Question

GHPT: \(\left\{{}\begin{matrix}\sqrt{x+5}+\sqrt{y-2}=7\\\sqrt{y+5}+\sqrt{x-2}=7\end{matrix}\right.\)

Answer 1

ĐKXĐ; ...

Đặt \(\left\{{}\begin{matrix}\sqrt{x+5}=a>0\\\sqrt{y-2}=b\ge0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}a+b=7\\\sqrt{a^2-7}+\sqrt{b^2+7}=7\end{matrix}\right.\)

\(\Leftrightarrow\sqrt{a^2-7}+\sqrt{\left(7-a\right)^2+7}=7\)

\(\Leftrightarrow\sqrt{a^2-14a+56}=7-\sqrt{a^2-7}\) (\(a\le\sqrt{56}\))

\(\Leftrightarrow a^2-14a+56=42+a^2-14\sqrt{a^2-7}\)

\(\Leftrightarrow\sqrt{a^2-7}=a-1\) 

\(\Leftrightarrow a^2-7=a^2-2a+1\Leftrightarrow a=4\Rightarrow b=3\)

\(\Rightarrow x;y\)