\(\left(\sqrt{x+1}+1\right)\left(5-x\right)=2x\left(1\right)\left(đk:x\ge-1\right)\)
\(đặt:\sqrt{x+1}=t\ge0\)
\(\Rightarrow\left(1\right)\Leftrightarrow\left(t+1\right)\left(6-t^2\right)=2t^2-2\)
\(-t^3+6t+6-t^2-2t^2+2=0\Leftrightarrow-t^3-3t^2+6t+8=0\Leftrightarrow-\left(t+1\right)\left(t-2\right)\left(t+4\right)=0\Leftrightarrow\left[{}\begin{matrix}t=-1\left(ktm\right)\\t=2\left(tm\right)\Rightarrow\sqrt{x+1}=2\Leftrightarrow x=3\left(tm\right)\\t=-4\left(ktm\right)\end{matrix}\right.\)