Điều kiện xác định: \(\left\{{}\begin{matrix}3x+1\ge0\\2-x\ge0\end{matrix}\right.\Leftrightarrow-\dfrac{1}{3}\le x\le2\)
\(\sqrt{3x+1}+x^2=2+\sqrt{2-x}\)
\(\Leftrightarrow\left(\sqrt{3x+1}-2\right)+\left(1-\sqrt{2-x}\right)+\left(x^2-1\right)=0\)
\(\Leftrightarrow\dfrac{3x-3}{\sqrt{3x+1}+2}+\dfrac{x-1}{1+\sqrt{2-x}}+\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\dfrac{3\left(x-1\right)}{\sqrt{3x+1}+2}+\dfrac{x-1}{\sqrt{2-x}+1}+\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left[\dfrac{3}{\sqrt{3x+1}+2}+\dfrac{1}{\sqrt{2-x}+1}+\left(x+1\right)\right]=0\)
\(\Leftrightarrow x-1=0\) (vì \(\dfrac{3}{\sqrt{3x+1}+2}+\dfrac{1}{\sqrt{2-x}+1}+\left(x+1\right)>0\))
\(\Leftrightarrow x=1\) (nhận)
Vậy phương trình có tập nghiệm là: \(S=\left\{1\right\}.\)
