problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
value | source stringclasses 8
values | synthetic bool 1
class | __index_level_0__ int64 0 742k |
|---|---|---|---|---|---|---|---|---|---|
6.182. $|x|^{3}+|x-1|^{3}=9$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly.
6.182. $|x|^{3}+|x-1|^{3}=9$. | ## Solution.
As in № 6.179, we have three cases
1) $\left\{\begin{array}{l}x<0, \\ -x^{3}-(x-1)^{3}=9\end{array} \Leftrightarrow\left\{\begin{array}{l}x<0, \\ 2 x^{3}-3 x^{2}+3 x+8=0 .\end{array}\right.\right.$
By checking, we find that $x_{1}=-1$, since $-2-3-3+8=0$. We divide the left side of the equation by $x+1:... | x_{1}=-1,x_{2}=2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,253 |
6.183. $\left\{\begin{array}{l}x y(x+1)(y+1)=72, \\ (x-1)(y-1)=2 .\end{array}\right.$ | ## Solution.
From the condition $\left\{\begin{array}{l}x y(x y+x+y+1)=72, \\ x y-(x+y)=1 .\end{array}\right.$
Let $\left\{\begin{array}{l}x y=u, \\ x+y=v .\end{array}\right.$ Relative to $u$ and $v$, the system takes the form
$\left\{\begin{array}{l}u(u+v+1)=72, \\ u-v=1\end{array} \Rightarrow v=u-1\right.$. Substi... | (2;3),(3;2),(\frac{-7+\sqrt{73}}{2};-\frac{7+\sqrt{73}}{2}),(-\frac{7+\sqrt{73}}{2};\frac{-7+\sqrt{73}}{2}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,254 |
6.184. $\left\{\begin{array}{l}2 x^{2}-3 x y+y^{2}=3 \\ x^{2}+2 x y-2 y^{2}=6\end{array}\right.$ | ## Solution.
Since the left-hand sides of the equations are homogeneous with respect to $x$ and $y$, let's introduce the substitution $y = t x$; the system of equations will take the form
$$
\left\{\begin{array}{l}
x^{2}\left(2-3 t+t^{2}\right)=3 \\
x^{2}\left(1+2 t-2 t^{2}\right)=6
\end{array}\right.
$$
Dividing th... | (2;1),(-2;-1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,255 |
6.185. $\left\{\begin{array}{l}x^{2}+2 y^{2}=17 \\ x^{2}-2 x y=-3\end{array}\right.$
The system of equations is:
\[
\left\{\begin{array}{l}
x^{2}+2 y^{2}=17 \\
x^{2}-2 x y=-3
\end{array}\right.
\] | ## Solution.
Since the left sides of the system are homogeneous, we make the substitution $y = t x$. Then the system will take the form $\left\{\begin{array}{l}x^{2}\left(1+2 t^{2}\right)=17, \\ x^{2}(1-2 t)=-3 .\end{array}\right.$
By dividing the left and right parts of the equations $(x \neq 0)$, we get
$$
\frac{1... | (3;2),(-3;-2),(\frac{\sqrt{3}}{3};\frac{5\sqrt{3}}{3}),(-\frac{\sqrt{3}}{3};\frac{-5\sqrt{3}}{3}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,256 |
6.186. $\left\{\begin{array}{l}a x+b y+c z=k, \\ a^{2} x+b^{2} y+c^{2} z=k^{2}, \\ a^{3} x+b^{3} y+c^{3} z=k^{3}, \quad a \neq b, b \neq c, c \neq a .\end{array}\right.$ | ## Solution.
We obtain the solution of the system using Cramer's rule. Let's compute the determinants of the system
$$
\Delta=\left|\begin{array}{ccc}
a & b & c \\
a^{2} & b^{2} & c^{2} \\
a^{3} & b^{3} & c^{3}
\end{array}\right|=a b^{2} c^{3}+a^{3} b c^{2}+a^{2} b^{3} c-a^{3} b^{2} c-a^{2} b c^{3}-a b^{3} c^{2}
$$
... | \frac{k(k-)(k-b)}{(-)(-b)};\frac{k(k-)(k-)}{b(b-)(b-)};\frac{k(k-)(k-b)}{(-)(-b)} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,257 |
6.187. $\left\{\begin{array}{l}(x+1)(y+1)=10 \\ (x+y)(xy+1)=25\end{array}\right.$ | ## Solution.
Rewrite the system of equations as $\left\{\begin{array}{l}x y+x+y+1=10, \\ (x+y)(x y+1)=25\end{array}\right.$ and introduce
the substitution $\left\{\begin{array}{l}x+y=u, \\ x y=v .\end{array}\right.$
With respect to $u$ and $v$, the system becomes $\left\{\begin{array}{l}v+u=9, \\ u(v+1)=25 .\end{arr... | (4;1),(1;4) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,258 |
6.188. $\left\{\begin{array}{l}x-a y+a^{2} z=a^{3}, \\ x-b y+b^{2} z=b^{3}, \\ x-c y+c^{2} z=c^{3}, \quad a \neq b, b \neq c, c \neq a .\end{array}\right.$ | ## Solution.
Subtracting the second equation of the system from the first, we have
$$
\begin{aligned}
& -(a-b) y+(a-b)(a+b) z=(a-b)\left(a^{2}+a b+b^{2}\right) \Leftrightarrow \\
& \Leftrightarrow-y+(a+b) z=a^{2}+a b+b^{2}
\end{aligned}
$$
Similarly, subtracting the third equation from the first, we get
$$
\begin{a... | ,++,+b+ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,259 |
6.189. $\left\{\begin{array}{l}(x-y)\left(x^{2}+y^{2}\right)=5, \\ (x+y)\left(x^{2}-y^{2}\right)=9 .\end{array}\right.$ | ## Solution.
Let's rewrite the system of equations as
$$
\left\{\begin{array}{l}
x^{3} + x y^{2} - x^{2} y - y^{3} = 5 \\
x^{3} - x y^{2} + x^{2} y - y^{3} = 9
\end{array}\right.
$$
By adding and subtracting the first and second equations, we get
$\left\{\begin{array}{l}2 x^{3} - 2 y^{3} = 14, \\ 2 x y^{2} - 2 x^{2... | (2;1),(-1;-2) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,260 |
6.190. $\left\{\begin{array}{l}x y=a, \\ y z=b, \quad a b c>0 . \\ z x=c,\end{array}\right.$ | Solution.
From the third equation, we find $z=\frac{c}{x}$ and substitute it into the second equation $\left\{\begin{array}{l}x y=a \\ y \cdot \frac{c}{x}=b, \\ z=\frac{c}{x} .\end{array}\right.$
From the second equation, we find $y=\frac{b}{c} x$ and substitute it into the first equation:
$$
\left\{\begin{array} { ... | x_{1,2}=\\sqrt{\frac{}{b}};y_{1,2}=\\sqrt{\frac{}{}};z_{1,2}=\\sqrt{\frac{}{}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,261 |
6.191. $\left\{\begin{array}{l}x^{2}+y=y^{2}+x \\ y^{2}+x=6\end{array}\right.$
The system of equations is:
\[
\left\{\begin{array}{l}
x^{2} + y = y^{2} + x \\
y^{2} + x = 6
\end{array}\right.
\] | ## Solution.
From the condition we have
$$
\begin{aligned}
& \left\{\begin{array} { l }
{ x ^ { 2 } - y ^ { 2 } - x + y = 0 , } \\
{ y ^ { 2 } + x = 6 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
(x-y)(x+y)-(x-y)=0, \\
y^{2}+x=6
\end{array}\right.\right. \\
& \Leftrightarrow\left\{\begin{array}{l}
(x-y)(x+y... | (2;2),(-3;-3),(\frac{1+\sqrt{21}}{2};\frac{1-\sqrt{21}}{2}),(\frac{1-\sqrt{21}}{2};\frac{1+\sqrt{21}}{2}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,262 |
6.192. $\left\{\begin{array}{l}\frac{4}{x+y}+\frac{4}{x-y}=3, \\ (x+y)^{2}+(x-y)^{2}=20 .\end{array}\right.$ | ## Solution.
Domain of definition: $x \neq \pm y$.
Let $\left\{\begin{array}{l}x+y=u, \\ x-y=v,\end{array}\right.$, where $u \neq 0$ and $v \neq 0$. Relative to $u$ and $v$, the system takes the form
$$
\left\{\begin{array} { l }
{ \frac { 4 } { u } + \frac { 4 } { v } = 3 , } \\
{ u ^ { 2 } + v ^ { 2 } = 2 0 }
\en... | (-\frac{5}{3};-\frac{\sqrt{65}}{3}),(-\frac{5}{3};\frac{\sqrt{65}}{3}),(3;-1),(3;1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,263 |
6.193. $\left\{\begin{array}{l}x+y z=2, \\ y+z x=2, \\ z+x y=2 .\end{array}\right.$ | ## Solution.
From the third equation of the system, we find $z=2-x y$ and substitute it into the first and second equations:
$$
\begin{aligned}
& \left\{\begin{array} { l }
{ x + y ( 2 - x y ) = 2 , } \\
{ y + ( 2 - x y ) x = 2 , } \\
{ z = 2 - x y }
\end{array} \Leftrightarrow \left\{\begin{array} { l }
{ x + 2 y ... | (1;1;1),(-2;-2;-2) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,264 |
6.194. $\left\{\begin{array}{l}\frac{5}{x^{2}-x y}+\frac{4}{y^{2}-x y}=-\frac{1}{6}, \\ \frac{7}{x^{2}-x y}-\frac{3}{y^{2}-x y}=\frac{6}{5} .\end{array}\right.$ | ## Solution.
