\(\left(2x+1\right)\left(x+1\right)^2\left(2x+3\right)=18\)\(\Leftrightarrow\left(2x+1\right).4.\left(x+1\right)^2.\left(2x+3\right)=72\)
\(\Leftrightarrow\left(2x+1\right)\left(2x+2\right)^2\left(2x+3\right)=72\)(1)
Đặt \(y=2x+2\) thay vào (1) ta có:
\(\left(1\right)\Leftrightarrow\left(y-1\right)y^2\left(y+1\right)=72\)\(\Leftrightarrow y^4-y^2-72=0\)
\(\Leftrightarrow\left(y^2-9\right)\left(y^2+8\right)=0\Leftrightarrow y^2=9\Leftrightarrow y=\pm3\)
*) \(y=3\Leftrightarrow2x+2=3\Leftrightarrow x=\dfrac{1}{2}\)
*)\(y=-3\Leftrightarrow2x+2=-3\Leftrightarrow x=-\dfrac{5}{2}\)
Vậy \(S=\left\{-\dfrac{5}{2};\dfrac{1}{2}\right\}\)
