Sửa lại \(\left(12x+7\right)^2.\left(3x+2\right).\left(2x+1\right)=3\)
\(\Leftrightarrow\left(12x+7\right)^2.4\left(3x+2\right).6\left(2x+1\right)=72\)
\(\Leftrightarrow\left(12x+7\right)^2.\left(12x+8\right).\left(12x+6\right)=72\)
Đặt \(12x+7=y\) , thế vào phương trình trên ta có:
\(y^2.\left(y+1\right).\left(y-1\right)=72\)\(\Leftrightarrow y^4-y^2=72\)
\(\Leftrightarrow\left(y^2-9\right)\left(y^2+8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}y^2-9=0\\y^2+8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}y=\pm3\\y^2=-8\end{matrix}\right.\Leftrightarrow y=\pm3\)vì \(y^2\ge0\)
Nếu \(y=3\Leftrightarrow12x+7=3\Leftrightarrow x=-\dfrac{1}{3}\)
Nếu \(y=-3\Leftrightarrow12x+7=-3\Leftrightarrow x=-\dfrac{5}{6}\)
