giải hộ mình vs mn ơi
sớm sớm 1 tí nhé mn 
ĐKXĐ: \(x\ge-1\)
Đặt \(\sqrt{x+1}=a\ge0\)
\(a^2=\left(2a^2-1\right)\sqrt{a+2}\)
\(\Leftrightarrow\left(2a^2-1\right)\left(\sqrt{a+2}-1\right)+a^2-1=0\)
\(\Leftrightarrow\frac{\left(2a^2-1\right)\left(a+1\right)}{\sqrt{a+2}+1}+\left(a-1\right)\left(a+1\right)=0\)
\(\Leftrightarrow\frac{2a^2-1}{\sqrt{a+2}+1}+a-1=0\)
\(\Leftrightarrow2a^2+a-2+\left(a-1\right)\sqrt{a+2}=0\)
Đặt \(\sqrt{a+2}=t\ge\sqrt{2}\)
\(\Rightarrow2\left(t^2-2\right)^2+t^2-4+\left(t^2-3\right)t=0\)
\(\Leftrightarrow2t^4+t^3-7t^2-3t+4=0\)
\(\Leftrightarrow\left(t^2+t-1\right)\left(2t^2-t-4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}t=\frac{-1+\sqrt{5}}{2}\left(l\right)\\t=\frac{-1-\sqrt{5}}{2}\left(l\right)\\t=\frac{1-\sqrt{33}}{4}\left(l\right)\\t=\frac{1+\sqrt{33}}{4}\end{matrix}\right.\) \(\Rightarrow a=t^2-2=\frac{1+\sqrt{33}}{8}\)
\(\Rightarrow x=a^2-1=\frac{\sqrt{33}-15}{32}\)
