Đáp án:
$a) x^2 -x-12=0$
$⇔x^2 -4x +3x -12=0$
$⇔x(x-4) +3(x-4)=0$
$⇔(x-4)(x+3)=0$
$⇔$\(\left[ \begin{array}{l}x-4=0\\x+3=0\end{array} \right.\)
$⇔$\(\left[ \begin{array}{l}x=4\\x=-3\end{array} \right.\)
$\text{Vậy phương trình có nghiệm S = {4 ; -3} }$
$b) x^3 -7x +6=0$
$⇔x^3 -x^2 +x^2 -x -6x +6=0$
$⇔x^2(x-1)+x(x-1) -6(x-1)=0$
$⇔(x-1)(x^2 +x-6)=0$
$⇔(x-1)(x^2+3x-2x-6)=0$
$⇔(x-1)[ x(x+3)-2(x+3)] =0$
$⇔(x-1)(x+3)(x-2)=0$
$⇔$\(\left[ \begin{array}{l}x-1=0\\x+3=0\\x-2=0\end{array} \right.\)
$⇔$\(\left[ \begin{array}{l}x=1\\x=-3\\x=2\end{array} \right.\)
$\text{Vậy phương trình có nghiệm S = {1 ; -3; 2}}$
