\(\sqrt{2x+3}-\sqrt{x+6}=3-x\left(1\right)\)
Đk: \(\frac{-3}{2}\le x\le3\)
\(\frac{\left(\sqrt{2x+3}-\sqrt{x+6}\right)\left(\sqrt{2x+3}+\sqrt{x-6}\right)}{\left(\sqrt{2x+3}+\sqrt{x+6}\right)}=3-x\)
\(\frac{x-3}{\left(\sqrt{2x+3}+\sqrt{x-6}\right)}=3-x\)
\(\left(x-3\right)\left(\frac{1}{\left(\sqrt{2x+3}+\sqrt{x+6}\right)}+1\right)=0\)
x-3=0 <=>x=3