Question

Giải hệ phương trình: \(\left\{{}\begin{matrix}\sqrt{x}+\sqrt{2-y}=\sqrt{2}\\\sqrt{2-x}+\sqrt{y}=\sqrt{2}\end{matrix}\right.\)

Answer 1

ĐKXĐ: ...

Trừ vế cho vế: \(\sqrt{x}-\sqrt{y}+\sqrt{2-y}-\sqrt{2-x}=0\)

\(\Leftrightarrow\frac{x-y}{\sqrt{x}+\sqrt{y}}+\frac{x-y}{\sqrt{2-y}+\sqrt{2-x}}=0\)

\(\Leftrightarrow\left(x-y\right)\left(\frac{1}{\sqrt{x}+\sqrt{y}}+\frac{1}{\sqrt{2-y}+\sqrt{2-x}}\right)=0\)

\(\Leftrightarrow x-y=0\Rightarrow x=y\)

Thay vào pt đầu: \(\sqrt{x}+\sqrt{2-x}=\sqrt{2}\)

\(\Leftrightarrow2+2\sqrt{x\left(2-x\right)}=2\)

\(\Leftrightarrow\sqrt{x\left(2-x\right)}=0\Rightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)