Domain of definition: $\left\{\begin{array}{l}x \neq 0, \\ y \neq 0, \\ x \neq y .\end{array}\right.$
Let $\frac{1}{x^{2}-x y}=u, \frac{1}{y^{2}-x y}=v$. Relative to $u$ and $v$, the system of equations
has the form: $\left\{\begin{array}{l}5 u+4 v=-\frac{1}{6}, \\ 7 u-3 v=\frac{6}{5} .\end{array} \Righ... | (-5,-3),(5,3) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,265 |
6.195. $\left\{\begin{array}{l}x^{2}+y^{2}-2 x+3 y-9=0 \\ 2 x^{2}+2 y^{2}+x-5 y-1=0\end{array}\right.$
The system of equations is:
\[
\left\{\begin{array}{l}
x^{2}+y^{2}-2 x+3 y-9=0 \\
2 x^{2}+2 y^{2}+x-5 y-1=0
\end{array}\right.
\] | Solution.
From the condition $\left\{\begin{array}{l}2 x^{2}+2 y^{2}-4 x+6 y-18=0, \\ 2 x^{2}+2 y^{2}+x-5 y-1=0 .\end{array}\right.$
Subtracting the second equation of the system from the first, we get:
$-5 x+11 y-17=0$.
From this, $y=\frac{5 x+17}{11}, x^{2}+\left(\frac{5 x+17}{11}\right)^{2}-2 x+3 \cdot \frac{5 x... | (-\frac{239}{146};\frac{117}{146}),(1;2) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,266 |
6.196. $\left\{\begin{array}{l}x^{2}+y^{2}=34 \\ x+y+x y=23\end{array}\right.$
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
6.196. $\left\{\begin{array}{l}x^{2}+y^{2}=34 \\ x+y+x y=23\end{array}\right.$ | ## Solution.
Rewrite the system of equations as $\left\{\begin{array}{l}(x+y)^{2}-2 x y=34, \\ (x+y)+x y=23\end{array}\right.$ and, by setting $\left\{\begin{array}{l}x+y=u, \\ x y=v,\end{array}\right.$, we get $\left\{\begin{array}{l}u^{2}-2 v=34, \\ u+v=23\end{array} \Rightarrow v=23-u\right.$,
$u^{2}-2(23-u)=34, u... | (3;5),(5;3) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,267 |
6.197. $\left\{\begin{array}{l}x^{2}+y^{4}=20, \\ x^{4}+y^{2}=20 .\end{array}\right.$ | Solution.
From the condition we have $x^{2}+y^{4}=x^{4}+y^{2}, y^{4}-x^{4}+x^{2}-y^{2}=0$,
$$
\left(y^{2}-x^{2}\right)\left(y^{2}+x^{2}\right)-\left(y^{2}-x^{2}\right)=0 \Leftrightarrow\left(y^{2}-x^{2}\right)\left(y^{2}+x^{2}-1\right)=0
$$
equivalent to two equations: $y^{2}-x^{2}=0$ or $y^{2}+x^{2}-1=0, y^{2}=x^{2... | (2;2),(-2;-2),(2;-2),(-2;2) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,268 |
6.198. $\left\{\begin{array}{l}x+y+\frac{1}{x-y}=\frac{a b+1}{b}, \\ x-y+\frac{1}{x+y}=\frac{a b+1}{a} .\end{array}\right.$ | ## Solution.
Domain of definition: $x \neq \pm y, a \neq 0, b \neq 0$.
Multiplying the first equation of the system by $x-y \neq 0$, and the second by $x+y \neq 0$, we get
$\left\{\begin{array}{l}x^{2}-y^{2}+1=\frac{a b+1}{b}(x-y), \\ x^{2}-y^{2}+1=\frac{a b+1}{a}(x+y) .\end{array} \Rightarrow 1=\frac{a}{b} \cdot \f... | x_{1}=\frac{+b}{2},y_{1}=\frac{-b}{2};x_{2}=\frac{+b}{2},y_{2}=\frac{-b}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,269 |
6.200. $\left\{\begin{array}{l}(x+y)^{2}+2 x=35-2 y, \\ (x-y)^{2}-2 y=3-2 x\end{array}\right.$ | ## Solution.
Let's rewrite the system of equations as
$$
\left\{\begin{array}{l}
x^{2}+2 x y+y^{2}+2 x=35-2 y \\
x^{2}-2 x y+y^{2}-2 y=3-2 x
\end{array}\right.
$$
Subtracting the second equation from the first, we get
$4 x y+2 x+2 y=32-2 y+2 x, x=\frac{8-y}{y}$.
Substituting this value of $x$ into the first equati... | (-5,-2),(3,2),(-3,-4),(1,4) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,271 |
6.201. $\left\{\begin{array}{l}\frac{4}{x+y-1}-\frac{5}{2 x-y+3}+\frac{5}{2}=0, \\ \frac{3}{x+y-1}+\frac{1}{2 x-y+3}+\frac{7}{5}=0 .\end{array}\right.$ | ## Solution.
Domain of definition: $\left\{\begin{array}{l}x+y \neq 1 \\ 2 x-y \neq-3\end{array}\right.$
Let $\left\{\begin{array}{l}\frac{1}{x+y-1}=u \\ \frac{1}{2 x-y+3}=v\end{array}\right.$
With respect to $u$ and $v$, the system takes the form
$\left\{\begin{array}{l}4 u-5 v=-\frac{5}{2}, \\ 3 u+v=-\frac{7}{5},... | (2,-3) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,272 |
6.202. $\left\{\begin{array}{l}\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=3, \\ \frac{1}{x y}+\frac{1}{y z}+\frac{1}{z x}=3, \\ \frac{1}{x y z}=1 .\end{array}\right.$ | ## Solution.
Domain of definition: $\left\{\begin{array}{l}x \neq 0, \\ y \neq 0, \\ z \neq 0 .\end{array}\right.$
Multiplying both sides of each equation in the system by $x y z \neq 0$, we get
$\left\{\begin{array}{l}y z+x z+x y=3 x y z, \\ z+x+y=3 x y z, \\ x y z=1\end{array} \Leftrightarrow\left\{\begin{array}{l... | (1;1;1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,273 |
6.203. $\left\{\begin{array}{l}x+y+z=0, \\ c x+a y+b z=0, \\ (x+b)^{2}+(y+c)^{2}+(z+a)^{2}=a^{2}+b^{2}+c^{2},\end{array} a \neq b \neq c\right.$.
6.203. $\left\{\begin{array}{l}x+y+z=0, \\ c x+a y+b z=0, \\ (x+b)^{2}+(y+c)^{2}+(z+a)^{2}=a^{2}+b^{2}+c^{2},\end{array} a \neq b \neq c\right.$.
The system of equations is... | ## Solution.
Rewrite the equation as
$$
\begin{aligned}
& \left\{\begin{array}{l}
x+y+z=0, \\
c x+a y+b z=0, \\
x^{2}+2 b x+b^{2}+y^{2}+2 c y+c^{2}+z^{2}+2 a z+a^{2}=a^{2}+b^{2}+c^{2}
\end{array} \Leftrightarrow\right. \\
& \Leftrightarrow\left\{\begin{array}{l}
x+y+z=0, \\
c x+a y+b z=0, \\
(x+y+z)^{2}-2 x y-2 x z-2... | (0;0;0),(-b--) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,274 |
6.204. $\left\{\begin{array}{l}x+y+\frac{x^{2}}{y^{2}}=7, \\ \frac{(x+y) x^{2}}{y^{2}}=12 .\end{array}\right.$ | ## Solution.
## Domain of definition: $y \neq 0$.
From the condition, we have
$\left\{\begin{array}{l}(x+y)+\left(\frac{x^{2}}{y^{2}}\right)=7, \\ (x+y) \cdot\left(\frac{x^{2}}{y^{2}}\right)=12 .\end{array}\right.$
Let $\left\{\begin{array}{l}x+y=u, \\ \frac{x^{2}}{y^{2}}=v .\end{array}\right.$
The system in terms... | (6,-3),(2,1),(6+2\sqrt{3},-2-2\sqrt{3}),(6-2\sqrt{3},-2+2\sqrt{3}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,275 |
6.205. $\left\{\begin{array}{l}\frac{3}{x^{2}+y^{2}-1}+\frac{2 y}{x}=1 \\ x^{2}+y^{2}+\frac{4 x}{y}=22\end{array}\right.$ | ## Solution.
Domain of definition: $\left\{\begin{array}{l}x \neq 0, \\ y \neq 0\end{array}\right.$
Let $\left\{\begin{array}{l}x^{2}+y^{2}=u, \\ \frac{y}{x}=v .\end{array}\right.$ Then, with respect to $u$ and $v$, the system of equations has the form
$$
\left\{\begin{array} { l }
{ \frac { 3 } { u - 1 } + 2 v = 1... | (\frac{-14\sqrt{106}}{53};\frac{-4\sqrt{106}}{53}),(\frac{14\sqrt{106}}{53};\frac{4\sqrt{106}}{53}),(3;1),(-3;-1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,276 |
6.206. $\left\{\begin{array}{l}x+y+x y=7 \\ x^{2}+y^{2}+x y=13 .\end{array}\right.$ | Solution.
From the condition, we get $\left\{\begin{array}{l}(x+y)+x y=7, \\ (x+y)^{2}-x y=13 .\end{array}\right.$
Let $\left\{\begin{array}{l}x+y=u, \\ x y=v,\end{array}\right.$ then we have $\left\{\begin{array}{l}u+v=7, \\ u^{2}-v=13 .\end{array}\right.$
By adding the equations of the system, we get $u^{2}-7+u=13... | (3;1),(1;3) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,277 |
6.207. $\left\{\begin{array}{l}x+y+z=6, \\ x(y+z)=5, \\ y(x+z)=8 .\end{array}\right.$ | ## Solution.
