=-22, S_{n}=3069$.
Using formula (4.5), we get
$$
\begin{aligned}
& \frac{2 \cdot 429-22(n-1)}{2} \cdot n=3069, (429-11(n-1)) n=3069 \\
& n^{2}-40 n+279=0 \Rightarrow n_{1}=9, n_{2}=31
\end{aligned}
$$
Answer: 9 or 31. | 9or31 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,662 |
4.025. The sum of an infinite geometric progression with a common ratio $|q|<1$ is 16, and the sum of the squares of the terms of this progression is 153.6. Find the fourth term and the common ratio of the progression. | ## Solution.
From the condition, we have $\left\{\begin{array}{l}b_{1}+b_{2}+b_{3}+\ldots=16, \\ b_{1}^{2}+b_{2}^{2}+b_{3}^{2}+\ldots=153.6,\end{array}\right.$
Using formula (4.6), we get
$$
\left\{\begin{array} { l }
{ b _ { 1 } + b _ { 1 } q + b _ { 1 } q ^ { 2 } + \ldots = 1 6 , } \\
{ b _ { 1 } ^ { 2 } + b _ { ... | \frac{3}{16},\frac{1}{4} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,663 |
4.026. Find the natural numbers forming an arithmetic progression if the product of the first three and the first four of its terms are 6 and 24, respectively. | Solution.
From the condition we have $\left\{\begin{array}{l}a_{1}\left(a_{1}+d\right)\left(a_{2}+d\right)=6, \\ a_{1}\left(a_{1}+d\right)\left(a_{1}+2 d\right)\left(a_{1}+3 d\right)=24\end{array} \Rightarrow\right.$
$$
\Rightarrow \frac{a_{1}\left(a_{1}+d\right)\left(a_{2}+d\right)}{a_{1}\left(a_{1}+d\right)\left(a_... | 1,2,3,4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,664 |
4.027. The sum of the third and ninth terms of an arithmetic progression is 6, their product is $135 / 16$. Find the sum of the first 15 terms of this progression. | ## Solution.
From the condition, we have $\left\{\begin{array}{l}a_{3}+a_{9}=6, \\ a_{3} \cdot a_{9}=\frac{135}{16}\end{array}\right.$.
Using formula (4.1), we get
$$
\begin{aligned}
& \left\{\begin{array} { l }
{ a _ { 1 } + 2 d + a _ { 1 } + 8 d = 6 , } \\
{ ( a _ { 1 } + 2 d ) ( a _ { 1 } + 8 d ) = \frac { 1 3 5... | 37.5or52.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,665 |
4.028. Find the number of terms in a finite geometric progression, where the first, second, and last terms are 3, 12, and 3072, respectively. | Solution.
From the condition we have $b_{1}=3, b_{2}=12, \ldots, b_{n}=3072$.
By formula (4.6) we get
$$
\left\{\begin{array} { l }
{ b _ { 1 } = 3 , } \\
{ b _ { 1 } q = 1 2 , } \\
{ b _ { 1 } q ^ { n - 1 } = 3 0 7 2 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
b_{1}=3, \\
q=4, \\
4^{n-1}=1024
\end{array}... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,666 |
4.029. Find the sum of all positive even two-digit numbers that are divisible by 3. | ## Solution.
From the condition, we have $a_{1}=12, a_{n}=96, d=12$.
Using formulas (4.4) and (4.5), we get
$$
n=\frac{a_{n}-a_{1}}{d}+1 ; n=\frac{96-12}{6}+1=15, S_{n}=\frac{12+96}{2} \cdot 15=810 \text {. }
$$
Answer: 810. | 810 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 50,667 |
4.030. Find the denominator $q$ of an infinite geometric progression $(|q|<1)$, where each term is four times the sum of all its subsequent terms. | ## Solution.
From the condition, we have $b_{1}=4\left(s-b_{1}\right)$.
By formula (4.12), we get
$$
b_{1}=4\left(\frac{b_{1}}{1-q}-b_{1}\right), b_{1}=\frac{4 b_{1}(1-1+q)}{1-q}, 1-q=4 q, q=\frac{1}{5}
$$
Answer: $\frac{1}{5}$. | \frac{1}{5} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,668 |
4.031. It is known that the interior angles of a certain convex polygon, the smallest of which is $120^{\circ}$, form an arithmetic progression with a difference of $5^{\circ}$. Determine the number of sides of the polygon. | Solution.
From the condition, we have $a_{1}=120^{\circ}, d=5^{\circ}$. Using the formulas for the sum of terms of an arithmetic progression (4.5) and the sum of the interior angles of an $n$-sided polygon $S_{n}=180^{\circ}(n-2)$, we get
$$
\frac{240^{\circ}+(n-1) 5^{\circ}}{2} \cdot n=180^{\circ}(n-2), n^{2}-25 n+1... | 9 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 50,669 |
4.032. The product of the third and sixth terms of an arithmetic progression is 406. When the ninth term of the progression is divided by its fourth term, the quotient is 2, and the remainder is 6. Find the first term and the common difference of the progression. | ## Solution.
From the condition, we have $\left\{\begin{array}{l}a_{3} \cdot a_{6}=406, \\ a_{9}=2 a_{4}+6 .\end{array}\right.$ Using formula (4.1), we get
$$
\begin{aligned}
& \left\{\begin{array}{l}
\left(a_{1}+2 d\right)\left(a_{1}+5 d\right)=406, \\
a_{1}+8 d=2\left(a_{1}+3 d\right)+6
\end{array}\right. \\
& \Rig... | 45 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,670 |
4.033. In an infinite geometric progression with positive terms and a common ratio $|q|<1$, the sum of the first three terms is 10.5, and the sum of the progression is 12. Find the progression. | Solution.
From formulas (4.6) and (4.12), we get:
$$
\left\{\begin{array} { l }
{ b _ { 1 } + b _ { 1 } q + b _ { 1 } q ^ { 2 } = 10.5 , } \\
{ b _ { 1 } = 12 ( 1 - q ) }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
b_{1}\left(1+q+q^{2}\right)=10.5 \\
b_{1}=12(1-q)
\end{array}\right.\right.
$$
From this, $12... | 6,3,\frac{3}{2},\ldots | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,671 |
4.034. Find the first three terms of an arithmetic progression, for which the sum of any number of terms is equal to three times the square of this number. | Solution.
Let $n=2$ and $n=3$. From the condition we have
$$
\left\{\begin{array}{l}
a_{1}+a_{2}=3 \cdot 2^{2} \\
a_{1}+a_{2}+a_{3}=3 \cdot 3^{2}
\end{array}\right.
$$
By formula (4.1) we get
$$
\left\{\begin{array} { l }
{ a _ { 1 } + a _ { 1 } + d = 1 2 , } \\
{ a _ { 1 } + a _ { 1 } + d + a _ { 1 } + 2 d = 2 7 ... | 3,9,15 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,672 |
4.035. When the thirteenth term of an arithmetic progression is divided by the third term, the quotient is 3, and when the eighteenth term is divided by the seventh term, the quotient is 2 and the remainder is 8. Determine the common difference and the first term of the progression. | Solution.
From the condition we have $\left\{\begin{array}{l}a_{13}=3 a_{3}, \\ a_{18}=2 a_{7}+8 .\end{array}\right.$
Using formula (4.1), we get
$$
\left\{\begin{array} { l }
{ a _ { 1 } + 1 2 d = 3 ( a _ { 1 } + 2 d ) } \\
{ a _ { 1 } + 1 7 d = 2 ( a _ { 1 } + 6 d ) + 8 }
\end{array} \Leftrightarrow \left\{\begin... | 4,12 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,673 |
6.001. $\frac{x^{2}+1}{x-4}-\frac{x^{2}-1}{x+3}=23$. | Solution.
