ĐKXĐ: \(x\le2\)
Đặt \(\sqrt{2-x}=a\ge0\Rightarrow2-x=a^2\Rightarrow2=a^2+x\), pt trở thành:
\(-x^2+a^2+x=a\Leftrightarrow\left(a-x\right)\left(a+x\right)-\left(a-x\right)=0\)
\(\Leftrightarrow\left(a-x\right)\left(a+x-1\right)=0\) \(\Rightarrow\left[{}\begin{matrix}a=x\\a=1-x\end{matrix}\right.\)
TH1: \(a=x\Rightarrow\sqrt{2-x}=x\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x^2+x-2=0\end{matrix}\right.\) \(\Rightarrow x=1\)
TH2: \(a=1-x\Rightarrow\sqrt{2-x}=1-x\Leftrightarrow\left\{{}\begin{matrix}1-x\ge0\\2-x=\left(1-x\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\le1\\x^2-x-1=0\end{matrix}\right.\) \(\Rightarrow x=\frac{1-\sqrt{5}}{2}\)
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