Đáp án+giải thích các bước giải:
1,
a)
$x(y-2)+2y-4$
$=x(y-2)+(2y-4)$
$=x(y-2)+2(y-2)$
$=(y-2)(x+2)$
b)
$x^2-4y^2$
$=(x-2y)(x+2y)$
2,
$5x(x-2)-2x+4=0$
$5x(x-2)-(2x-4)=0$
$5x(x-2)-2(x-2)=0$
$(x-2)(5x-2)=0$
$⇒$\(\left[ \begin{array}{l}x-2=0\\5x-2=0\end{array} \right.\)
$⇒$\(\left[ \begin{array}{l}x=2\\x=\dfrac{2}{5}\end{array} \right.\)
Vậy \(\left[ \begin{array}{l}x=2\\x=\dfrac{2}{5}\end{array} \right.\)