Domain of definition: $x \neq-3, x \neq 4$.
$\frac{x^{2}+1}{x-4}-\frac{x^{2}-1}{x+3}=23 \Leftrightarrow \frac{16 x^{2}-25 x-275}{(x-4)(x+3)}=0 \Rightarrow$
$\Rightarrow 16 x^{2}-25 x-275=0 \Rightarrow x_{1}=-\frac{55}{16}, x_{2}=5$.
Answer: $x_{1}=-\frac{55}{16}, x_{2}=5$. | x_{1}=-\frac{55}{16},x_{2}=5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,674 |
6.003. $\frac{x^{2}+x-5}{x}+\frac{3 x}{x^{2}+x-5}+4=0$.
6.003. $\frac{x^{2}+x-5}{x}+\frac{3 x}{x^{2}+x-5}+4=0$. | Solution.
Domain of definition: $x \neq 0, x \neq \frac{-1 \pm \sqrt{21}}{2}$.
Let $\frac{x^{2}+x-5}{x}=z$, then $z+\frac{1}{z}+4=0 \Rightarrow$
$\Rightarrow z^{2}+4 z+3=0, \Rightarrow z_{1}=-3, z_{2}=-1$.
To find $x$, solve the two equations:
$\frac{x^{2}+x-5}{x}=-3$ or $\frac{x^{2}+x-5}{x}=-1$.
Solving each of ... | x_{1}=-5,x_{2}=1,x_{3,4}=-1\\sqrt{6} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,675 |
6.004. $x^{4}-\frac{50}{2 x^{4}-7}=14$. | Solution.
Domain of definition: $x \neq \pm 4 \sqrt{\frac{7}{2}}$.
Let $2 x^{4}-7=z$, then $\frac{z+2}{2}-\frac{50}{z}=14 \Rightarrow$
$\Rightarrow z^{2}-21 z-100=0 \Leftrightarrow z_{1}=-4, z_{2}=25$.
To find $x$, solve the two equations $2 x^{4}-7=-4$ or $2 x^{4}-7=25$, solving which, we get $x_{1}=-\sqrt[4]{\fra... | x_{1,2}=\\sqrt[4]{\frac{3}{2}},x_{3,4}=\2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,676 |
6.005. $\frac{1}{x(x+2)}-\frac{1}{(x+1)^{2}}=\frac{1}{12}$. | Solution.
Domain of definition: $x \neq 0, x \neq-1, x \neq-2$.
$\frac{1}{x(x+2)}-\frac{1}{(x+1)^{2}}=\frac{1}{12} \Leftrightarrow \frac{1}{x^{2}+2 x}-\frac{1}{x^{2}+2 x+1}=\frac{1}{12}$.
Let $x^{2}+2 x=z$, then
$\frac{z^{2}+z+12}{z(z+1)}=0 \Rightarrow z^{2}+z+12=0 \Rightarrow z_{1}=-4, z_{2}=3$.
To find $x$, solv... | x_{1,2}\in\varnothing,x_{3}=-3,x_{4}=1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,677 |
6.006. $x+\frac{1}{x}=2 \frac{m^{2}+n^{2}}{m^{2}-n^{2}}$.
6.006. $x+\frac{1}{x}=2 \frac{m^{2}+n^{2}}{m^{2}-n^{2}}$. | ## Solution.
Domain of definition: $x \neq 0, m \neq \pm n$.
$$
x+\frac{1}{x}=2 \frac{m^{2}+n^{2}}{m^{2}-n^{2}} \Leftrightarrow x^{2}-2 \frac{m^{2}+n^{2}}{m^{2}-n^{2}} x+1=0,
$$
roots $x_{1}=\frac{m+n}{m-n}, x_{2}=\frac{m-n}{m+n}$.
Answer: $x_{1}=\frac{m+n}{m-n}, x_{2}=\frac{m-n}{m+n}$, where $m \neq|n|$. | x_{1}=\frac{+n}{-n},x_{2}=\frac{-n}{+n} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,678 |
6.007. $\frac{x^{2}}{a^{3}}+\frac{b^{3}}{x^{2}}=\frac{b}{a}+\frac{b^{2}}{a^{2}}$. | ## Solution.
Domain of definition: $x \neq 0, a \neq 0$.
$$
\frac{x^{2}}{a^{3}}+\frac{b^{3}}{x^{2}}=\frac{b}{a}+\frac{b^{2}}{a^{2}} \Leftrightarrow \frac{x^{4}-\left(a^{2} b+a b^{2}\right) x^{2}+a^{3} b^{3}}{a^{3} x^{2}}=0
$$
or $x^{4}-\left(a^{2} b+a b^{2}\right) x^{2}+a^{3} b^{3}=0, a x \neq 0$.
The equation is a... | x_{1,2}=\b\sqrt{},where>0;x_{3,4}=\\sqrt{b},whereb\geq0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,679 |
6.009. $\frac{5 a}{y+a}+\frac{4 a}{y+2 a}+\frac{3 a}{y+3 a}=8$. | Solution.
Domain of definition: $y \neq -a, y \neq -2a, y \neq -3a$.
$$
\frac{5a}{y+a} + \frac{4a}{y+2a} + \frac{3a}{y+3a} = 8 \Leftrightarrow \frac{8y^3 + 36ay^2 + 38a^2y}{(y+a)(y+2a)(y+3a)} = 0
$$
or
$$
\begin{aligned}
& \left\{\begin{array}{l}
y(4y^2 + 18ay + 19a^2) = 0, \\
y \neq -a, \\
y \neq -2a, \\
y \neq -3... | y_1=0,y_{2,3}=\frac{(-9\\sqrt{5})}{4} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,680 |
6.010. $\frac{1}{x^{3}+2}-\frac{1}{x^{3}+3}=\frac{1}{12}$. | Solution.
Domain of definition: $x \neq-\sqrt[3]{2}, x \neq-\sqrt[3]{3}$.
$\frac{1}{x^{3}+2}-\frac{1}{x^{3}+3}=\frac{1}{12} \Leftrightarrow \frac{x^{6}+5 x^{3}-6}{12\left(x^{3}+2\right)\left(x^{3}+3\right)}=0 \Leftrightarrow$
$\Leftrightarrow\left\{\begin{array}{l}x^{6}+5 x^{3}-6=0, \\ \left(x^{3}+2\right)\left(x^{3... | x_{1}=-\sqrt[3]{6},x_{2}=1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,681 |
6.012. $(x-1)\left(x^{2}-3\right)+(2 x-1)\left(x^{2}+2\right)=3$. | Solution.
Domain of definition: $x \in R$.
## We have
$$
\begin{aligned}
& x^{3}-x^{2}-3 x+3+2 x^{3}-x^{2}+4 x-2=3 \Leftrightarrow \\
& \Leftrightarrow 3 x^{3}-2 x^{2}+x-2=0 \Leftrightarrow 3 x^{3}-3 x^{2}+x^{2}-x+2 x-2=0 \Leftrightarrow \\
& \Leftrightarrow 3 x^{2}(x-1)+x(x-1)+2(x-1)=0 \Leftrightarrow(x-1)\left(3 x... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,682 |
6.013. $3\left(x+\frac{1}{x^{2}}\right)-7\left(1+\frac{1}{x}\right)=0$. | ## Solution.
