\(\left(6x+7\right)^2\left(3x+4\right)\left(x+1\right)=1\left(1\right)\)
Đặt \(y=3x+\dfrac{7}{2}\)
\(\Rightarrow\left\{{}\begin{matrix}6x+7=2y\\3x+4=y+\dfrac{1}{2}\\x+1=\dfrac{2y-1}{6}\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\left(2y\right)^2.\left(y+\dfrac{1}{2}\right).\left(\dfrac{2y-1}{6}\right)=1\)
\(\Leftrightarrow4y^2.\dfrac{\left(2y+1\right)\left(2y-1\right)}{12}=1\)
\(\Leftrightarrow y^2\left(4y^2-1\right)=3\)
\(\Leftrightarrow4y^4-y^2-3=0\)
\(\Leftrightarrow\left[{}\begin{matrix}y^2=1\\y^2=-\dfrac{3}{4}\left(loại\right)\end{matrix}\right.\)
\(\Leftrightarrow y=\pm1\)
\(TH_1:y=1\Rightarrow3x+\dfrac{7}{2}=1\Rightarrow3x=-\dfrac{5}{2}\Rightarrow x=-\dfrac{5}{6}\)
\(TH_2:y=-1\Rightarrow3x+\dfrac{7}{2}=-1\Rightarrow3x=-\dfrac{9}{2}\Rightarrow x=-\dfrac{3}{2}\)
Vậy \(x\in\left\{-\dfrac{3}{2};-\dfrac{5}{6}\right\}\)
