\(\left(12x+7\right)^2\left(3x+2\right)\left(2x+1\right)=3\)
\(\Leftrightarrow\left(12x+7\right)^2\cdot4\left(3x+2\right)\cdot6\left(2x+1\right)=3\cdot4\cdot6\)
\(\Leftrightarrow\left(12x+7\right)^2\left(12x+8\right)\left(12x+6\right)=72\) (1)
Đặt 12x + 7 = a
(1) \(\Leftrightarrow a^2\left(a+1\right)\left(a-1\right)=72\)
\(\Leftrightarrow a^2\left(a^2-1\right)=72\) (2)
Đặt \(a^2=b\)
(2) \(\Leftrightarrow b\left(b-1\right)=72\)
\(\Leftrightarrow b^2-b-72=0\)
\(\Leftrightarrow b^2+8b-9b-72=0\)
\(\Leftrightarrow b\left(b+8\right)-9\left(b+8\right)=0\)
\(\Leftrightarrow\left(b-9\right)\left(b+8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}b-9=0\\b+8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}b=9\\b=-8\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}a^2=9\Leftrightarrow a=\pm3\\a^2=-8\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}12x+7=3\\12x+7=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}12x=-4\\12x=-10\end{matrix}\right.\Leftrightarrow}\left[{}\begin{matrix}x=-\dfrac{1}{3}\\x=-\dfrac{5}{6}\end{matrix}\right.\)