Domain of definition: $x \neq 0$.
$$
\begin{aligned}
& 3\left(x+\frac{1}{x^{2}}\right)-7\left(1+\frac{1}{x}\right)=0 \Leftrightarrow \frac{3\left(x^{3}+1\right)-7(x+1) x}{x^{2}}=0 \Leftrightarrow \\
& \Leftrightarrow \frac{3(x+1)\left(x^{2}-x+1\right)-7(x+1) x}{x^{2}}=0 \Leftrightarrow \frac{(x+1)\left(3... | x_{1}=-1,x_{2}=\frac{1}{3},x_{3}=3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,683 |
6.014. $\frac{4}{x^{2}+4}+\frac{5}{x^{2}+5}=2$. | ## Solution.
Domain: $x \in R$.
$\frac{2 x^{4}+9 x^{2}}{\left(x^{2}+4\right)\left(x^{2}+5\right)}=0 \Leftrightarrow 2 x^{4}+9 x^{2}=0 \Leftrightarrow x^{2}\left(2 x^{2}+9\right)=0$,
$x^{2}=0, x_{1}=0$ or $2 x^{2}+9=0, x_{2,3} \in \varnothing$.
Answer: $x=0$. | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,684 |
6.015. $\frac{7(x-2)(x-3)(x-4)}{(2 x-7)(x+2)(x-6)}=-2$. | ## Solution.
Domain of definition: $x \neq \frac{7}{2}, x \neq-2, x \neq 6$.
From the condition, we get $\frac{11 x^{3}-93 x^{2}+190 x}{(2 x-7)(x+2)(x-6)}=0$. Considering the domain of definition, this equation is equivalent to
$$
\begin{aligned}
& 11 x^{3}-93 x^{2}+190 x=0 \Leftrightarrow x\left(11 x^{2}-93 x+190\r... | x_{1}=0,x_{2}=5,x_{3}=\frac{38}{11} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,685 |
6.016. $\frac{x^{2}+1}{x}+\frac{x}{x^{2}+1}=2.9$. | ## Solution.
Domain of definition: $x \neq 0$.
$$
\begin{aligned}
& \text { Let } \frac{x^{2}+1}{x}=y, \text { then } y+\frac{1}{y}-2.9=0 \Leftrightarrow \frac{y^{2}-2.9 y+1}{y}=0 \Leftrightarrow \\
& \Leftrightarrow\left\{\begin{array}{l}
y^{2}-2.9 y+1=0, \\
y \neq 0
\end{array} \Rightarrow y_{1}=\frac{5}{2}, y_{2}=... | x_{1}=2,x_{2}=\frac{1}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,686 |
6.017. $\frac{x+n}{m+n}-\frac{m-n}{x-n}=\frac{x+p}{m+p}-\frac{m-p}{x-p}$. | ## Solution.
Domain of definition: $x \neq n, x \neq p, m \neq -n, m \neq -p$.
From the condition, we get
$$
\begin{aligned}
& \frac{(x+n)(x-n)-(m-n)(m+n)}{(m+n)(x-n)}=\frac{(x+p)(x-p)-(m-p)(m+p)}{(m+p)(x-p)} \Leftrightarrow \\
& \Leftrightarrow \frac{x^{2}-n^{2}-m^{2}+n^{2}}{(m+n) x-n(m+n)}=\frac{x^{2}-p^{2}-m^{2}+... | x_{1}=,x_{2}=-,x_{3}=+n+pifn\neqp;x\inRexceptnpifn=p | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,687 |
6.018. $x^{2}+x+x^{-1}+x^{-2}=4$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
6.018. $x^{2}+x+x^{-1}+x^{-2}=4$. | ## Solution.
Domain of definition: $x \neq 0$.
From the condition, we have $\left(x^{2}+\frac{1}{x^{2}}\right)+\left(x+\frac{1}{x}\right)-4=0$.
Let $x+\frac{1}{x}=y \Rightarrow x^{2}+2+\frac{1}{x^{2}}=y^{2}$ or $x^{2}+\frac{1}{x^{2}}=y^{2}-2$. Our equation becomes $y^{2}-2+y-4=0 \Leftrightarrow y^{2}+y-6=0$, from wh... | x_{1,2}=\frac{-3\\sqrt{5}}{2},x_{3,4}=1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,688 |
6.019. $\frac{21}{x^{2}-4 x+10}-x^{2}+4 x=6$. | ## Solution.
Domain: $x \in R$.
From the condition, we have $\frac{21}{x^{2}-4 x+10}-\left(x^{2}-4 x+10\right)+4=0$.
Let $x^{2}-4 x+10=y \neq 0: \frac{21}{y}-y+4=0 \Leftrightarrow \frac{y^{2}-4 y-21}{y}=0$. The equation $y^{2}-4 y-21=0$ has roots $y_{1}=-3, y_{2}=7$. Relative to $x$, we get two equations: $x^{2}-4 x... | x_{1}=3,x_{2}=1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,689 |
6.020. $\frac{x-a}{x-b}+\frac{x-b}{x-a}=2.5$. | ## Solution.
Domain of definition: $x \neq b, x \neq a$.
$$
\text { Let } \frac{x-a}{x-b}=y: y+\frac{1}{y}-2.5=0 ; \frac{y^{2}-2.5 y+1}{y}=0
$$
The equation $y^{2}-2.5 y+1=0$ has roots $y_{1}=\frac{1}{2}, y_{2}=2$. We obtain
two equations in terms of $x: \frac{x-a}{x-b}=\frac{1}{2}$, from which $x_{1}=2 a-b$, or $\... | x_{1}=2a-b,x_{2}=2b- | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,690 |
6.021. $8 x^{4}+x^{3}+64 x+8=0$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
6.021. $8 x^{4}+x^{3}+64 x+8=0$. | Solution.
Domain: $x \in R$.
From the condition we have
$$
\begin{aligned}
& \left(8 x^{4}+x^{3}\right)+(64 x+8)=0 \Leftrightarrow x^{3}(8 x+1)+8(8 x+1)=0 \Leftrightarrow \\
& \Leftrightarrow(8 x+1)\left(x^{3}+8\right)=0 \Leftrightarrow(8 x+1)(x+2)\left(x^{2}-2 x+4\right)=0
\end{aligned}
$$
From this, $8 x+1=0, x_{... | x_{1}=-\frac{1}{8},x_{2}=-2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,691 |
6.022. $(x+3)^{3}-(x+1)^{3}=56$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
6.022. $(x+3)^{3}-(x+1)^{3}=56$. | ## Solution.
Domain: $x \in R$.
From the condition we have
$$
\begin{aligned}
& (x+3-x-1)\left((x+3)^{2}+(x+3)(x+1)+(x+1)^{2}\right)=56 \Leftrightarrow \\
& \Leftrightarrow 2\left(x^{2}+6 x+9+x^{2}+4 x+3+x^{2}+2 x+1\right)=56 \Leftrightarrow \\
& \Leftrightarrow x^{2}+4 x-5=0, x_{1}=-5, x_{2}=1
\end{aligned}
$$
Ans... | x_{1}=-5,x_{2}=1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,692 |
6.023. $\frac{x+2}{x+1}+\frac{x+6}{x+3}+\frac{x+10}{x+5}=6$. | Solution.
Domain of definition: $x \neq-1, x \neq-3, x \neq-5$.
From the condition we have:
$$
\begin{aligned}
& \frac{(x+1)+1}{x+1}+\frac{(x+3)+3}{x+3}+\frac{(x+5)+5}{x+5}=6 \Leftrightarrow \\
& \Leftrightarrow \frac{x+1}{x+1}+\frac{1}{x+1}+\frac{x+3}{x+3}+\frac{3}{x+3}+\frac{x+5}{x+5}+\frac{5}{x+5}=6 \Leftrightarr... | x_{1}=0,x_{2}=\frac{-9+\sqrt{12}}{3},x_{3}=\frac{-9-\sqrt{12}}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,693 |
6.024. $4 x^{2}+12 x+\frac{12}{x}+\frac{4}{x^{2}}=47$.
