Để PT có nghiệm thì
\(\left\{{}\begin{matrix}3-x\ge0\\x+2\ge0\end{matrix}\right.\)\(\Leftrightarrow3\ge x\ge-2\)
Ta năng 2 vế lên bình phương để khử căn:
\(\left(3-x\right)+\left(x+2\right)+2\sqrt{\left(3-x\right)\left(x+2\right)}=9\)
\(\Leftrightarrow2\sqrt{-x^2+x+6}=4\)
\(\Leftrightarrow\sqrt{-x^2+x+6}=2\)
\(\Leftrightarrow-x^2+x+6=4\)
\(\Leftrightarrow-\left(x-\frac{1}{2}\right)^2+\frac{9}{4}=0\)
\(\Leftrightarrow x-\frac{1}{2}=\frac{3}{2}\Rightarrow x=1\)( TMĐK)