ĐKXĐ: \(x\ge-1\)
Đặt \(\sqrt{x+1}=t\ge0\Rightarrow x=t^2-1\)
\(\Rightarrow\left(t^2+1\right)t=2\left(t^2-1\right)+1\)
\(\Leftrightarrow t^3-2t^2+t+1=0\)
\(\Leftrightarrow t\left(t-1\right)^2+1=0\)
Do \(t\ge0\Rightarrow t\left(t-1\right)^2\ge0\Rightarrow t\left(t-1\right)^2+1>0\)
Pt vô nghiệm