Let's rewrite the system of equations as
$\left\{\begin{array}{l}x+y+z=6, \\ x y+x z=5, \\ y x+y z=8 .\end{array} \Rightarrow z=6-x-y\right.$.
$\left\{\begin{array}{l}x y+x(6-x-y)=5, \\ y x+y(6-x-y)=8\end{array} \Leftrightarrow\left\{\begin{array}{l}x^{2}-6 x+5=0, \\ y^{2}-6 y+8=0,\end{array} \quad\righ... | (5;2;-1),(5;4;-3),(1;2;3),(1;4;1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,278 |
6.208. $\left\{\begin{array}{l}(x-y)\left(x^{2}-y^{2}\right)=3 a^{3}, \\ (x+y)\left(x^{2}+y^{2}\right)=15 a^{3} .\end{array}\right.$
The system of equations is:
\[
\left\{\begin{array}{l}
(x-y)\left(x^{2}-y^{2}\right)=3 a^{3}, \\
(x+y)\left(x^{2}+y^{2}\right)=15 a^{3} .
\end{array}\right.
\] | Solution.
Rewrite the system of equations as
$$
\left\{\begin{array}{l}
x^{3}-x y^{2}-x^{2} y+y^{3}=3 a^{3} \\
x^{2}+x y^{2}+x^{2} y+y^{3}=15 a^{3}
\end{array}\right.
$$
Add and subtract the second and first equations:
$$
\left\{\begin{array} { l }
{ 2 x ^ { 3 } + 2 y ^ { 3 } = 1 8 a ^ { 3 } } \\
{ 2 x y ^ { 2 } +... | (2),(2) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,279 |
6.209. $\left\{\begin{array}{l}x^{3}+y^{3}=19 \\ x^{2} y+x y^{2}=-6\end{array}\right.$ | ## Solution.
From the condition we have $\left\{\begin{array}{l}(x+y)\left((x+y)^{2}-3 x y\right)=19, \\ x y(x+y)=-6 .\end{array}\right.$
Let $\left\{\begin{array}{l}x+y=u, \\ x y=v .\end{array}\right.$ Relative to $u$ and $v$, the system takes the form
$$
\left\{\begin{array}{l}
u\left(u^{2}-3 v\right)=19, \\
u v=-... | (-2,3),(3,-2) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,280 |
6.210. $\left\{\begin{array}{l}x^{4}+y^{4}=17 \\ x^{2}+y^{2}=5\end{array}\right.$ | ## Solution.
Let's rewrite the system of equations as follows:
$$
\begin{aligned}
& \left\{\begin{array} { l }
{ ( x ^ { 2 } + y ^ { 2 } ) ^ { 2 } - 2 x ^ { 2 } y ^ { 2 } = 1 7 , } \\
{ x ^ { 2 } + y ^ { 2 } = 5 }
\end{array} \Leftrightarrow \left\{\begin{array} { l }
{ 2 5 - 2 x ^ { 2 } y ^ { 2 } = 1 7 , } \\
{ x ... | (2;1),(1;2),(-2;1),(1;-2),(2;-1),(-1;2),(-2;-1),(-1;-2) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,281 |
6.211. $\left\{\begin{array}{l}x y-\frac{x}{y}=\frac{16}{3}, \\ x y-\frac{y}{x}=\frac{9}{2} .\end{array}\right.$ | ## Solution.
Domain of definition: $\left\{\begin{array}{l}x \neq 0, \\ y \neq 0 .\end{array}\right.$
Subtracting the second equation of the system from the first, we get
$$
-\frac{x}{y}+\frac{y}{x}=\frac{5}{6}
$$
Let $\frac{x}{y}=t$. Then we have $6 t^{2}+5 t-6=0$, from which $t_{1}=-\frac{3}{2}, t_{2}=\frac{2}{3}... | (2;3),(-2;-3) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,282 |
6.212. $\left\{\begin{array}{l}\frac{3}{uv}+\frac{15}{vw}=2, \\ \frac{15}{vw}+\frac{5}{wu}=2, \\ \frac{5}{wu}+\frac{3}{uv}=2 .\end{array}\right.$ | Solution.
Domain of definition: $\left\{\begin{array}{l}u \neq 0, \\ v \neq 0, \\ w \neq 0 .\end{array}\right.$
Let $\frac{3}{u v}=x, \frac{15}{v w}=y, \frac{5}{w u}=z$.
With respect to $x, y$ and $z$, the system of equations becomes
$\left\{\begin{array}{l}x+y=2, \\ y+z=2, \\ z+x=2 .\end{array} \Rightarrow z=2-x\r... | (-1,-3,-5),(1,3,5) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,283 |
6.213. $\left\{\begin{array}{l}x^{6}+y^{6}=65, \\ x^{4}-x^{2} y^{2}+y^{4}=13 .\end{array}\right.$ | ## Solution.
By the formula for the sum of cubes, we get
$$
\begin{aligned}
& \left\{\begin{array} { l }
{ ( x ^ { 2 } + y ^ { 2 } ) ( x ^ { 4 } - x ^ { 2 } y ^ { 2 } + y ^ { 4 } ) = 6 5 , } \\
{ x ^ { 4 } - x ^ { 2 } y ^ { 2 } + y ^ { 4 } = 1 3 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
\left(x^{2}+y^{2}... | (2;1),\quad(-2;-1),\quad(2;-1),\quad(-2;1),\quad(1;2),\quad(-1;-2),\quad(1;-2),\quad(-1;2) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,284 |
6.214. $\left\{\begin{array}{l}x+y+z=0, \\ 2 x+3 y+z=0, \\ (x+1)^{2}+(y+2)^{2}+(z+3)^{2}=14 .\end{array}\right.$ | ## Solution.
From the first equation of the system, we find $z=-x-y$ and substitute it into the second and third equations:
$$
\begin{aligned}
& \left\{\begin{array}{l}
2 x+3 y-x-y=0 \\
(x+1)^{2}+(y+2)^{2}+(-x-y+3)^{2}=14
\end{array} \Leftrightarrow\right. \\
& \Leftrightarrow\left\{\begin{array}{l}
x+2 y=0 \\
(x+1)^... | (0;0;0),(2;-1;-1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,285 |
6.215. $\left\{\begin{array}{l}x^{3}+3 x y^{2}=158 \\ 3 x^{2} y+y^{3}=-185 .\end{array}\right.$ | Solution.
Add and subtract the first and second equations:
$$
\begin{aligned}
& \left\{\begin{array} { l }
{ x ^ { 3 } + 3 x y ^ { 2 } + 3 x ^ { 2 } y + y ^ { 3 } = - 2 7 , } \\
{ x ^ { 3 } + 3 x y ^ { 2 } - 3 x ^ { 2 } y - y ^ { 3 } = 3 4 3 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
(x+y)^{3}=-27, \\
(x-... | (2,-5) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,286 |
6.216. $\left\{\begin{array}{l}x^{2}+y-20=0 \\ x+y^{2}-20=0 .\end{array}\right.$
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
6.216. $\left\{\begin{array}{l}x^{2}+y-20=0 \\ x+y^{2}-20=0 .\end{array}\right.$ | ## Solution.
From the condition $\left\{\begin{array}{l}x^{2}+y=20, \\ x+y^{2}=20,\end{array}\right.$ we get $x^{2}+y=x+y^{2}, x^{2}-y^{2}-x+y=0$, $(x-y)(x+y)-(x-y)=0,(x-y)(x+y-1)=0$.
Therefore, the given system is equivalent to two systems of equations:
$$
\left\{\begin{array}{l}
x-y=0 \\
x+y^{2}-20=0
\end{array}\r... | (-5;-5),(4;4),(\frac{1+\sqrt{77}}{2};\frac{1-\sqrt{77}}{2}),(\frac{1-\sqrt{77}}{2};\frac{1+\sqrt{77}}{2}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,287 |
6.217. $\left\{\begin{array}{l}x^{4}+x^{2} y^{2}+y^{4}=91 \\ x^{2}+x y+y^{2}=13\end{array}\right.$ | ## Solution.
Let's rewrite the system of equations as follows:
$$
\begin{aligned}
& \left\{\begin{array} { l }
{ ( x ^ { 4 } + y ^ { 4 } ) + x ^ { 2 } y ^ { 2 } = 9 1 , } \\
{ x ^ { 2 } + x y + y ^ { 2 } = 1 3 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
\left(x^{2}+y^{2}\right)^{2}-x^{2} y^{2}=91, \\
x^{2}... | (1;3),(3;1),(-1;-3),(-3;-1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,288 |
6.218. $\left\{\begin{array}{l}x^{3}+y^{3}=9 a^{3}, \\ x^{2} y+x y^{2}=6 a^{3} .\end{array}\right.$ | ## Solution.
Let's rewrite the system of equations as
$$
\left\{\begin{array}{l}
(x+y)\left((x+y)^{2}-3 x y\right)=9 a^{3} \\
x y(x+y)=6 a^{3}
\end{array}\right.
$$
Let $\left\{\begin{array}{l}x+y=u, \\ x y=v .\end{array}\right.$ Then, in terms of $u$ and $v$, the system becomes
$$
\left\{\begin{array}{l}
u\left(u^... | (2),(2) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,289 |
6.219. $\left\{\begin{array}{l}x+y+z=3, \\ x+2 y-z=2, \\ x+y z+z x=3 .\end{array}\right.$ | ## Solution.
