7.138. $\left\{\begin{array}{l}x^{2 y^{2}-1}=5 \\ x^{y^{2}+2}=125 .\end{array}\right.$
## Solution.

Domain of definition: $\quad 0<x \neq 1$.

By logarithmizing the first and second equations of the system with base 5, we get

$$
\left\{\begin{array} { l } 
{ \operatorname { l o g } _ { 5 } x ^ { 2 y ^ { 2 } - 1 } = \operatorname { l o g } _ { 5 } 5 , } \\
{ \operatorname { l o g } _ { 5 } x ^ { y ^ { 2 } + 2 } = \operatorname { l o g } _ { 5 } 1 2 5 }
\end{array} \Rightarrow \left\{\begin{array}{l}
\left(2 y^{2}-1\right) \log _{5} x=1, \\
\left(y^{2}+2\right) \log _{5} x=3
\end{array} \Rightarrow \log _{5} x=\frac{1}{2 y^{2}-1}\right.\right.
$$

From the second equation of the system, we have $\frac{y^{2}+2}{2 y^{2}-1}=3, y^{2}=1$, from which $y= \pm 1$. Then $\log _{5} x=1$, i.e., $x=5$.

Answer: $(5 ; 1)(5 ;-1)$.