ĐKXĐ : \(-1\le x\le9\)
\(\sqrt{x+1}=2+\sqrt{9-x}\)
\(\Leftrightarrow x+1=4+9-x+4\sqrt{9-x}\)
\(\Leftrightarrow\left(x-6\right)^2=4\left(9-x\right)\)
\(\Leftrightarrow x^2-12x+36=36-4x\)
\(\Leftrightarrow x\left(x-8\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=8\end{array}\right.\)
Thử lại được x = 8 thỏa mãn pt.