Tìm x: x^3 +27 +(x+3)(x -9)=0

Phương pháp:

Phân tích vế trái thành nhân tử chung.

Cách giải:

$x^3+27+\left(x+3\right)\left(x-9\right)=0$

$\iff \left(x^3+3^3\right)+\left(x+3\right)\left(x-9\right)=0$

$\iff \left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0$

$\iff \left(x+3\right)\left(x^2-3x+9+x-9\right)=0$

$\iff \left(x+3\right)\left(x^2-2.x\right)=0\iff x\left(x+3\right)\left(x-2\right)=0\iff \left[\begin{array}{l}x=0\\ x+3=0\\ x-2=0\end{array}\right.\iff \left[\begin{array}{l}x=0\\ x=-3\\ x=2\end{array}\right.$