ĐK:\(x\ge-1\)
\(PT\Leftrightarrow2\left(x^2-4x+4\right)+2\left(x+1\right)-5\left(x-2\right)\sqrt{x+1}\)\(=0\)
\(\Leftrightarrow2\left(x-2\right)^2+2\left(x+1\right)-5\left(x-2\right)\sqrt{x+1}=0\)
Đặt \(\left\{{}\begin{matrix}u=x-2\\v=\sqrt{x+1}\end{matrix}\right.\) (\(v\ge0\))
\(PT\Leftrightarrow2u^2-5uv+2v^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}u=2v\\u=\frac{1}{2}v\end{matrix}\right.\)
\(u=2v\) \(\Leftrightarrow x-2=2\sqrt{x+1}\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ge2\\x^2-4x+4=4x+4\end{matrix}\right.\) \(\Leftrightarrow x=8\)
\(2u=v\Leftrightarrow2\left(x-2\right)=\sqrt{x+1}\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ge2\\4\left(x^2-4x+4\right)=x+1\end{matrix}\right.\) \(\Leftrightarrow x=3\)
Vậy..