From the first equation of the system, we find $z=3-x-y$. Then we get
$$
\begin{aligned}
& \left\{\begin{array}{l}
x+2 y-3+x+y=2 \\
x+y(3-x-y)+(3-x-y) x=3, \\
x+y z+z x=3
\end{array}\right. \\
& x+\frac{5-2 x}{3}\left(3-x-\frac{5-2 x}{3}\right)+\left(3-x-\frac{5-2 x}{3}\right.
\end{aligned}
$$
from whic... | (1;1;1),(7;-3;-1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,290 |
6.220. $\left\{\begin{array}{l}\frac{x^{3}}{y}+x y=40 \\ \frac{y^{3}}{x}+x y=10 .\end{array}\right.$ | ## Solution.
Domain of definition: $\left\{\begin{array}{l}x \neq 0, \\ y \neq 0\end{array}\right.$
Multiplying the first equation of the system by $y \neq 0$, and the second by $x \neq 0$, we get $\left\{\begin{array}{l}x^{3}+x y^{2}=40 y, \\ y^{3}+x^{2} y=10 x\end{array} \Leftrightarrow\left\{\begin{array}{l}x\left... | (4;2),(-4;-2) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,291 |
6.221. $\left\{\begin{array}{l}x+y+z=2, \\ 2 x+3 y+z=1, \\ x^{2}+(y+2)^{2}+(z-1)^{2}=9 .\end{array}\right.$ | ## Solution.
From the first equation of the system, we find $z=2-x-y$.
$$
\left\{\begin{array} { l }
{ 2 x + 3 y + 2 - x - y = 1 , } \\
{ x ^ { 2 } + ( y + 2 ) ^ { 2 } + ( 2 - x - y - 1 ) ^ { 2 } = 9 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x+2 y=-1, \\
x^{2}+(y+2)^{2}+(1-x-y)^{2}=9
\end{array} \Rightar... | (-1;0;3),(3;-2;1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,292 |
6.222.
$$
\left\{\begin{array}{l}
\sqrt{\frac{x+1}{x+y}}+\sqrt{\frac{x+y}{x+1}}=2 \\
\sqrt{\frac{x+1}{y+2}}-\sqrt{\frac{y+2}{x+1}}=1.5
\end{array}\right.
$$ | Solution.
Domain of definition: $\left\{\begin{array}{l}\frac{x+1}{x+y}>0, \\ \frac{x+1}{y+2}>0 .\end{array}\right.$
Rewrite the system of equations as
$\left\{\begin{array}{l}\sqrt{\frac{x+1}{x+y}}+\frac{1}{\sqrt{\frac{x+1}{x+y}}}-2=0, \\ \sqrt{\frac{x+1}{y+2}}-\frac{1}{\sqrt{\frac{x+1}{y+2}}}-1.5=0 .\end{array}\ri... | (11;1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,293 |
6.223. $\left\{\begin{array}{l}x^{2}+2 y+\sqrt{x^{2}+2 y+1}=1, \\ 2 x+y=2 .\end{array}\right.$ | ## Solution.
Homework: $x^{2}+2 y+1 \geq 0$.
Let $\sqrt{x^{2}+2 y+1}=t$, where $t \geq 0$; then $x^{2}+2 y=t^{2}-1 \Leftrightarrow t^{2}-1+t=1$, or $t^{2}+t-2=0$, from which $t_{1}=-2, t_{2}=1 ; t_{1}=-2<0$ is not suitable. Then $\sqrt{x^{2}+2 y+1}=1, x^{2}+2 y+1=1, x^{2}+2 y=0$. We have $\left\{\begin{array}{l}x^{2}... | (2,-2) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,294 |
6.224. $\left\{\begin{array}{l}\sqrt{x+y}+\sqrt{y+z}=3, \\ \sqrt{y+z}+\sqrt{z+x}=5, \\ \sqrt{z+x}+\sqrt{x+y}=4 .\end{array}\right.$ | ## Solution.
Domain of definition: $\left\{\begin{array}{l}x+y \geq 0, \\ y+z \geq 0, \\ z+x \geq 0 .\end{array}\right.$
By adding all three equations of the system, we get
$$
2 \sqrt{x+y}+2 \sqrt{y+z}+2 \sqrt{z+x}=12 \Leftrightarrow \sqrt{x+y}+\sqrt{y+z}+\sqrt{z+x}=6
$$
By subtracting the first, second, and third ... | (3,-2,6) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,295 |
6.225. $\left\{\begin{array}{l}\sqrt[4]{x}+\sqrt[4]{y}=3 \\ x+y=17\end{array}\right.$
The system of equations is:
$\left\{\begin{array}{l}\sqrt[4]{x}+\sqrt[4]{y}=3 \\ x+y=17\end{array}\right.$ | Solution.
Let $\left\{\begin{array}{l}\sqrt[4]{x}=u \geq 0, \\ \sqrt[4]{y}=v \geq 0 .\end{array}\right.$ With respect to $u$ and $v$, the system takes the form
$\left\{\begin{array}{l}u+v=3, \\ u^{4}+v^{4}=17\end{array} \Leftrightarrow\left\{\begin{array}{l}u+v=3, \\ \left((u+v)^{2}-2 u v\right)^{2}-2 u^{2} v^{2}=17\... | (1;16),(16;1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,296 |
6.226. $\left\{\begin{array}{l}\sqrt{x+\frac{1}{y}}+\sqrt{y+\frac{1}{x}}=2 \sqrt{2}, \\ \left(x^{2}+1\right) y+\left(y^{2}+1\right) x=4 x y\end{array}\right.$
The system of equations is:
\[
\left\{\begin{array}{l}
\sqrt{x+\frac{1}{y}}+\sqrt{y+\frac{1}{x}}=2 \sqrt{2}, \\
\left(x^{2}+1\right) y+\left(y^{2}+1\right) x=4 ... | ## Solution.
Domain of definition: $\left\{\begin{array}{l}x>0, \\ y>0 .\end{array}\right.$
## From the condition we have
$$
\left\{\begin{array} { l }
{ \sqrt { \frac { x y + 1 } { y } } + \sqrt { \frac { x y + 1 } { x } } = 2 \sqrt { 2 } , } \\
{ x ^ { 2 } y + y + y ^ { 2 } x + x = 4 x y }
\end{array} \Leftrighta... | (1;1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,297 |
6.227. $\left\{\begin{array}{l}\sqrt[3]{u+v}+\sqrt[3]{v+w}=3 \\ \sqrt[3]{v+w}+\sqrt[3]{w+u}=1 \\ \sqrt[3]{w+u}+\sqrt[3]{u+v}=0\end{array}\right.$ | ## Solution.
$$
\text { Let }\left\{\begin{array}{l}
\sqrt[3]{u+v}=x \\
\sqrt[3]{v+w}=y \\
\sqrt[3]{w+u}=z
\end{array}\right.
$$
With respect to $x, y, z$, the system takes the form $\left\{\begin{array}{l}x+y=3, \\ y+z=1, \\ z+x=0\end{array} \Leftrightarrow\left\{\begin{array}{l}x+y=3, \\ y-x=1, \\ z=-x .\end{array}... | (-4;5;3) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,298 |
6.228. $\left\{\begin{array}{l}x \sqrt{y}+y \sqrt{x}=30 \\ x \sqrt{x}+y \sqrt{y}=35\end{array}\right.$ | Solution.
Let $\left\{\begin{array}{l}\sqrt{x}=u \geq 0, \\ \sqrt{y}=v \geq 0 .\end{array}\right.$
With respect to $u$ and $v$, the system of equations becomes
$\left\{\begin{array}{l}u^{2} v+v^{2} u=30, \\ u^{3}+v^{3}=35\end{array} \Leftrightarrow\left\{\begin{array}{l}u v(u+v)=30, \\ \left.(u+v)(u+v)^{2}-3 u v\rig... | (4,9),(9,4) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,299 |
6.229. $\left\{\begin{array}{l}x+\sqrt{y}-56=0 \\ \sqrt{x}+y-56=0\end{array}\right.$ | Solution.
Domain of definition: $\left\{\begin{array}{l}x \geq 0, \\ y \geq 0 .\end{array}\right.$
Subtracting the second equation of the system from the first, we get
$$
\begin{aligned}
& x+\sqrt{y}-\sqrt{x}-y=0, \quad(x-y)-(\sqrt{x}-\sqrt{y})=0 \\
& (\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y})-(\sqrt{x}-\sqrt{y})=0 \Lef... | (49;49) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,300 |
6.230. $\left\{\begin{array}{l}\sqrt[3]{x+2 y}+\sqrt[3]{x-y+2}=3, \\ 2 x+y=7\end{array}\right.$ | Solution.
Raising the first equation of the system to the third power, we get
$$
\begin{gathered}
x+2 y+3 \sqrt[3]{(x+2 y)^{2}(x-y+2)}+3 \sqrt[3]{(x+2 y)(x-y+2)^{2}}+x-y+ \\
+2=27 \Leftrightarrow 2 x+7+3 \sqrt[3]{(x+2 y)(x-y+2)}(\sqrt[3]{x+2 y}+3 \sqrt[3]{x-y+2})=25
\end{gathered}
$$
Using the equations of the syste... | (2;3),(\frac{13}{3};-\frac{5}{3}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,301 |
6.231. $\left\{\begin{array}{l}\sqrt{x+y}+\sqrt{2 x+y+2}=7, \\ 3 x+2 y=23 .\end{array}\right.$ | Solution.
