ĐKXĐ: ...
\(\left(x+1\right)^2+1-3x\sqrt{x+1}=0\)
Đặt \(\sqrt{x+1}=t\ge0\Rightarrow x=t^2-1\)
\(t^4+1-3t\left(t^2-1\right)=0\)
\(\Leftrightarrow t^4-3t^3+3t+1=0\)
\(\Leftrightarrow t^4-2t^3-t^2-t^3+2t^2+t-t^2+2t+1=0\)
\(\Leftrightarrow t^2\left(t^2-2t-1\right)-t\left(t^2-2t-1\right)-\left(t^2-2t-1\right)=0\)
\(\Leftrightarrow\left(t^2-2t-1\right)\left(t^2-t-1\right)=0\)