\(x^2=2+\sqrt{2-x}\) ĐK : \(x\le2\)
Đặt \(\sqrt{2-x}=y\ge0\rightarrow y^2=2-x\) . Ta có hệ :
\(\left\{{}\begin{matrix}x^2=y+2\\y^2=2-x\end{matrix}\right.\Rightarrow x^2-y^2=x+y\)
\(\Leftrightarrow\left(x+y\right).\left(x-y-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-y\\x=y+1\end{matrix}\right.\)
Với x = -y ta có pt : \(\sqrt{2+y}=y\)
\(\Leftrightarrow2+y=y^2\)
\(\Leftrightarrow y^2-y-2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}y=-1\left(l\right)\\y=2\left(tm\right)\Rightarrow x=-2\left(tm\right)\end{matrix}\right.\)
Với x = y + 1 ta có : \(\sqrt{1-y}=y\)
\(\Leftrightarrow\left\{{}\begin{matrix}y\ge0\\y^2+y-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y\ge0\\\left[{}\begin{matrix}y=\dfrac{-1-\sqrt{5}}{2}\left(l\right)\\y=\dfrac{-1+\sqrt{5}}{2}\Rightarrow x=\dfrac{1+\sqrt{5}}{2}\end{matrix}\right.\end{matrix}\right.\)