6.024. $4 x^{2}+12 x+\frac{12}{x}+\frac{4}{x^{2}}=47$. | ## Solution.
Domain of definition: $x \neq 0$.
Grouping, we get:
$$
\begin{aligned}
& \left(4 x^{2}+\frac{4}{x^{2}}\right)+\left(12 x+\frac{12}{x}\right)-47=0 \Leftrightarrow \\
& \Leftrightarrow 4\left(x^{2}+\frac{1}{x^{2}}\right)+12\left(x+\frac{1}{x}\right)-47=0
\end{aligned}
$$
Let $x+\frac{1}{x}=y \Rightarrow ... | x_{1,2}=\frac{-11\\sqrt{105}}{4},x_{3}=\frac{1}{2},x_{4}=2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,694 |
6.025. $(x-a)^{3}-(x-b)^{3}=b^{3}-a^{3}$. | ## Solution.
Domain of definition: $x \in R$.
Let's factorize the left and right parts of the equation as differences of cubes:
$$
(x-a-x+b)\left((x-a)^{2}+(x-a)(x-b)+(x-b)^{2}\right)=(b-a)\left(b^{2}+b a+a^{2}\right)
$$
$(b-a)\left(x^{2}-2 a x+a^{2}+x^{2}-(a+b) x+a b+x^{2}-2 b x+b^{2}\right)-$ $-(b-a)\left(b^{2}+b... | x_{1}=0,x_{2}=+b | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,695 |
6.026. $\frac{a x^{2}}{x-1}=(a+1)^{2}$. | ## Solution.
Domain of definition: $x \neq 1$.
We bring the equation to a common denominator:
$$
\frac{a x^{2}-(a+1)^{2}(x-1)}{x-1}=0 \Leftrightarrow \frac{a x^{2}-(a+1)^{2} x+(a+1)^{2}}{x-1}=0
$$
Considering the domain of definition, $a x^{2}-(a+1)^{2} x+(a+1)^{2}=0$, from which
$$
\begin{aligned}
& x_{1,2}=\frac... | x_{1}=\frac{+1}{},\,x_{2}=+1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,696 |
6.028. $\frac{x}{a+b}+\frac{2a-x}{a-b}-\frac{a+b}{x}=1$. | Solution.
Domain of definition: $a \neq \pm b, x \neq 0$.
From the given, we have
$$
\begin{aligned}
& \frac{x}{a+b}+\frac{2 a-x}{a-b}-\frac{a+b}{x}-1=0 \Leftrightarrow \\
& \Leftrightarrow \frac{(a-b) x^{2}+(2 a-x)(a+b) x-(a+b)^{2}(a-b)-(a+b)(a-b) x}{(a+b)(a-b) x}=0 \Leftrightarrow \\
& \Leftrightarrow \frac{(a-b) ... | x_{1}=+b,x_{2}=\frac{^{2}-b^{2}}{2b} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,697 |
6.029. $\frac{a^{2}-1}{a x-1}+\frac{a-x}{a}=1$.
6.029. $\frac{a^{2}-1}{a x-1}+\frac{a-x}{a}=1$. | ## Solution.
Domain of definition: $a \neq 0, x \neq \frac{1}{a}$.
From the condition we have:
$$
\begin{aligned}
& \frac{a^{2}-1}{a x-1}+\frac{a}{a}-\frac{x}{a}=1 \Leftrightarrow \frac{a^{2}-1}{a x-1}+1-\frac{x}{a}=1 \Leftrightarrow \frac{a^{2}-1}{a x-1}-\frac{x}{a}=0 \Leftrightarrow \\
& \Leftrightarrow \frac{a x^... | x_{1}=\frac{1-^{2}}{},x_{2}= | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,698 |
6.030. $\left(\frac{x^{2}+6}{x^{2}-4}\right)^{2}=\left(\frac{5 x}{4-x^{2}}\right)^{2}$. | ## Solution.
Domain of definition: $x \neq \pm 2$.
Rewrite this equation as
$$
\begin{aligned}
& \left(\frac{x^{2}+6}{x^{2}-4}\right)^{2}=\left(\frac{5 x}{x^{2}-4}\right)^{2} \Rightarrow\left|\frac{x^{2}+6}{x^{2}-4}\right|=\sqrt{\left(\frac{5 x}{x^{2}-4}\right)^{2}} \Rightarrow\left|\frac{x^{2}+6}{x^{2}-4}\right|=\l... | x_{1}=-3,x_{2}=3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,699 |
6.031. $\sqrt{3 x+4}+\sqrt{x-4}=2 \sqrt{x}$. | ## Solution.
Domain of definition: $3 x+4 \geq 0, x-4 \geq 0, x \geq 0 \Rightarrow x \geq 4$.
Squaring both sides of the equation, we get
$$
\begin{aligned}
& 3 x+4+2 \sqrt{(3 x+4)(x-4)}+x-4=4 x \Leftrightarrow \\
& \Leftrightarrow 2 \sqrt{(3 x+4)(x-4)}=0
\end{aligned}
$$
Squaring again, we get: $(3 x+4)(x-4)=0$. F... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,700 |
6.032. $\sqrt{x+\sqrt{x+11}}+\sqrt{x-\sqrt{x+11}}=4$.
6.032. $\sqrt{x+\sqrt{x+11}}+\sqrt{x-\sqrt{x+11}}=4$. | Solution.
Let $\sqrt{x+11}=y \geq 0$ or $x+11=y^{2}$, i.e., $x=y^{2}-11$. Then
$$
\sqrt{y^{2}+y-11}+\sqrt{y^{2}-y-11}=4 \text { or } \sqrt{y^{2}+y-11}=4-\sqrt{y^{2}-y-11}
$$
Squaring both sides of the equation, we get
$$
y^{2}+y-11=16-8 \sqrt{y^{2}-y-11}+y^{2}-y-11, 8 \sqrt{y^{2}-y-11}=16-2 y
$$
or $4 \sqrt{y^{2}-... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,701 |
6.034. $1+\sqrt{1+x \sqrt{x^{2}-24}}=x$.
6.034. $1+\sqrt{1+x \sqrt{x^{2}-24}}=x$. | Solution.
Write the equation as $\sqrt{1+x \sqrt{x^{2}-24}}=x-1$. Squaring both sides of the equation, we get
$$
\begin{aligned}
& \left\{\begin{array} { l }
{ 1 + x \sqrt { x ^ { 2 } - 2 4 } = x ^ { 2 } - 2 x + 1 , } \\
{ x - 1 \geq 0 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x \sqrt{x^{2}-24}=x^{2}-2 x... | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,702 |
6.036. $\sqrt{3 x+7}-\sqrt{x+1}=2$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly.
6.036. $\sqrt{3 x+7}-\sqrt{x+1}=2$. | Solution.
Domain of definition: $\left\{\begin{array}{l}3 x+7 \geq 0, \\ x+1 \geq 0\end{array} \Leftrightarrow x \geq-1\right.$.
By squaring both sides of the equation, we get
$$
\begin{aligned}
& 3 x+7-2 \sqrt{(3 x+7)(x+1)}+x+1=4 \Leftrightarrow 4 x+4=2 \sqrt{(3 x+7)(x+1)} \Leftrightarrow \\
& \Leftrightarrow 2(x+1... | x_{1}=-1,x_{2}=3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,703 |
6.037. $\sqrt[3]{1+\sqrt{x}}+\sqrt[3]{1-\sqrt{x}}=2$. | Solution.