Domain of definition: $\left\{\begin{array}{l}x+y \geq 0, \\ 2 x+y+2 \geq 0 .\end{array}\right.$
Squaring the first equation of the system, we get
$$
\begin{aligned}
& x+y+2 \sqrt{(x+y)(2 x+y+2)}+2 x+y+2=49 \Leftrightarrow \\
& \Leftrightarrow 3 x+2 y+2 \sqrt{(x+y)(2 x+y+2)}=47 \Leftrightarrow \\
& \Leftri... | (-9;25),(5;4) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,302 |
6.232.
$$
\left\{\begin{array}{l}
\sqrt{\frac{20 y}{x}}=\sqrt{x+y}+\sqrt{x-y} \\
\sqrt{\frac{16 x}{5 y}}=\sqrt{x+y}-\sqrt{x-y}
\end{array}\right.
$$ | ## Solution.
Domain of definition: $\left\{\begin{array}{l}x+y \geq 0, \\ x-y \geq 0, \\ x>0, \\ y>0 .\end{array}\right.$
Dividing the first equation by the second, we get
$$
\sqrt{\frac{20 y}{x} \cdot \frac{5 y}{16 x}}=\frac{\sqrt{x+y}+\sqrt{x-y}}{\sqrt{x+y}-\sqrt{x-y}} \Leftrightarrow \sqrt{\frac{25 y^{2}}{4 x^{2}... | (5;4) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,303 |
6.233. $\left\{\begin{array}{l}\sqrt{2 x+y+1}-\sqrt{x+y}=1, \\ 2 x+2 y=4\end{array}\right.$ | ## Solution.
Domain of definition: $\left\{\begin{array}{l}2 x+y+1 \geq 0 \\ x+y \geq 0\end{array}\right.$
Squaring the first equation, we have
$$
\begin{aligned}
& 2 x+y+1-2 \sqrt{(2 x+y+1)(x+y)}+x+y=1 \Leftrightarrow \\
& \Leftrightarrow 3 x+2 y-2 \sqrt{(2 x+y+1)(x+y)}=0
\end{aligned}
$$
Since $3 x+2 y=4$, from t... | (2,-1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,304 |
6.234. $\left\{\begin{array}{l}u^{-\frac{1}{2}} \cdot \sqrt[3]{u}+v^{-\frac{1}{2}} \cdot \sqrt[3]{v}=1.5 \\ u v=64\end{array}\right.$ | ## Solution.
Rewrite the system of equations as $\left\{\begin{array}{l}\frac{1}{\sqrt[6]{u}}+\frac{1}{\sqrt[6]{v}}=1.5 . \\ u v=64 .\end{array}\right.$
Let $\left\{\begin{array}{l}\sqrt[6]{u}=x>0, \\ \sqrt[6]{v}=y>0 .\end{array}\right.$ The system in terms of $x$ and $y$ becomes
$$
\left\{\begin{array}{l}
\frac{1}{... | (1;64),(64;1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,305 |
6.235. $\left\{\begin{array}{l}\sqrt[3]{\frac{y+1}{x}}-2 \sqrt[3]{\frac{x}{y+1}}=1, \\ \sqrt{x+y+1}+\sqrt{x-y+10}=5 .\end{array}\right.$ | Solution.
Domain of definition: $\left\{\begin{array}{l}x \neq 0, \\ y \neq-1, \\ x+y+1 \geq 0, \\ x-y+10 \geq 0 .\end{array}\right.$
Rewrite the first equation of the system as
$\sqrt[3]{\frac{y+1}{x}}-\frac{2}{\sqrt[3]{\frac{y+1}{x}}}-1=0 \Rightarrow\left(\sqrt[3]{\frac{y+1}{x}}\right)^{2}-\sqrt[3]{\frac{y+1}{x}}-... | (7,-8),(\frac{49}{64},\frac{41}{8}),(1,7) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,306 |
6.236. $\left\{\begin{array}{l}\sqrt{x^{2}+y^{2}}+\sqrt{x^{2}-y^{2}}=6, \\ x y^{2}=6 \sqrt{10} .\end{array}\right.$ | Solution.
Domain of definition: $\left\{\begin{array}{l}x>0, \\ x^{2}-y^{2} \geq 0 .\end{array}\right.$
We square the first equation of the system, then
$$
\begin{aligned}
& x^{2}+y^{2}+2 \sqrt{x^{4}-y^{4}}+x^{2}-y^{2}=36 \Leftrightarrow \sqrt{x^{4}-y^{4}}=18-x^{2} \Rightarrow \\
\Rightarrow & x^{4}-y^{4}=324-36 x^{... | (\sqrt{10};-\sqrt{6}),(\sqrt{10};\sqrt{6}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,307 |
6.237. $\left\{\begin{array}{l}\sqrt{x}+\sqrt{y}=3 \\ \sqrt{x+5}+\sqrt{y+3}=5\end{array}\right.$ | ## Solution.
Domain of definition: $\left\{\begin{array}{l}x \geq 0, \\ y \geq 0 .\end{array}\right.$
Rewrite the system of equations as $\left\{\begin{array}{l}\sqrt{x}=3-\sqrt{y} \\ \sqrt{x+5}=5-\sqrt{y+3}\end{array} \Rightarrow\right.$
$$
\begin{aligned}
& \Rightarrow\left\{\begin{array} { l }
{ x = 9 - 6 \sqrt ... | (4;1),(\frac{121}{64};\frac{169}{64}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,308 |
6.238. $\left\{\begin{array}{l}\sqrt{\left(\frac{x^{2}-y^{2}}{x^{2}+y^{2}}-1\right)^{2}}=1.6, \\ x y=2 .\end{array}\right.$ | ## Solution.
Domain of definition: $\left\{\begin{array}{l}x \neq 0, \\ y \neq 0 .\end{array}\right.$
We will transform the first equation of the system
$$
\sqrt{\left(\frac{x^{2}-y^{2}-x^{2}-y^{2}}{x^{2}+y^{2}}\right)^{2}}=1.6 \Leftrightarrow \sqrt{\left(\frac{-2 y^{2}}{x^{2}+y^{2}}\right)^{2}}=1.6 \Rightarrow \fra... | (-1,-2),(1,2) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,309 |
6.240. $\left\{\begin{array}{l}\sqrt{\frac{x}{y}+2+\frac{y}{x}}=\frac{5}{2}, \\ |x+y|=5 .\end{array}\right.$ | Solution.
Domain of definition: $\left\{\begin{array}{l}x \neq 0, \\ y \neq 0 .\end{array}\right.$
Transform the first equation of the system
$\sqrt{\frac{x^{2}+2 x y+y^{2}}{x y}}=\frac{5}{2} \Leftrightarrow \frac{\sqrt{(x+y)^{2}}}{\sqrt{x y}}=\frac{5}{2} \Leftrightarrow \frac{|x+y|}{\sqrt{x y}}=\frac{5}{2}$.
Thus,... | (-4,-1),(-1,-4),(1,4),(4,1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,310 |
6.241. $\left\{\begin{array}{l}u-v+\sqrt{\frac{u-v}{u+v}}=\frac{12}{u+v} \\ u^{2}+v^{2}=41 .\end{array}\right.$ | ## Solution.
Domain of definition: $\left\{\begin{array}{l}\frac{u-v}{u+v} \geq 0, \\ u+v \neq 0 .\end{array}\right.$
This system is equivalent to two systems of equations:
1) $\left\{\begin{array}{l}u-v \geq 0, \\ u+v>0, \\ u-v+\sqrt{\frac{u-v}{u+v}}=\frac{12}{u+v}, \\ u^{2}+v^{2}=41\end{array}\right.$
2) $\left\{\... | (5,-4),(5,4),(-\sqrt{28.5},-\sqrt{12.5}),(-\sqrt{28.5},\sqrt{12.5}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,311 |
6.242. $\left\{\begin{array}{l}\sqrt{\frac{3 x-2 y}{2 x}}+\sqrt{\frac{2 x}{3 x-2 y}}=2, \\ x^{2}-18=2 y(4 y-9) .\end{array}\right.$ | ## Solution.
Domain of definition: $\frac{3 x-2 y}{2 x}>0$.
Transform the first equation of the system
$$
\sqrt{\frac{3 x-2 y}{2 x}}+\frac{1}{\sqrt{\frac{3 x-2 y}{2 x}}}-2=0 \Leftrightarrow\left(\sqrt{\frac{3 x-2 y}{2 x}}\right)^{2}-2 \sqrt{\frac{3 x-2 y}{2 x}}+1=0
$$
where $\sqrt{\frac{3 x-2 y}{2 x}} \neq 0,\left(... | (3;\frac{3}{2}),(6;3) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,312 |
6.243. $\left\{\begin{array}{l}5 \sqrt{x^{2}-3 y-88}+\sqrt{x+6 y}=19 \\ 3 \sqrt{x^{2}-3 y-88}=1+2 \sqrt{x+6 y} .\end{array}\right.$ | ## Solution.
Let $\left\{\begin{array}{l}\sqrt{x^{2}-3 y-88}=u \geq 0, \\ \sqrt{x+6 y}=v \geq 0 .\end{array}\right.$
With respect to $u$ and $v$, the system of equations has the form
$$
\left\{\begin{array}{l}
5 u+v=19, \\
3 u=1+2 v
\end{array} \Rightarrow u=\frac{1+2 v}{3}, 5\left(\frac{1+2 v}{3}\right)+v=19 \Right... | (10;1),(-\frac{21}{2};\frac{53}{12}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,313 |
### 6.244. Solve the equation
$$
\begin{aligned}
& x(x+1)+(x+1)(x+2)+(x+2)(x+3)+(x+3)(x+4)+\ldots \\
& \ldots+(x+8)(x+9)+(x+9)(x+10)=1 \cdot 2+2 \cdot 3+\ldots+8 \cdot 9+9 \cdot 10
\end{aligned}
$$ | ## Solution.
