ĐKXĐ : \(x>-\frac{3}{2}\)
pt \(\Leftrightarrow2\left(x+1\right)\left(2x+3\right)=8x^2+18x+11\)
\(\Leftrightarrow2x^2+10x+6=8x^2+18x+11\)
\(\Leftrightarrow6x^2+8x+5=0\)
\(\Leftrightarrow6\left(x^2+\frac{4}{3}x+\frac{5}{6}\right)=0\)
\(\Leftrightarrow6\left(x+\frac{2}{3}\right)^2+\frac{7}{3}=0\) ( ***** )
Vậy pt vô nghiệm