\(\left(2x+1\right)\left(2x+3\right)\left(x+1\right)^2=18\)

\(\Leftrightarrow\left(2x+2-1\right)\left(2x+2+1\right)\left(x+1\right)^2=18\)

\(\Leftrightarrow\left(\left(2x+2\right)^2-1\right)\left(x+1\right)^2=18\)

\(\Leftrightarrow4\left(x+1\right)^4-\left(x+1\right)^2-18=0\)

Đặt t = \(\left(x+1\right)^2\) \(\left(t\ge0\right)\)

pt \(\Leftrightarrow4t^2-t-18=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t=\dfrac{9}{4}\left(nh\right)\\t=-2\left(l\right)\end{matrix}\right.\)

\(\Leftrightarrow\left(x+1\right)^2-\dfrac{9}{4}=0\)

\(\Leftrightarrow\left(x+1-\dfrac{3}{2}\right)\left(x+1+\dfrac{3}{2}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=\dfrac{5}{2}\end{matrix}\right.\)