Domain of definition: $x \geq 0$.
Raising both sides of the equation to the third power, we get
$$
\begin{aligned}
& 1+\sqrt{x}+3 \sqrt[3]{(1+\sqrt{x})^{2}(1-\sqrt{x})}+3 \sqrt[3]{(1+\sqrt{x})(1-\sqrt{x})^{2}}+1-\sqrt{x}=8 \Leftrightarrow \\
& \Leftrightarrow 3 \sqrt[3]{(1+\sqrt{x})^{2}(1-\sqrt{x})}+3 \sqr... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,704 |
6.039. $\left(\frac{x+5}{x}\right)^{\frac{1}{2}}+4\left(\frac{x}{x+5}\right)^{\frac{1}{2}}=4$. | ## Solution.
From the condition we have
$$
\sqrt{\frac{x+5}{x}}+4 \sqrt{\frac{x}{x+5}}-4=0 \Leftrightarrow \sqrt{\frac{x+5}{x}}+\frac{4}{\sqrt{\frac{x}{x+5}}}-4=0
$$
Let
$$
\begin{aligned}
& \sqrt{\frac{x+5}{x}}=y > 0: y+\frac{4}{y}-4=0 \Leftrightarrow y^{2}-4 y+4=0 \Leftrightarrow \\
& \Leftrightarrow(y-2)^{2}=0 \... | \frac{5}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,706 |
6.040. $\sqrt[3]{24+\sqrt{x}}-\sqrt[3]{5+\sqrt{x}}=1$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
6.040. $\sqrt[3]{24+\sqrt{x}}-\sqrt[3]{5+\sqrt{x}}=1$. | ## Solution.
Domain of definition: $x \geq 0$.
Raising both sides of the equation to the third power, we get
$$
\begin{aligned}
& 24+\sqrt{x}-3 \sqrt[3]{(24+\sqrt{x})^{2}(5+\sqrt{x})}+3 \sqrt[3]{(24+\sqrt{x})(5+\sqrt{x})^{2}}-5-\sqrt{x}=1 \Leftrightarrow \\
& \Leftrightarrow-3 \sqrt[3]{(24+\sqrt{x})}(5+\sqrt{x})(\sq... | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,707 |
6.042. $x^{2}+3 x-18+4 \sqrt{x^{2}+3 x-6}=0$. | ## Solution.
Let $\sqrt{x^{2}+3 x-6}=y \geq 0$. Then $x^{2}+3 x-6=y^{2}$
or
$$
x^{2}+3 x=y^{2}+6
$$
and the equation becomes:
$$
y^{2}+6-18+4 y=0 \Leftrightarrow y^{2}+4 y-12=0, y_{1}=-6 \text { (not suitable), } y_{2}=2 \text {. }
$$
Then
$$
\begin{aligned}
& \sqrt{x^{2}+3 x-6}=2 \Leftrightarrow x^{2}+3 x-6=4 \... | x_{1}=-5,x_{2}=2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,708 |
6.043. $\sqrt{x^{2}+32}-2 \sqrt[4]{x^{2}+32}=3$. | ## Solution.
Let $\sqrt[4]{x^{2}+32}=y>0$. With respect to $y$, we obtain the equation $y^{2}-2 y-3=0$, from which $y_{1}=-1$ (not suitable), $y_{2}=3$. Then
$$
\sqrt[4]{x^{2}+32}=3 \Leftrightarrow x^{2}+32=81 \Leftrightarrow x^{2}=49
$$
This expression satisfies the given equation. Hence, $|x|=7$ or $x_{1,2}= \pm 7... | x_{1}=-7,x_{2}=7 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,709 |
6.046. $3 \sqrt[3]{x}-5 \sqrt[3]{x^{-1}}=2 x^{-1}$. | Solution.
From the condition we have $3 \sqrt[3]{x}-\frac{5}{\sqrt[3]{x}}=\frac{2}{x}$.
Let $\sqrt[3]{x}=y, y \neq 0$, and the equation takes the form
$$
3 y-\frac{5}{y}-\frac{2}{y^{3}}=0(y \neq 0) \Leftrightarrow 3 y^{4}-5 y^{2}-2=0 \Leftrightarrow 3\left(y^{2}\right)^{2}-5\left(y^{2}\right)-2=0
$$
from which $y^{... | x_{1}=-2\sqrt{2},x_{2}=2\sqrt{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,710 |
6.049. $\sqrt{x^{3}+8}+\sqrt[4]{x^{3}+8}=6$.
6.049. $\sqrt{x^{3}+8}+\sqrt[4]{x^{3}+8}=6$. | ## Solution.
Domain of definition: $x^{3}+8 \geq 0 \Leftrightarrow x^{3} \geq-8 \Leftrightarrow x \geq-2$.
Let $\sqrt[4]{x^{3}+8}=y, y>0$, and the equation becomes $y^{2}+y=6 \Leftrightarrow$ $\Leftrightarrow y^{2}+y-6=0$, from which $y_{1}=-3, y_{2}=2 ; y_{1}=-3$ is not suitable. Then $\sqrt[4]{x^{3}+8}=2, x^{3}+8=1... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,711 |
6.050. $\frac{(5-x) \sqrt{5-x}+(x-3) \sqrt{x-3}}{\sqrt{5-x}+\sqrt{x-3}}=2$. | ## Solution.
Domain of definition: $\left\{\begin{array}{l}5-x \geq 0, \\ x-3 \geq 0,\end{array} \Leftrightarrow 3 \leq x \leq 5\right.$.
Rewrite the equation as $\frac{(\sqrt{5-x})^{3}+(\sqrt{x-3})^{3}}{\sqrt{5-x}+\sqrt{x-3}}=2$ and factor the numerator of the left side as the sum of cubes:
$$
\frac{(\sqrt{5-x}+\sq... | x_{1}=5,x_{2}=3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,712 |
6.052.
$$
\frac{1}{x-\sqrt{x^{2}-x}}-\frac{1}{x+\sqrt{x^{2}-x}}=\sqrt{3}
$$ | ## Solution.
Domain of definition: $\left\{\begin{array}{l}x^{2}-x \geq 0, \\ x \neq 0\end{array} \Leftrightarrow\left\{\begin{array}{l}x(x-1) \geq 0, \\ x \neq 0\end{array} \Leftrightarrow x \in(-\infty ; 0) \cup[1 ;+\infty)\right.\right.$.
From the condition we get
$$
\begin{aligned}
& \frac{x+\sqrt{x^{2}-x}-x+\sq... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,713 |
6.053. $\frac{\sqrt[3]{x^{4}}-1}{\sqrt[3]{x^{2}}-1}-\frac{\sqrt[3]{x^{2}}-1}{\sqrt[3]{x}+1}=4$ | ## Solution.
Domain of definition: $x \neq \pm 1$.
Let $\sqrt[3]{x}=y, y \neq \pm 1$. The equation in terms of $y$ becomes
$$
\begin{aligned}
& \frac{y^{4}-1}{y^{2}-1}-\frac{y^{2}-1}{y+1}=4 \Leftrightarrow \frac{\left(y^{2}-1\right)\left(y^{2}+1\right)}{y^{2}-1}-\frac{(y-1)(y+1)}{y+1}=4 \Leftrightarrow \\
& \Leftrig... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,714 |
6.054. $\sqrt{5+\sqrt[3]{x}}+\sqrt{5-\sqrt[3]{x}}=\sqrt[3]{x}$. | Solution.
