\(\sqrt{2-x}=3-\sqrt{3x+1}\left(ĐK:-\frac{1}{3}\le x\le2\right)\)
\(\Leftrightarrow\sqrt{2-x}+\sqrt{3x+1}=3\)
\(\Leftrightarrow2-x+3x+1+2\sqrt{\left(2-x\right)\left(3x+1\right)}=9\)
\(\Leftrightarrow2\sqrt{\left(2-x\right)\left(3x+1\right)}=6-2x\)
\(\Leftrightarrow\sqrt{\left(2-x\right)\left(3x+1\right)}=3-x\left(ĐK:x\le3\right)\)
\(\Leftrightarrow\left(2-x\right)\left(3x+1\right)=9-6x+x^2\)
\(\Leftrightarrow6x+2-3x^2-x=9-6x+x^2\)
\(\Leftrightarrow4x^2-11x+7=0\)
\(\Leftrightarrow\left(4x-7\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[\begin{matrix}x=\frac{7}{4}\left(tm\right)\\x=1\left(tm\right)\end{matrix}\right.\)
Vậy pt đã cho có nghiệm là \(S=\left\{\frac{7}{4};1\right\}\)