Let's rewrite the equation as
$$
\begin{aligned}
& \left(x^{2}+x\right)+\left(x^{2}+3 x+2\right)+\left(x^{2}+5 x+6\right)+\left(x^{2}+7 x+12\right)+\ldots \\
& \ldots+\left(x^{2}+17 x+72\right)+\left(x^{2}+19 x+90\right)=2+6+12+\ldots+72+90 \Leftrightarrow \\
& \Leftrightarrow\left(x^{2}+x\right)+\left(x... | x_{1}=0,x_{2}=-10 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,314 |
6.245. Find the coefficients $m$ and $n$ of the quadratic trinomial $x^{2}+m x+n$, given that the remainders when it is divided by the binomials $x-m$ and $x-n$ are respectively $m$ and $\boldsymbol{n}$. | Solution.
Dividing $x^{2}+m x+n$ by $x-m$, we have the solution
$$
\begin{aligned}
& \begin{array}{c||c}
x^{2}+m x+n & x-m \\
-x^{2}-m x & \frac{x+2 m}{}
\end{array} \\
& \begin{array}{l}
2 m x+n \\
\frac{2 m x+2 m^{2}}{2 m^{2}}+n
\end{array}
\end{aligned}
$$
Thus, $2 m^{2}+n=m$.
Dividing $x^{2}+m x+n$ by $x-n$, we... | (0;0),(\frac{1}{2};0),(1;-1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,315 |
6.246. The quadratic equation $a x^{2}+b x+c=0$ has two roots. Form a new quadratic equation, one of whose roots is one unit less than the larger root, and the other is one unit more than the smaller root of the given equation. | ## Solution.
The smaller root of the quadratic equation \(a x^{2}+b x+c=0\).
Then \(y_{1}=\frac{-b+\sqrt{b^{2}-4 a c}}{2 a}-1, y_{2}=\frac{-b-\sqrt{b^{2}-4 a c}}{2 a}+1\) are the roots o... | ^{2}++(-+\sqrt{b^{2}-4})=0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,316 |
6.247. Determine the values of $m$ for which one of the roots of the equation $z^{3}-\left(m^{2}-m+7\right) z-\left(3 m^{2}-3 m-6\right)=0$ is -1. Find the other two roots of the equation for these values of $m$. | Solution.
Let $z_{1}=-1$, then $(-1)^{3}-\left(m^{2}-m+7\right)(-1)-\left(3 m^{2}-3 m-6\right)=0$, $m^{2}-m-6=0$, from which $m_{1}=-2, m_{2}=3$. For $m_{1}=-2$ and $m_{2}=3$, the equation becomes $z^{3}-13 z-12=0$. Dividing the left side of the equation by $z+1$: $\frac{z^{3}-13 z-12}{z+1}=z^{2}-z-12, z^{2}-z-12=0$, ... | m_{1}=-2,m_{2}=3;z_{1}=-1,z_{2}=-3,z_{3}=4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,317 |
6.248. Show that if the coefficients $a, b$ and $c$ of the equation $a x^{2}+b x+c=0$ are related by the condition $2 b^{2}-9 a c=0$, then the ratio of the roots of the equation is 2. | Solution.
$$
\text { Let } \frac{x_{1}}{x_{2}}=2 \text {. }
$$
From Vieta's formulas, we get:
$$
\left\{\begin{array}{l}
x_{1}+x_{2}=-\frac{b}{a}, \\
x_{1} \cdot x_{2}=\frac{c}{a}, \\
\frac{x_{1}}{x_{2}}=2
\end{array} \Rightarrow x_{1}=2 x_{2},\left\{\begin{array} { l }
{ 2 x _ { 2 } + x _ { 2 } = - \frac { b } { a... | proof | Algebra | proof | Yes | Yes | olympiads | false | 48,318 |
6.249. Show that if $a$ and $b$ are the roots of the equation $x^{2}+p x+1=0$, and $b$ and $c$ are the roots of the equation $x^{2}+q x+2=0$, then $(b-a)(b-c)=p q-6$. | Solution.
By Vieta's theorem, we have $\left\{\begin{array}{l}a+b=-p, \\ a b=1, \\ b+c=-q, \\ b c=2 .\end{array}\right.$
From this, multiplying the first equation of the system by the third, $(a+b)(b+c)=p q, b^{2}+a b+b c+a c=p q$, $b^{2}+a b+b c+a c-2 a b-2 b c=p q-2 a b-2 b c$, $b^{2}-a b-b c+a c=p q-2 a b-2 b c$, ... | proof | Algebra | proof | Yes | Yes | olympiads | false | 48,319 |
6.250. For what values of $a$ do the equations $x^{2}+a x+1=0$ and $x^{2}+x+a=0$ have a common root? | Solution.
If for some $a$ and $x$ the left parts of the equations are equal to 0, then they are equal to each other:
$$
\left\{\begin{array}{l}
x^{2}+a x+1=0 \\
x^{2}+x+a=0
\end{array}\right.
$$
Subtracting the second equation of the system from the first, we get
$a x-x+1-a=0,(a-1) x-(a-1)=0,(a-1)(x-1)=0$, from whi... | -2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,320 |
6.251. For what positive $p$ are the roots of the equation
$5 x^{2}-4(p+3) x+4=p^{2}$
of opposite signs? Find these roots. | Solution.
Consider the quadratic equation $5 x^{2}-4(p+3) x+4-p^{2}=0$.
The roots of the quadratic equation are of opposite signs when the following system of inequalities is satisfied:
1) $\left\{\begin{array}{l}16(p+3)^{2}-4 \cdot 5\left(4-p^{2}\right)>0, \\ p>0, \\ (p+3)>0, \\ 4-p^{2}<0 ;\end{array}\right.$
2) $\... | p>2;x_{1}=\frac{-p+2}{5},x_{2}=p+2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,321 |
6.252. Find the coefficients of the equation $x^{2}+p x+q=0$ given that the difference between the roots of the equation is 5, and the difference between their cubes is 35. | ## Solution.
From the condition we have
$$
\begin{aligned}
& \left\{\begin{array}{l}
x_{1}-x_{2}=5, \\
x_{1}^{3}-x_{2}^{3}=35
\end{array} \Leftrightarrow\left\{\begin{array}{l}
x_{1}-x_{2}=5, \\
\left(x_{1}-x_{2}\right.
\end{array}\right)\left(\left(x_{1}-x_{2}\right)^{2}+3 x_{1} x_{2}\right)=35\right. \\
& \Leftrigh... | p_{1}=1,q_{1}=-6;p_{2}=-1,q_{2}=-6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,322 |
6.253. Form a quadratic equation with roots $(a+b)^{2}$ and $(a-b)^{2}$, if $a$ and $b$ are the roots of the equation $x^{2}+p x+q=0$. | ## Solution.
Let's form the quadratic equation $y^{2}+P y+Q=0$.
By Vieta's theorem and the condition, we have
$$
\begin{aligned}
& \left\{\begin{array}{l}
P=-\left((a-b)^{2}+(a+b)^{2}\right)=-\left(a^{2}-2 a b+b^{2}+a^{2}+2 a b+b^{2}\right), \\
Q=(a-b)^{2} \cdot(a+b)^{2}
\end{array} \Leftrightarrow\right. \\
& \Left... | y^{2}-2(p^{2}-2q)y+(p^{2}-4q)p^{2}=0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,323 |
6.254. Let $\alpha$ and $\beta$ be the roots of the equation $3 x^{2}+7 x+4=0$. Without solving this equation, form a new quadratic equation with numerical coefficients, whose roots are $\frac{\alpha}{\beta-1}$ and $\frac{\beta}{\alpha-1}$. | Solution.
By Vieta's theorem, we have $\left\{\begin{array}{l}\alpha+\beta=-\frac{7}{3} \\ \alpha \cdot \beta=\frac{4}{3} .\end{array}\right.$
Let $y^{2}+p x+q=0$ be the new quadratic equation. From the condition by Vieta's theorem, we get
$$
\begin{aligned}
& \left\{\begin{array}{l}
p=-\left(\frac{\alpha}{\beta-1}+... | 21y^{2}-23y+6=0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,324 |
7.150. $\left(b^{\frac{\log _{100} a}{\lg a}} \cdot a^{\frac{\log _{100} b}{\lg b}}\right)^{2 \log _{a b}(a+b)}$. | Solution.
$$
\begin{aligned}
& \left(b^{\frac{\log _{100} a}{\lg a}} \cdot a^{\frac{\log _{100} b}{\lg b}}\right)^{2 \log _{a b}(a+b)}=\left(b^{\frac{1 \lg a}{2 \lg b}} \cdot a^{\frac{1 \lg ^{2} b}{2 \lg b}}\right)^{2 \log _{a b}(a+b)}= \\
& =\left((a b)^{\frac{1}{2}}\right)^{2 \log _{a b}(a+b)}=(a b)^{\log _{a b}(a+b... | +b | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,326 |
7.151. $\left(\left(\log _{b}^{4} a+\log _{a}^{4} b+2\right)^{1 / 2}+2\right)^{1 / 2}-\log _{b} a-\log _{a} b$. | ## Solution.
$$
\begin{aligned}
& \left(\left(\log _{b}^{4} a+\log _{a}^{4} b+2\right)^{1 / 2}+2\right)^{1 / 2}-\log _{b} a-\log _{a} b= \\
& \left(\left(\log _{b}^{4} a+\frac{1}{\log _{b}^{4} a}+2\right)^{1 / 2}+2\right)^{1 / 2}-\log _{b} a-\frac{1}{\log _{b} a}=\sqrt{\sqrt{\frac{\log _{b}^{8} a+2 \log _{b}^{4} a+1}{... | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,327 | |
7.154. $\frac{\log _{a} b-\log _{\sqrt{a} / b^{3}} \sqrt{b}}{\log _{a / b^{4}} b-\log _{a / b^{6}} b}: \log _{b}\left(a^{3} b^{-12}\right)$. | ## Solution.
