Đáp án:
$a) x(x-2)+x-2=0$
$⇔(x-2)(x+1)=0$
$⇔$\(\left[ \begin{array}{l}x-2=0\\x+1=0\end{array} \right.\)
$⇔$\(\left[ \begin{array}{l}x=2\\x=-1\end{array} \right.\)
$\text{Vậy x ∈ {2 ; -1}}$
$b) x^2-4x+5=0$
$⇔x^2 -4x +4 +1 =0$
$⇔(x-2)^2 +1=0$
Vì $(x-2)² ≥ 0$
Nên $(x-2)² +1 > 0$
$\text{Vậy x ∈ { ∅ }}$