Domain of definition: $\left\{\begin{array}{l}5+\sqrt[3]{x} \geq 0 \\ 5-\sqrt[3]{x} \geq 0\end{array} \Leftrightarrow -125 \leq x \leq 125\right.$.
By squaring both sides of the equation, we get the equation
$$
\begin{aligned}
& 5+\sqrt[3]{x}+2 \sqrt{(5+\sqrt[3]{x})(5-\sqrt[3]{x})}+5-\sqrt[3]{x}=\sqrt[3]{x... | 64 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,715 |
6.055. $\sqrt{x \sqrt[5]{x}}+\sqrt[5]{x \sqrt{x}}=56$.
6.055. $\sqrt{x \sqrt[5]{x}}+\sqrt[5]{x \sqrt{x}}=56$. | ## Solution.
Domain of definition: $x \geq 0$.
From the condition we have
$$
x^{\frac{6}{10}}-x^{\frac{3}{10}}=56 \Leftrightarrow\left(x^{\frac{3}{10}}\right)^{2}-x^{\frac{3}{10}}-56=0
$$
Let $x^{\frac{3}{10}}=y \geq 0$. The equation in terms of $y$ becomes $y^{2}-y-56=0$, from which $y_{1}=-7$ or $y_{2}=8 ; y_{1}=... | 1024 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,716 |
6.058. $\sqrt[3]{\frac{5-x}{x+3}}+\sqrt[7]{\frac{x+3}{5-x}}=2$. | Solution.
Domain of definition: $\left\{\begin{array}{l}x \neq-3, \\ x \neq 5\end{array}\right.$
Let $\sqrt[7]{\frac{5-x}{x+3}}=z, z \neq 0$. The equation in terms of $z$ becomes $z+\frac{1}{z}=2 \Leftrightarrow z^{2}-2 z+1=0 \Leftrightarrow(z-1)^{2}=0 \Leftrightarrow z-1=0, z=1$.
Then $\sqrt[7]{\frac{5-x}{x+3}}=1 \... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,717 |
6.059. $\sqrt[5]{\frac{16 z}{z-1}}+\sqrt[5]{\frac{z-1}{16 z}}=2.5$. | Solution.
Domain of definition: $\left\{\begin{array}{l}z \neq 1, \\ z \neq 0 .\end{array}\right.$
Let $\sqrt[5]{\frac{16 z}{z-1}}=y, y \neq 0$. The equation in terms of $y$ becomes $y+\frac{1}{y}=2.5 \Leftrightarrow y^{2}-2.5 y+1=0$, from which $y_{1}=\frac{1}{2}, y_{2}=2$. Then either
$\sqrt[5]{\frac{16 z}{z-1}}=\... | z_{1}=-\frac{1}{511},z_{2}=2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,718 |
6.060. $\sqrt[3]{5 x+7}-\sqrt[3]{5 x-12}=1$.
Solve the equation $\sqrt[3]{5 x+7}-\sqrt[3]{5 x-12}=1$. | ## Solution.
Let's rewrite the equation as $\sqrt[3]{5 x+7}=\sqrt[3]{5 x-12}+1$ and cube both sides:
$$
\begin{aligned}
& 5 x+7=5 x-12+3(\sqrt[3]{5 x-12})^{2}+3 \sqrt[3]{5 x-12}+1 \Leftrightarrow \\
& \Leftrightarrow(\sqrt[3]{5 x-12})^{2}+\sqrt[3]{5 x-12}-6=0
\end{aligned}
$$
Let $\sqrt[3]{5 x-12}=t$. The equation i... | x_{1}=-3,x_{2}=4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,719 |
6.061. $2 \sqrt[3]{x}+5 \sqrt[6]{x}-18=0$. | ## Solution.
Domain of definition: $x \geq 0$.
Let $\sqrt[6]{x}=y \geq 0$. The equation in terms of $y$ becomes $2 y^{2}+5 y-18=0$, from which we find $y_{1}=-\frac{9}{2}, y_{2}=2 ; y_{1}=-\frac{9}{2}<0$ is not valid.
Then $\sqrt[6]{x}=2, x=2^{6}=64$.
Answer: $x=64$. | 64 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,720 |
6.062. $\sqrt{3 x^{2}+1}+\sqrt{x^{2}+3}=\sqrt{6 x^{2}+10}$. | ## Solution.
By squaring both sides of the equation, we have
$$
\begin{aligned}
& 3 x^{2}+1+2 \sqrt{\left(3 x^{2}+1\right)\left(x^{2}+3\right)}+x^{2}+3=6 x^{2}+10 \Leftrightarrow \\
& \Leftrightarrow \sqrt{\left(3 x^{2}+1\right)\left(x^{2}+3\right)}=x^{2}+3 \Rightarrow \\
& \Rightarrow\left(3 x^{2}+1\right)\left(x^{2... | x_{1,2}=\1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,721 |
6.063. $\frac{\sqrt{x}+\sqrt[3]{x}}{\sqrt{x}-\sqrt[3]{x}}=3$. | ## Solution.
Domain of definition: $0<x \neq 1$.
Rewrite the equation as
$$
\frac{\sqrt[6]{x^{3}}+\sqrt[6]{x^{2}}}{\sqrt[6]{x^{3}}-\sqrt[6]{x^{2}}}=3 \Leftrightarrow \frac{\sqrt[6]{x^{3}}(\sqrt[6]{x}+1)}{\sqrt[6]{x^{3}}(\sqrt[6]{x}-1)}=3 \Leftrightarrow \frac{\sqrt[6]{x}+1}{\sqrt[6]{x}-1}=3 \Leftrightarrow
$$
$$
\b... | 64 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,722 |
6.064. $\sqrt{x+2}+\sqrt{3 x+8}=\sqrt{2 x+6}$.
6.064. $\sqrt{x+2}+\sqrt{3 x+8}=\sqrt{2 x+6}$. | Solution.
Domain of definition: $\left\{\begin{array}{l}x+2 \geq 0, \\ 3 x+8 \geq 0, \\ 2 x+6 \geq 0\end{array} \Leftrightarrow x \geq-2\right.$.
Write the equation in the form $\sqrt{x+2}-\sqrt{2 x+6}=-\sqrt{3 x+8}$ and square both sides:
$$
\begin{aligned}
& x+2-2 \sqrt{(x+2)(2 x+6)}+2 x+6=3 x+8 \Leftrightarrow \\... | -2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,723 |
6.065. $\sqrt{2 x+5}+\sqrt{5 x+6}=\sqrt{12 x+25}$. | Solution.
Domain of definition: $\left\{\begin{array}{l}2 x+5 \geq 0, \\ 5 x+6 \geq 0, \\ 12 x+25 \geq 0\end{array} \Leftrightarrow x \geq-\frac{6}{5}\right.$.
By squaring both sides of the equation, we have
$$
\begin{aligned}
& 2 x+5+2 \sqrt{(2 x+5)(5 x+6)}+5 x+6=12 x+25 \Leftrightarrow \\
& \Leftrightarrow 2 \sqrt... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,724 |
6.066. $x^{2}-4 x-6=\sqrt{2 x^{2}-8 x+12}$. | Solution.
By squaring both sides of the equation, we have
\[
\left(x^{2}-4 x-6\right)^{2}=2 x^{2}-8 x+12 \Leftrightarrow\left(x^{2}-4 x-6\right)^{2}-2\left(x^{2}-4 x-6+12\right)=0
\]
Let \(x^{2}-4 x-6=y, y \geq 0\). The equation in terms of \(y\) becomes \(y^{2}-2 y-24=0\), from which \(y_{1}=-4, y_{2}=6 ; y_{1}=-4\... | x_{1}=-2,x_{2}=6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,725 |
6.067. $\left\{\begin{array}{l}(x+0.2)^{2}+(y+0.3)^{2}=1, \\ x+y=0.9 .\end{array}\right.$ | ## Solution.