$$
\begin{aligned}
& \frac{\log _{a} b-\log _{\sqrt{a} / b^{3}} \sqrt{b}}{\log _{a / b^{4}} b-\log _{a / b^{6} b}}: \log _{b}\left(a^{3} b^{-12}\right)=\frac{\log _{a} b-\frac{\log _{a} \sqrt{b}}{\log _{a} \frac{\sqrt{a}}{b^{3}}}}{\frac{\log _{a} b}{\log _{a} \frac{a}{b^{4}}}-\frac{\log _{a} b}{\log _{a} ... | \log_{}b | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,330 |
7.155. $\left(6\left(\log _{b} a \cdot \log _{a^{2}} b+1\right)+\log _{a} b^{-6}+\log _{a}^{2} b\right)^{1 / 2}-\log _{a} b$ for $a>1$. | Solution.
$$
\begin{aligned}
& \left(6\left(\log _{b} a \cdot \log _{a^{2}} b+1\right)+\log _{a} b^{-6}+\log _{a}^{2} b\right)^{1 / 2}-\log _{a} b= \\
& =\left(6\left(\frac{1}{2}+1\right)-6 \log _{a} b+\log _{a}^{2} b\right)^{1 / 2}-\log _{a} b=\sqrt{9-6 \log _{a} b+\log _{a}^{2} b}- \\
& -\log _{a} b=\sqrt{\left(3-\l... | -3ifb\geq^{3},3-2\log_{}bif0<b<^{3},b\neq1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,331 |
7.156. $\frac{\log _{a} b+\log _{a}\left(b^{1 / 2 \log _{b} a^{2}}\right)}{\log _{a} b-\log _{a b} b} \cdot \frac{\log _{a b} b \cdot \log _{a} b}{b^{2 \log _{b} \log _{a} b}-1}$. | Solution.
$$
\frac{\log _{a} b+\log _{a}\left(b^{1 / 2 \log _{b} a^{2}}\right)}{\log _{a} b-\log _{a b} b} \cdot \frac{\log _{a b} b \cdot \log _{a} b}{b^{2 \log _{b} \log _{a} b}-1}=\frac{\log _{a} b+\log _{a} a}{\log _{a} b-\frac{\log _{a} b}{1+\log _{a} b}} \times
$$
$$
\times \frac{\frac{\log _{a} b}{1+\log _{a} ... | \frac{1}{\log_{}b-1} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,332 |
7.157. It is known that $\log _{a} x=\alpha, \log _{b} x=p, \log _{c} x=\gamma, \log _{d} x=\delta$ and $x \neq 1$. Find $\log _{a b c d} x$. | Solution.
$$
\begin{aligned}
& \log _{a b c d} x=\frac{\log _{x} x}{\log _{x} a b c d}=\frac{1}{\log _{x} a+\log _{x} b+\log _{x} c+\log _{x} d}= \\
& =\frac{1}{\frac{1}{\log _{a} x}+\frac{1}{\log _{b} x}+\frac{1}{\log _{c} x}+\frac{1}{\log _{d} x}}= \\
& =\frac{1}{\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}+\fr... | \frac{\alpha\beta\gamma\delta}{\beta\gamma\delta+\alpha\gamma\delta+\alpha\beta\delta+\alpha\beta\gamma} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,333 |
7.158. It is known that $\beta=10^{\frac{1}{1-\lg \alpha}}$ and $\gamma=10^{\frac{1}{1-\lg \beta}}$. Find the dependence of $\alpha$ on $\gamma$. | ## Solution.
$\lg \beta=\frac{1}{1-\lg \alpha} ; \lg \gamma=\frac{1}{1-\lg \beta}=\frac{1}{1-\frac{1}{1-\lg \alpha}}=\frac{1-\lg \alpha}{-\lg \alpha}=-\frac{1}{\lg \alpha}+1$;
$\frac{1}{\lg \alpha}=1-\lg \gamma ; \lg \alpha=\frac{1}{1-\lg \gamma} ; \alpha=10^{1 /(1-\lg \gamma)}$.
Answer: $\alpha=10^{1 /(1-\lg \gamma... | \alpha=10^{1/(1-\lg\gamma)} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,334 |
7.159. Prove that $\log _{a b} c=\frac{\log _{a} c \cdot \log _{b} c}{\log _{a} c+\log _{b} c}$. | ## Solution.
$$
\begin{aligned}
& \log _{a b} c=\frac{\log _{a} c}{\log _{a} a b}=\frac{\log _{a} c}{1+\log _{a} b}=\frac{\log _{a} c \cdot \frac{\log _{a} c}{\log _{a} b}}{\left(1+\log _{a} b\right) \frac{\log _{a} c}{\log _{a} b}}= \\
& =\frac{\log _{a} c \cdot \log _{b} c}{\frac{\log _{a} c}{\log _{a} b}+\log _{a} ... | proof | Algebra | proof | Yes | Yes | olympiads | false | 48,335 |
7.160. Simplify the expression $\log _{a+b} m+\log _{a-b} m-2 \log _{a+b} m \cdot \log _{a-b} m$, given that $m^{2}=a^{2}-b^{2}$. | Solution.
$\log _{a+b} m+\log _{a-b} m-2 \log _{a+b} m \cdot \log _{a-b} m=\log _{a+b} m+\frac{\log _{a+b} m}{\log _{a+b}(a-b)}-$
$-2 \frac{\log _{a+b} m \cdot \log _{a+b} m}{\log _{a+b}(a-b)}=\log _{a+b} m \cdot\left(1+\frac{1}{\log _{a+b}(a-b)}-\frac{2 \log _{a+b} m}{\log _{a+b}(a-b)}\right)=$
$=\frac{\log _{a+b} ... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,336 |
7.161. Find $\log _{30} 8$, given that $\lg 5=a, \lg 3=b$. | Solution.
$$
\begin{aligned}
& \log _{30} 8=\frac{\log _{2} 8}{\log _{2} 30}=\frac{3}{\log _{2}(2 \cdot 5 \cdot 3)}=\frac{3}{1+\log _{2} 5+\log _{2} 3} . \\
& \lg 5=\frac{\log _{2} 5}{\log _{2} 10}=\frac{\log _{2} 5}{\log _{2}(2 \cdot 5)}=\frac{\log _{2} 5}{1+\log _{2} 5}=a ; \log _{2} 5=\frac{a}{1-a} . \\
& \lg 3=\fr... | \frac{3(1-)}{1+b} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,337 |
7.162. Prove that $\frac{\log _{a} x}{\log _{a b} x}=1+\log _{a} b$. | Solution.
$\frac{\log _{a} x}{\log _{a b} x}=\frac{\log _{a} x}{\frac{\log _{a} x}{\log _{a} a b}}=\frac{\log _{a} x \cdot \log _{a} a b}{\log _{a} x}=\log _{a} a b=\log _{a} a+\log _{a} b=1+\log _{a} b$.
This is what we needed to prove. | 1+\log_{}b | Algebra | proof | Yes | Yes | olympiads | false | 48,338 |
7.163. Knowing that $\lg 2=a$ and $\log _{2} 7=b$, find $\lg 56$. | ## Solution.
$\lg 56=\lg (7 \cdot 8)=\lg 7+\lg 8=\lg 7+3 \lg 2=\frac{\log _{2} 7}{\log _{2} 10}+3 \lg 2=\log _{2} 7 \cdot \lg 2+3 \lg 2=$
$=a b+3 a=a(b+3)$.
Answer: $a(b+3)$. | (b+3) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,339 |
7.165. $3 \cdot 4^{x}+\frac{1}{3} \cdot 9^{x+2}=6 \cdot 4^{x+1}-\frac{1}{2} \cdot 9^{x+1}$. | Solution.
From the condition we have
$3 \cdot 4^{x}+\frac{1}{3} \cdot 81 \cdot 9^{x}=6 \cdot 4 \cdot 4^{x}-\frac{1}{2} \cdot 9 \cdot 9^{x} \Rightarrow 3 \cdot 9^{x}=2 \cdot 4^{x} \Rightarrow\left(\frac{9}{4}\right)^{x}=\frac{2}{3}$,
$\left(\frac{3}{2}\right)^{2 x}=\left(\frac{3}{2}\right)^{-1}$, from which $x=-\frac... | -\frac{1}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,340 |
7.166. $\sqrt{\log _{0.04} x+1}+\sqrt{\log _{0.2} x+3}=1$. | ## Solution.
Domain of definition: $\left\{\begin{array}{l}\frac{1}{2} \log _{0.2} x+1 \geq 0, \\ \log _{0.2}+3 \geq 0, \quad \Leftrightarrow 00\end{array} \quad\right.$
Switch to base 0.2. We have
$$
\begin{aligned}
& \sqrt{\frac{1}{2} \log _{0.2} x+1}+\sqrt{\log _{0.2} x+3}=1 \Leftrightarrow \\
& \Leftrightarrow \... | 25 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,341 |
7.167. $\sqrt{\log _{x} \sqrt{5 x}}=-\log _{x} 5$. | Solution.
Domain of definition: $\left\{\begin{array}{l}\log _{x} \sqrt{5 x} \geq 0, \\ -\log _{x} 5 \geq 0, \\ 0<x \neq 1\end{array}\right.$ or $0<x \leq \frac{1}{5}$.
By squaring both sides of the equation, we have or
$\log _{x} \sqrt{5 x}=\log _{x}^{2} 5 \Leftrightarrow 2 \log _{x}^{2} 5-\log _{x} 5-1=0 \Rightarr... | \frac{1}{25} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,342 |
7.168. $\log _{4 x+1} 7+\log _{9 x} 7=0$. | Solution.
