ĐK: \(x\ge-2\)
Đặt \(a=\sqrt{x+2},b=\sqrt{x^2-2x+4}\Rightarrow\left\{{}\begin{matrix}2a^2=2x+4\\b^2=x^2-2x+4\end{matrix}\right.\Rightarrow2a^2+b^2-9=x^2-1\)
\(pt\Leftrightarrow9\left(ab-2\right)=2\left(2a^2+b^2-9\right)\\ \Leftrightarrow9ab-18=4a^2+2b^2-18\\ \Leftrightarrow4a^2+2b^2-9ab=0\\ \Leftrightarrow\left(a-2b\right)\left(4a-b\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}a=2b\\4a=b\end{matrix}\right.\)
\(a=2b\Rightarrow\sqrt{x+2}=2\sqrt{x^2-2x+4}\Leftrightarrow x+2=4x^2-8x+16\Leftrightarrow4x^2-9x+14=0\)vô nghiệm
\(4a=b\Rightarrow4\sqrt{x+2}=\sqrt{x^2-2x+4}\Leftrightarrow16x+32=x^2-2x+4\Leftrightarrow x^2-18x-28=0\\ \Leftrightarrow\left[{}\begin{matrix}x=9+\sqrt{109}\left(TM\right)\\x=9-\sqrt{109}\left(TM\right)\end{matrix}\right.\)
