Question

\(\left\{{}\begin{matrix}\sqrt{x+2}-2\sqrt{y-1}=-4\\3\sqrt{x+2}+y=16\end{matrix}\right.\)

Answer 1

ĐKXĐ: ...

Đặt \(\left\{{}\begin{matrix}\sqrt{x+2}=a\ge0\\\sqrt{y-1}=b\ge0\end{matrix}\right.\) ta được hệ:

\(\left\{{}\begin{matrix}a-2b=-4\\3a+b^2+1=16\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=2b-4\\b^2+3a-15=0\end{matrix}\right.\)

\(\Rightarrow b^2+3\left(2a-4\right)-15=0\)

\(\Leftrightarrow b^2+6b-27=0\Rightarrow\left[{}\begin{matrix}b=3\Rightarrow a=2\\b=-9\left(l\right)\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\sqrt{x+2}=2\\\sqrt{y-1}=3\end{matrix}\right.\) \(\Rightarrow...\)