Domain of definition: $\left\{\begin{array}{l}0<4 x+1 \neq 1, \\ 0<9 x \neq 1\end{array} \Leftrightarrow 0<x \neq \frac{1}{9}\right.$.
Switch to base 7. We have
$\frac{1}{\log _{7}(4 x+1)}+\frac{1}{\log _{7} 9 x}=0 \Rightarrow \log _{7} 9 x=-\log _{7}(4 x+1) \Leftrightarrow$
$\Leftrightarrow 9 x=\frac{1}{... | \frac{1}{12} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,343 |
7.169. $\left(16^{\sin x}\right)^{\cos x}+\frac{6}{4^{\sin ^{2}\left(x-\frac{\pi}{4}\right)}}-4=0$. | Solution.
Transform the denominator of the second term of the equation:
$4^{\sin ^{2}\left(x-\frac{\pi}{4}\right)}=4^{\left(\sin x \cos \frac{\pi}{4}-\cos x \sin \frac{\pi}{4}\right)^{2}}=4^{\left(\frac{\sqrt{2}}{2}\right)^{2}\left(\sin ^{2} x-2 \sin x \cos x+\cos ^{2} x\right)}=4^{\frac{1}{2}(1-\sin 2 x)}=$
$=4^{\fr... | \frac{\pin}{2};n\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,344 |
7.170. $\log _{2}(2-x)-\log _{2}(2-\sqrt{x})=\log _{2} \sqrt{2-x}-0.5$. | Solution.
Domain of Definition (DOD): $\left\{\begin{array}{l}2-x>0, \\ 2-\sqrt{x}>0,\end{array}, 0 \leq x<2\right.$.
From the condition, we have $\log _{2} \frac{2-x}{2-\sqrt{x}}=\log _{2} \frac{\sqrt{2-x}}{\sqrt{2}} \Leftrightarrow \frac{2-x}{2-\sqrt{x}}=\frac{\sqrt{2-x}}{\sqrt{2}} \Leftrightarrow$ $\Leftrightarrow... | 0;\frac{16}{9} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,345 |
7.171. $5^{1+\log _{4} x}+5^{\log _{0.25} x-1}=\frac{26}{5}$. | Solution.
Domain of definition: $x>0$.
Let's switch to base 4. We have $5 \cdot 5^{\log _{4} x}+\frac{1}{5 \cdot 5^{\log _{4} x}}-\frac{26}{5}=0 \Leftrightarrow$ $\Leftrightarrow 25 \cdot\left(5^{\log _{4} x}\right)^{2}-26 \cdot 5^{\log _{4} x}+1=0 \Rightarrow\left(5^{\log _{4} x}\right)_{1}=5^{-2},\left(5^{\log _{4}... | \frac{1}{16};1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,346 |
7.174. $3^{\log _{3}^{2} x}+x^{\log _{3} x}=162$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
7.174. $3^{\log _{3}^{2} x}+x^{\log _{3} x}=162$. | ## Solution.
Domain of definition: $0<x \neq 1$.
Rewrite the equation as
$$
\begin{aligned}
& \left(3^{\log _{3} x}\right)^{\log _{3} x}+x^{\log _{3} x}=162 \Leftrightarrow x^{\log _{3} x}+x^{\log _{3} x}=162 \Leftrightarrow \\
& \Leftrightarrow x^{\log _{3} x}=81 \Leftrightarrow \log _{3}^{2} x=4
\end{aligned}
$$
... | \frac{1}{9};9 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,349 |
7.175. $\lg \left(x^{3}+8\right)-0.5 \lg \left(x^{2}+4 x+4\right)=\lg 7$. | ## Solution.
Domain of definition: $x+2>0, x>-2$.
Rewrite the equation as
$$
\begin{aligned}
& \lg (x+2)\left(x^{2}-2 x+4\right)-0.5 \lg (x+2)^{2}=\lg 7 \Leftrightarrow \\
& \Leftrightarrow \lg (x+2)\left(x^{2}-2 x+4\right)-\lg (x+2)=\lg 7 \Leftrightarrow \lg \frac{(x+2)\left(x^{2}-2 x+4\right)}{x+2}=\lg 7 \Leftrigh... | -1;3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,350 |
7.176. $2^{\log _{5} x^{2}}-2^{1+\log _{5} x}+2^{\log _{5} x-1}-1=0$.
7.176. $2^{\log _{5} x^{2}}-2^{1+\log _{5} x}+2^{\log _{5} x-1}-1=0$. | Solution.
Domain of definition: $x>0$.
Rewrite the equation as $2^{2 \log _{5} x}-2 \cdot 2^{\log _{5} x}+\frac{2^{\log _{5} x}}{2}-1=0 \Leftrightarrow$ $\Leftrightarrow 2 \cdot 2^{2 \log _{5} x}-3 \cdot 2^{\log _{5} x}-2=0$. Solving this equation as a quadratic equation in terms of $2^{\log _{5} x}$, we find $2^{\lo... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,351 |
7.177. $\frac{\log _{2}\left(9-2^{x}\right)}{3-x}=1$.
Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly.
7.177. $\frac{\log _{2}\left(9-2^{x}\right)}{3-x}=1$. | ## Solution.
Domain of definition: $\left\{\begin{array}{l}9-2^{x}>0, \\ 3-x \neq 0,\end{array} \Leftrightarrow 3 \neq x<\log _{2} 9\right.$.
## From the condition
$$
\log _{2}\left(9-2^{x}\right)=3-x \Leftrightarrow 9-2^{x}=2^{3-x} \Leftrightarrow 2^{2 x}-9 \cdot 2^{x}+8=0
$$
Solving it as a quadratic equation in ... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,352 |
7.178. $\log _{5} x+\log _{25} x=\log _{1 / 5} \sqrt{3}$. | ## Solution.
Domain of definition: $x>0$.
Let's switch to base 5. We have $\log _{5} x+\frac{1}{2} \log _{5} x=-\frac{1}{2} \log _{5} 3 \Leftrightarrow$
$\Leftrightarrow 2 \log _{5} x+\log _{5} x=\log _{5} \frac{1}{3} \Leftrightarrow \log _{5} x^{3}=\log _{5} \frac{1}{3}$. From this, we have $x^{3}=\frac{1}{3}$, $x=\... | \frac{1}{\sqrt[3]{3}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,353 |
7.179. $\log _{a^{2}} x^{2}+\log _{a}(x-1)=\log _{a} \log _{\sqrt{5}} 5$. | Solution.
Domain of definition: $\left\{\begin{array}{l}x>1, \\ 0<a \neq 1 .\end{array}\right.$
From the condition we have
$$
\log _{a} x+\log _{a}(x-1)=\log _{a} 2 \Rightarrow \log _{a} x(x-1)=\log _{a} 2
$$
from which $x^{2}-x-2=0 \Rightarrow x_{1}=2, x_{2}=-1 ; x_{2}=-1$ does not satisfy the domain of definition... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,354 |
7.181. $\log _{x} 3+\log _{3} x=\log _{\sqrt{x}} 3+\log _{3} \sqrt{x}+0.5$. | Solution.
Domain of definition: $0<x \neq 1$.
Switch to base 3. We have
$\frac{1}{\log _{3} x}+\log _{3} x=\frac{2}{\log _{3} x}+\frac{1}{2} \log _{3} x+\frac{1}{2} \Leftrightarrow \log _{3}^{2} x-\log _{3} x-2=0 \Rightarrow$
$\Rightarrow\left(\log _{3} x\right)_{1}=-1$ or $\left(\log _{3} x\right)_{2}=2$, from whi... | \frac{1}{3};9 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,356 |
7.182. $\log _{\sqrt{x}} a \cdot \log _{a^{2}} \frac{a^{2}}{2 a-x}=1$. | ## Solution.
Domain of definition: $\left\{\begin{array}{l}0<a \neq 1 \\ x \neq 2 a \\ 0<x \neq 1\end{array}\right.$
Switch to base $a \cdot$ We have
$\frac{\log _{a} a}{\log _{a} \sqrt{x}} \cdot \frac{\log _{a} \frac{a^{2}}{2 a-x}}{\log _{a} a^{2}}=1 \Leftrightarrow \log _{a}(2 a-x)+\log _{a} x=2 \Leftrightarrow$
... | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,357 | |
7.183. $5^{-2 \log _{0.04}\left(3-4 x^{2}\right)}+1.5 \log _{1 / 8} 4^{x}=0$. | ## Solution.
Domain of definition: $3-4 x^{2}>0 \Leftrightarrow-\frac{\sqrt{3}}{2}<x<\frac{\sqrt{3}}{2}$.
From the condition
$$
5^{\log _{5}\left(3-4 x^{2}\right)}+1.5 x \log _{2^{-3}} 2^{2}=0 \Leftrightarrow 3-4 x^{2}-x=0 \Leftrightarrow 4 x^{2}+x-3=0
$$
from which $x_{1}=-1, x_{2}=\frac{3}{4} ; x_{1}=-1$ does not... | \frac{3}{4} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,358 |
7.184. $\log _{a} x+\log _{a^{2}} x+\log _{a^{3}} x=11$. | ## Solution.
Domain of definition: $\left\{\begin{array}{l}x>0, \\ 0<a \neq 1 .\end{array}\right.$
Switch to base $a$. We have $\log _{a} x+\frac{1}{2} \log _{a} x+\frac{1}{3} \log _{a} x=11 \Leftrightarrow$ $\Leftrightarrow \log _{a} x=6$, from which $x=a^{6}$.
Answer: $a^{6}$, where $0<a \neq 1$. | ^6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 48,359 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.