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