\(\Leftrightarrow\left(4x^2+8x+3\right)\left(x+1\right)^2-18=0\)
\(\Leftrightarrow\left[4\left(x+1\right)^2-1\right]\left(x+1\right)^2-18=0\)
Đặt \(\left(x+1\right)^2=t\ge0\)
\(\left(4t-1\right)t-18=0\Leftrightarrow4t^2-t-18=0\Rightarrow\left[{}\begin{matrix}t=\frac{9}{4}\\t=-2\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\left(x+1\right)^2=\frac{9}{4}\Rightarrow\left[{}\begin{matrix}x+1=\frac{3}{2}\\x+1=-\frac{3}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{1}{2}\\x=-\frac{5}{2}\end{matrix}\right.\)
