\(\left(12x+7\right)^2\left(3x+2\right)\left(2x+1\right)=3\)
\(\Leftrightarrow\left(144x^2+168x+49\right)\left(6x^2+7x+2\right)=3\)
\(\Leftrightarrow\left(144x^2+168x+49\right)\left(144x^2+168+48\right)=72\)
Đặt \(144x^2+168x+48=u\)
\(\Rightarrow144x^2+168x+49=u+1\left(1\right)\)
Do đó: \(u\left(u+1\right)=72\Leftrightarrow u^2+u-72=0\)
\(\Leftrightarrow\left(u-8\right)\left(u+9\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}u-8=0\\u+9=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}u=8\\u=-9\end{matrix}\right.\)
Với \(u=8;u=-9\) bạn thay vào (1) và tìm x nha.