ĐKXĐ: \(x\le2\)
\(pt\Leftrightarrow\sqrt{x^2+2x+4}+\sqrt{2-x}=\sqrt{\left(x^2+2x+4\right)\left(2-x\right)}+1\)
\(\Leftrightarrow\sqrt{\left(x^2+2x+4\right)\left(2-x\right)}-\sqrt{2-x}-\sqrt{x^2+2x+4}+1=0\)
\(\Leftrightarrow\sqrt{2-x}\left(\sqrt{x^2+2x+4}-1\right)-\left(\sqrt{x^2+2x+4}-1\right)=0\)
\(\Leftrightarrow\left(\sqrt{2-x}-1\right)\left(\sqrt{x^2+2x+4}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{2-x}-1=0\\\sqrt{x^2+2x+4}-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x^2+2x+3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\\left(x+1\right)^2=-1\left(\text{vô nghiệm}\right)\end{matrix}\right.\Leftrightarrow x=1\left(tm\right)\)
