Đáp án: $x∈\{2;-2\}$
Giải thích các bước giải:
$(x-2)(x^2+2x+5)+2(x-2)(x+2)-5(x-2)=0$
$⇔(x-2)(x^2+2x+5)+(x-2)(2x+4)-5(x-2)=0$
$⇔(x-2)(x^2+2x+5+2x+4-5)=0$
$⇔(x-2)(x^2+4x+4)=0$
$⇔(x-2)(x+2)^2=0$
$⇔\left[ \begin{array}{l}x-2=0\\(x+2)^2=0\end{array} \right.$
$⇔\left[ \begin{array}{l}x-2=0\\x+2=0\end{array} \right.$
$⇔\left[ \begin{array}{l}x=2\\x=-2\end{array} \right.$
Vậy $x∈\{2;-2\}$