ĐKXĐ: \(x\ge-1\)
Đặt \(\sqrt{x+1}=t\ge0\Rightarrow x=t^2-1\)
\(\Leftrightarrow t^2-\left(2t^2-1\right)\sqrt{t+2}=0\)
\(\Leftrightarrow-4t^3+t^2+2t+\left(2t^2-1\right)\left(2t-\sqrt{t+2}\right)=0\)
\(\Leftrightarrow-t\left(4t^2-t-2\right)+\frac{\left(2t^2-1\right)\left(4t^2-t-2\right)}{2t+\sqrt{t+2}}=0\)
\(\Leftrightarrow\left(4t^2-t-2\right)\left(-t+\frac{2t^2-1}{2t+\sqrt{t+2}}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}4t^2-t-2=0\\\frac{2t^2-1}{2t+\sqrt{t+2}}=t\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow2t^2-1=2t^2+t\sqrt{t+2}\)
\(\Leftrightarrow t\sqrt{t+2}=-1\) (vô nghiệm do vế trái không âm, vế phải âm)
