a) \(x^2+x=6\)
\(\Rightarrow x^2+x-6=0\)
\(\Rightarrow x^2+3x-2x-6=0\)
\(\Rightarrow x\left(x+3\right)-2\left(x+3\right)=0\)
\(\Rightarrow\left(x+3\right)\left(x-2\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}x+3=0\\x-2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\)
b) \(6x^3+x^2=2x\)
\(\Leftrightarrow6x^3+x^2-2x=0\)
\(\Leftrightarrow x\left(6x^2+x-2\right)=0\)
\(\Leftrightarrow x\left(6x^2+4x-3x-2\right)=0\)
\(\Leftrightarrow x\left[\left(6x^2+4x\right)-\left(3x+2\right)\right]=0\)
\(\Leftrightarrow x\left[2x\left(3x+2\right)-\left(3x+2\right)\right]=0\)
\(\Leftrightarrow x\left(3x-2\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\3x-2=0\\2x-1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\x=-\dfrac{2}{3}\\x=\dfrac{1}{2}\end{matrix}\right.\)
Vậy ...