Rewrite the system as
$$
\left\{\begin{array} { l }
{ ( x + 0.2 ) ^ { 2 } + ( y + 0.3 ) ^ { 2 } = 1 , } \\
{ x + 0.2 - 0.2 + y + 0.3 - 0.3 = 0.9 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
(x+0.2)^{2}+(y+0.3)^{2}=1 \\
(x+0.2)+(y+0.3)=1.4
\end{array}\right.\right.
$$
Let $\left\{\begin{array}{... | (0.4;0.5),(0.6;0.3) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,726 |
6.068. $\left\{\begin{array}{l}x^{3}+y^{3}=7 \\ x^{3} y^{3}=-8\end{array}\right.$ | ## Solution.
According to Vieta's theorem, only the following options are possible:
1) $\left\{\begin{array}{l}x^{3}=8, \\ y^{3}=-1,\end{array} \quad\left\{\begin{array}{l}x_{1}=2 \\ y_{1}=-1\end{array}\right.\right.$
or $\quad 2)\left\{\begin{array}{l}x^{3}=-1, \\ y^{3}=8,\end{array},\left\{\begin{array}{l}x_{2}=-1,... | (2,-1),(-1,2) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,727 |
6.069. $\left\{\begin{array}{l}x^{-1}+y^{-1}=5 \\ x^{-2}+y^{-2}=13\end{array}\right.$ | ## Solution.
Rewrite the system of equations as $\left\{\begin{array}{l}\frac{1}{x}+\frac{1}{y}=5, \\ \frac{1}{x^{2}}+\frac{1}{y^{2}}=13 .\end{array}\right.$
Domain of definition: $\left\{\begin{array}{l}x \neq 0, \\ y \neq 0 .\end{array}\right.$
By bringing to a common denominator, we get
$\left\{\begin{array}{l}x... | (\frac{1}{3};\frac{1}{2})(\frac{1}{2};\frac{1}{3}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,728 |
6.070. $\left\{\begin{array}{l}\frac{x}{y}+\frac{y}{x}=\frac{13}{6}, \\ x+y=5\end{array}\right.$ | Solution.
Domain of definition: $\left\{\begin{array}{l}x \neq 0, \\ y \neq 0 .\end{array}\right.$
Multiplying both sides of the first equation by $6 x y \neq 0$, we get
$$
\begin{aligned}
& \left\{\begin{array} { l }
{ 6 ( x ^ { 2 } + y ^ { 2 } ) = 1 3 x y , } \\
{ x + y = 5 }
\end{array} \Leftrightarrow \left\{\b... | (2;3),(3;2) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,729 |
6.071. $\left\{\begin{array}{l}x-y=1, \\ x^{3}-y^{3}=7 .\end{array}\right.$ | ## Solution.
Transform the second equation of the system
$\left\{\begin{array}{l}x-y=1, \\ (x-y)\left(x^{2}+x y+y^{2}\right)=7\end{array} \Leftrightarrow\left\{\begin{array}{l}x-y=1, \\ (x-y)^{2}+3 x y=7\end{array}\right.\right.$
$\left\{\begin{array}{l}x-y=1, \\ 1+3 x y=7\end{array} \Leftrightarrow\left\{\begin{arra... | (2;1)(-1;-2) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,730 |
6.072. $\left\{\begin{array}{l}\frac{1}{y-1}-\frac{1}{y+1}=\frac{1}{x}, \\ y^{2}-x-5=0 .\end{array}\right.$ | Solution.
Domain of definition: $\left\{\begin{array}{l}y \neq \pm 1, \\ x \neq 0 .\end{array}\right.$
Transform the first equation of the system
$$
\left\{\begin{array} { l }
{ x ( y + 1 ) - x ( y - 1 ) = y ^ { 2 } - 1 , } \\
{ y ^ { 2 } - x = 5 }
\end{array} \Leftrightarrow \left\{\begin{array} { l }
{ x y + x -... | (4,3),(4,-3) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,731 |
6.073. $\left\{\begin{array}{l}y^{2}-x y=-12, \\ x^{2}-x y=28\end{array}\right.$ | ## Solution.
From the condition, we have $\left\{\begin{array}{l}y(y-x)=-12, \\ x(x-y)=28 .\end{array}\right.$
Dividing the first equation by the second, we get
$$
\left\{\begin{array} { l }
{ \frac { y ( x - y ) } { x ( x - y ) } = \frac { 12 } { 28 } , } \\
{ x ( x - y ) = 28 }
\end{array} \Leftrightarrow \left\{... | (-7,-3),(7,3) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,732 |
6.074. $\left\{\begin{array}{l}x+y+\frac{x}{y}=9, \\ \frac{(x+y) x}{y}=20 .\end{array}\right.$ | ## Solution.
Domain of definition: $y \neq 0$.
Let $\left\{\begin{array}{l}x+y=u, \\ \frac{x}{y}=v .\end{array}\right.$ We have $\left\{\begin{array}{l}u+v=9, \\ u v=20 .\end{array}\right.$ By Vieta's theorem, the only possible options are:
1) $\left\{\begin{array}{l}u_{1}=4 \\ v_{1}=5\end{array}\right.$
or
2) $\lef... | (\frac{10}{3};\frac{2}{3})(4;1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,733 |
6.075. $\left\{\begin{array}{l}x^{2} y+x y^{2}=6, \\ x y+x+y=5 .\end{array}\right.$ | ## Solution.
From the condition we have $\left\{\begin{array}{l}x y(x+y)=6, \\ x y+(x+y)=5 .\end{array}\right.$
Let $\left\{\begin{array}{l}x y=u, \\ x+y=v .\end{array}\right.$
For $u$ and $v$, we get the system $\left\{\begin{array}{l}u v=6, \\ u+v=5\end{array}\right.$, from which, by Vieta's theorem, we find $u_{1... | (1;2),(2;1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,734 |
6.076. $\left\{\begin{array}{l}x^{2} y^{3}+x^{3} y^{2}=12 \\ x^{2} y^{3}-x^{3} y^{2}=4\end{array}\right.$
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
6.076. $\left\{\begin{array}{l}x^{2} y^{3}+x^{3} y^{2}=12 \\ x^{2} y^{3... | Solution.
From the condition we have $\left\{\begin{array}{l}x^{2} y^{2}(y+x)=12, \\ x^{2} y^{2}(y-x)=4 .\end{array}\right.$
Dividing the first equation by the second, we get
$\frac{x^{2} y^{2}(y+x)}{x^{2} y^{2}(y-x)}=\frac{12}{4} \Leftrightarrow \frac{y+x}{y-x}=3 \Leftrightarrow y+x=3 y-3 x \Leftrightarrow y=2 x$.
... | (1;2) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,735 |
6.077. $\left\{\begin{array}{l}x^{4}+y^{4}=82, \\ x y=3\end{array}\right.$ | ## Solution.
Rewrite the system as
$$
\left\{\begin{array} { l }
{ ( ( x + y ) ^ { 2 } - 2 x y ) ^ { 2 } - 2 x ^ { 2 } y ^ { 2 } = 82, } \\
{ x y = 3 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
\left((x+y)^{2}-6\right)^{2}-18=82, \\
x y=3
\end{array} \Leftrightarrow\right.\right.
$$
$\Leftrightarrow\left\... | (1,3),(3,1),(-1,-3),(-3,-1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,736 |
6.078. $\left\{\begin{array}{l}x^{3}+y^{3}=35 \\ x+y=5\end{array}\right.$ | ## Solution.
