Đáp án:
m) $x+6x^2=0\\\Rightarrow x.(1+6x)=0\\\Rightarrow\left[\begin{matrix}x=0\\1+6x=0\end{matrix} \right.\\\Rightarrow\left[\begin{matrix}x=0\\x=\dfrac{-1}{6}\end{matrix} \right.$
Vậy $x\in\left\{0;\dfrac{-1}{6}\right\}$.
n) $x+1=(x+1)^2\\\Rightarrow (x+1)^2-(x+1)=0\\\Rightarrow (x+1).\!(x+1-1)=0\\\Rightarrow \left[\begin{matrix}x+1=0\\x=0\end{matrix}\right.\\\Rightarrow \left[\begin{matrix}x=-1\\x=0\end{matrix}\right.$
Vậy $x\in\left\{-1;0\right\}$.
