\(\Leftrightarrow3\left(x+1\right)\left(\sqrt{x^2+x+3}-\left(x+1\right)\right)+2x-4=0\)
\(\Leftrightarrow\frac{-\left(3x+1\right)\left(x-2\right)}{\sqrt{x^2+x+3}+x+1}+2\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\\frac{3\left(x+1\right)}{\sqrt{x^2+x+3}+x+1}=2\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow3x+3=2\sqrt{x^2+x+3}+2x+2\)
\(\Leftrightarrow2\sqrt{x^2+x+3}=x+1\) (\(x\ge-1\))
\(\Leftrightarrow4\left(x^2+x+3\right)=x^2+2x+1\)
\(\Leftrightarrow3x^2+2x+11=0\left(vn\right)\)
Vậy pt đã cho có nghiệm duy nhất \(x=2\)
