ĐKXĐ: \(x\le2\)
Đặt \(\sqrt{2-x}=a\ge0\Rightarrow2-x=a^2\Rightarrow2=a^2+x\)
Phương trình trở thành:
\(a=a^2+x-x^2\Leftrightarrow a^2-x^2+x-a=0\)
\(\Leftrightarrow\left(a-x\right)\left(a+x\right)-\left(a-x\right)=0\)
\(\Leftrightarrow\left(a-x\right)\left(a+x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=x\left(x\ge0\right)\\a=1-x\left(x\le1\right)\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\sqrt{2-x}=x\\\sqrt{2-x}=1-x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+x-2=0\\x^2-x-1=0\end{matrix}\right.\)
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