Giải thích các bước giải:
Bài 1:
a.$\dfrac{xy}{x^2-y^2}=\dfrac{xy}{(x-y)(x+y)}$
b.$\dfrac{y}{xy-x^2}=\dfrac{y}{x(y-x)}$
c.$\dfrac{xy}{y^2-xy}=\dfrac{xy}{y(y-x)}=\dfrac{x}{y-x}$
$\to$Mẫu chung là $(x-y)(x+y)x$
Bài 2:
$2x^2=3x\to 2x^2-3x=0\to x(2x-3)=0\to x\in\{0,\dfrac 32\}$
Bài 3:
$\dfrac{2}{x+4}+\dfrac{3}{x^2-16}$
$=\dfrac{2(x-4)}{(x+4)(x-4)}+\dfrac{3}{(x+4)(x-4)}$
$=\dfrac{2(x-4)+3}{(x+4)(x-4)}$
$=\dfrac{2x-5}{(x+4)(x-4)}$
Bài 4:
$\dfrac{5x-4}{3x^2}:\dfrac{10x-8}{x^2y}$
$=\dfrac{5x-4}{3x^2}.\dfrac{x^2y}{10x-8}$
$=\dfrac{5x-4}{3x^2}.\dfrac{x^2y}{2(5x-4)}$
$=\dfrac{3}{2y}$