Tìm x biết x^2 – 16 + x(x – 4) = 0 x = 2 hoặc x = - 4

* Hướng dẫn giải

$\begin{array}{l}{x}^2-16+x\left(x-4\right)=0\\ \left(x^2-16\right)+x\left(x-4\right)=0\\ \left(x+4\right)\left(x-4\right)+x\left(x-4\right)=0\\ \left(x-4\right)\left(x+4+x\right)=0\\ \left(x-4\right)\left(2x+4\right)=0\\ \Rightarrow \left[\begin{array}{l}x-4=0\\ 2x+4=0\end{array}\right.\\ \Rightarrow \left[\begin{array}{l}x=4\\ x=-2\end{array}\right.\end{array}$

Vậy $x=-2$ hoặc $x=4$.