a) \(\left(x+6\right).\left(3x-1\right)+x+6=0\Leftrightarrow3x^2-x+18x-6+x+6=0\)
\(3x^2+18x=0\Leftrightarrow3x\left(x+6\right)\Leftrightarrow\left\{{}\begin{matrix}3x=0\\x+6=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=-6\end{matrix}\right.\)
vậy \(x=0;x=-6\)
\(\left(x+6\right)\left(3x-1\right)+\left(x+6\right)=0\)
\(\Leftrightarrow\left(x+6\right)\left(3x-1+1\right)=0\)
\(\Leftrightarrow\left(x+6\right)3x=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+6=0\\x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-6\\x=0\end{matrix}\right.\)
Vậy x = -6 hoặc x = 0
