\(\left(4x+3\right)^2\left(x+1\right)\left(2x+1\right)=810\)
\(\Leftrightarrow\left(16x^2+24x+9\right)\left(2x^2+x+2x+1\right)=810\)
\(\Leftrightarrow\left(16x^2+24x+9\right)\left(2x^2+3x+1\right)=810\)
Đặt \(2x^2+3x+1=a\)
\(\Leftrightarrow\left(8a+1\right)a=810\)
\(\Leftrightarrow8a^2+a-810=0\)
\(\Delta=1^2-4.\left(-810\right).8=1+25920=25921>0\)
--> pt có 2 nghiệm
\(\left\{{}\begin{matrix}x_1=\dfrac{-1+\sqrt{25921}}{16}=10\\x_2=\dfrac{-1-\sqrt{25921}}{16}=-\dfrac{81}{8}\end{matrix}\right.\)
\(\rightarrow\left[{}\begin{matrix}2x^2+3x+1=10\\2x^2+3x+1=-\dfrac{81}{8}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}2x^2+3x-9=0\\2x^2+3x+\dfrac{89}{8}=0\left(vô.lí\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-3\end{matrix}\right.\)
Vậy \(S=\left\{\dfrac{3}{2};-3\right\}\)
