ĐKXĐ : \(x\ne-\frac{1}{3}\)
Ta có : \(\sqrt{x^2+x+2}=\frac{3x^2+3x+2}{3x+1}\)
\(\Leftrightarrow\sqrt{x^2+x+2}-2=\frac{3x^2+3x+2}{3x+1}-2\)
\(\Leftrightarrow\frac{x^2+x+2-4}{\sqrt{x^2+x+2}+2}=\frac{3x^2+3x+2-6x-2}{3x+1}\)
\(\Leftrightarrow\frac{x^2+x-2}{\sqrt{x^2+x+2}+2}=\frac{3x^2-3x}{3x+1}\)
\(\Leftrightarrow\frac{\left(x-1\right)\left(x+2\right)}{\sqrt{x^2+x+2}+2}-\frac{3x\left(x-1\right)}{3x+1}=0\)
\(\Leftrightarrow\left(x-1\right)\left[\frac{x+2}{\sqrt{x^2+x+2}+2}-\frac{3x}{3x+1}\right]=0\)
\(\Leftrightarrow x=1\)( Thỏa mãn )
