ĐK: \(x\le2\)
\(\sqrt{x+10}=2-x\)
\(\Leftrightarrow x+10=\left(2-x\right)^2\)
\(\Leftrightarrow x^2-4x+4-x-10=0\)
\(\Leftrightarrow x^2-5x-6=0\)
\(\Leftrightarrow\left(x-6\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=6\left(loai\right)\\x=-1\left(chon\right)\end{matrix}\right.\)
Vậy pt có nghiệm duy nhất \(x=-1\)