\(\left(2x+1\right)\left(2x+3\right)\left(x+1\right)^2-18=0\)
\(\Leftrightarrow\left(x^2+2x+1\right)\left(4x^2+8x+3\right)-18=0\)
Đặt \(x^2+2x+1=a\ge0\)
\(\Rightarrow a\left(4a-1\right)-18=0\)
\(\Leftrightarrow4a^2-a-18=0\)
\(\Leftrightarrow\left(4a^2+8a\right)+\left(-9a-18\right)=0\)
\(\Leftrightarrow\left(a+2\right)\left(4a-9\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}a=-2\left(l\right)\\a=\frac{9}{4}\end{cases}}\)
\(\Rightarrow x^2+2x+1=\frac{9}{4}\)
\(\Leftrightarrow4x^2+8x-5=0\)
\(\Leftrightarrow\left(4x^2-2x\right)+\left(10x-5\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(2x+5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=-\frac{5}{2}\end{cases}}\)
