ĐK: \(-2\le x\le3\)
PT \(\Leftrightarrow\sqrt{3-x}=3-\sqrt{x+2}\)
\(\Leftrightarrow3-x=9+x+2-6\sqrt{x+2}\)
\(\Leftrightarrow6\sqrt{x+2}=2x+8\)
\(\Leftrightarrow36x+72=4x^2+32x+64\)
\(\Leftrightarrow4x^2-4x-8=0\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\) (TM)
Vậy phương trình có tập nghiệm \(S=\left\{-1;2\right\}\)
\(=\sqrt{3-x}=3-\sqrt{x+2}\Leftrightarrow3-x=9-6\sqrt{x+2}+x+2\)
\(\Leftrightarrow2x+8-6\sqrt{x+2}=0\Leftrightarrow x+4-3\sqrt{x+2}\)
\(\Leftrightarrow\left(x+2\right)-2.\frac{3}{2}\sqrt{x+2}+\frac{9}{4}-\frac{1}{4}=0\)
\(\Leftrightarrow\left(\sqrt{x+2}-\frac{3}{2}\right)^2=\frac{1}{4}\Rightarrow\left[{}\begin{matrix}\sqrt{x+2}-\frac{3}{4}=\frac{1}{2}\\\sqrt{x+2}-\frac{3}{4}=-\frac{1}{2}\end{matrix}\right.\)