By the formula for the sum of cubes, we get
$$
\begin{aligned}
& \left\{\begin{array} { l }
{ ( x + y ) ( x ^ { 2 } - x y + y ^ { 2 } ) = 3 5 , } \\
{ x + y = 5 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
(x+y) \\
x+y=5
\end{array}\left((x+y)^{2}-3 x y\right)=35,\right.\right. \\
& \Leftrighta... | (2,3)(3,2) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,737 |
6.079. $\left\{\begin{array}{l}x^{3}+y^{3}=9, \\ x y=2 .\end{array}\right.$ | ## Solution.
Rewrite the system as
$$
\left\{\begin{array} { l }
{ x ^ { 3 } + \frac { 8 } { x ^ { 3 } } = 9 , } \\
{ y = \frac { 2 } { x } }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x^{6}-9 x^{3}+8=0, \text { where } x \neq 0 \\
y=\frac{2}{x}
\end{array}\right.\right.
$$
From the first equation, we get ... | (1;2)(2;1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,738 |
6.080. $\left\{\begin{array}{l}u^{2}+u v=15, \\ v^{2}+u v=10\end{array}\right.$ | ## Solution.
Rewrite the system as $\left\{\begin{array}{l}u(u+v)=15, \\ v(u+v)=10\end{array}\right.$ and divide the first equation by the second:
$$
\frac{u(u+v)}{v(u+v)}=\frac{15}{10} \Leftrightarrow \frac{u}{v}=\frac{3}{2} \Leftrightarrow u=\frac{3 v}{2}
$$
Substituting $u=\frac{3 v}{2}$ into the second equation ... | (-3,-2),(3,2) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,739 |
6.081. $\left\{\begin{array}{l}x^{3}+y^{3}=65 \\ x^{2} y+x y^{2}=20\end{array}\right.$ | ## Solution.
By expanding the left sides, we can represent the system as
$$
\left\{\begin{array} { l }
{ ( x + y ) ( x ^ { 2 } - x y + y ^ { 2 } ) = 6 5 , } \\
{ x y ( x + y ) = 2 0 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
(x+y)\left((x+y)^{2}-3 x y\right)=65 \\
x y(x+y)=20
\end{array}\right.\right.
$$
... | (4;1)(1;4) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,740 |
6.082. $\left\{\begin{array}{l}x^{2}+y^{4}=5 \\ x y^{2}=2\end{array}\right.$ | ## Solution.
From the second equation of the system $x=\frac{2}{y^{2}}$. Then from the first equation we get $\left(\frac{2}{y^{2}}\right)^{2}+y^{4}=5, y^{8}-5 y^{4}+4=0$. Hence, $y_{1}^{4}=1$ or $y_{2}^{4}=4$, from which $y_{1}=-1, y_{2}=1, y_{3}=-\sqrt{2}, y_{4}=\sqrt{2}$. Then $x_{1,2}=\frac{2}{1}=2$;
$$
x_{3,4}=\... | (2;1)(2;-1)(1;\sqrt{2})(1;-\sqrt{2}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,741 |
6.083. $\left\{\begin{array}{l}12(x+y)^{2}+x=2.5-y, \\ 6(x-y)^{2}+x=0.125+y .\end{array}\right.$ | Solution.
Rewrite the system as $\left\{\begin{array}{l}12(x+y)^{2}+(x+y)-2.5=0, \\ 6(x-y)^{2}+(x-y)-0.125=0 .\end{array}\right.$
Then the first equation will be quadratic in terms of $x+y$, and the second in terms of $x-y$. Solving the specified equations, we get
$$
\begin{gathered}
(x+y)_{1,2}=\frac{-1 \pm \sqrt{1... | (\frac{1}{4};\frac{1}{6}),(\frac{1}{12};\frac{1}{3}),(-\frac{5}{24};-\frac{7}{24}),(-\frac{3}{8};-\frac{1}{8}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,742 |
6.084. $\left\{\begin{array}{l}\frac{2}{x}+\frac{y}{3}=3 \\ \frac{x}{2}+\frac{3}{y}=\frac{3}{2}\end{array}\right.$ | ## Solution.
Domain of definition: $\left\{\begin{array}{l}x \neq 0 \\ y \neq 0\end{array}\right.$
By bringing to a common denominator, we get
$$
\left\{\begin{array}{l}
6+x y=9 x, \\
x y+6=3 y
\end{array} \Rightarrow 9 x-3 y=0\right.
$$
from which $y=3 x$. From this, we get $\frac{2}{x}+\frac{3 x}{3}=3, x^{2}-3 x+... | (1;3)(2;6) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,743 |
6.086. $\left\{\begin{array}{l}(x-y)\left(x^{2}-y^{2}\right)=45, \\ x+y=5\end{array}\right.$
Solve the system of equations:
\[
\left\{\begin{array}{l}
(x-y)\left(x^{2}-y^{2}\right)=45, \\
x+y=5
\end{array}\right.
\] | Solution.
Rewrite this system in the form
$\left\{\begin{array}{l}(x-y)(x-y)(x+y)=45, \\ x+y=5\end{array} \Leftrightarrow\left\{\begin{array}{l}(x-y)^{2}(x+y)=45, \\ x+y=5 .\end{array} \Rightarrow\right.\right.$
$\Rightarrow(x-y)^{2}=9$, from which $x-y=-3$ or $x-y=3$. We obtain a combination of two systems:
1) $\l... | (4,1),(1,4) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,744 |
6.087. $\left\{\begin{array}{l}x^{4}-y^{4}=15 \\ x^{3} y-x y^{3}=6 .\end{array}\right.$ | ## Solution.
Let $t=\frac{y}{x}$, then $y=t x$ and the system becomes
$\left\{\begin{array}{l}x^{4}-t^{4} x^{4}=15, \\ x^{4} t-x^{4} t^{3}=6\end{array} \Leftrightarrow\left\{\begin{array}{l}x^{4}\left(1-t^{4}\right)=15, \\ x^{4}\left(t-t^{3}\right)=6 .\end{array}\right.\right.$
After dividing, we get $\frac{1-t^{4}}... | (-2,-1),(2,1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,745 |
6.089. $\left\{\begin{array}{l}u^{3}+v^{3}+1=m, \\ u^{3} v^{3}=-m .\end{array}\right.$ | ## Solution.
From the first equation of the system $v^{3}=m-1-u^{3}$. Substituting this value of $v^{3}$ into the second equation, we get
$$
u^{3}\left(m-1-u^{3}\right)=-m \Leftrightarrow\left(u^{3}\right)^{2}-(m-1) u^{3}-m=0,
$$
from which
$$
\begin{aligned}
& u_{1,2}^{3}=\frac{m-1 \pm \sqrt{(m-1)^{2}+4 m}}{2}=\fr... | (-1;\sqrt[3]{})(\sqrt[3]{};-1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,746 |
6.090. $\left\{\begin{array}{l}a x+\frac{b}{y}=2, \\ \frac{b}{x}+a y=2 a b .\end{array}\right.$ | ## Solution.
Domain of definition: $\left\{\begin{array}{l}x \neq 0 \\ y \neq 0\end{array}\right.$
Rewrite the system as $\left\{\begin{array}{l}a x y+b=2 y, \\ b+a x y=2 a b x\end{array} \Rightarrow 2 y=2 a b x, y=a b x\right.$.
From the first equation of the system, we get $a^{2} x^{2}-2 a x+1=0$ (for $b \neq 0$) ... | \frac{1}{},b | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,747 |
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