Giải thích các bước giải:
Câu 1:
Ta có:
$\dfrac{a^2b-2ab^2+b^3}{2a^2-ab-b^2}$
$=\dfrac{b(a^2-2ab+b^2)}{2a^2-2ab+ab-b^2}$
$=\dfrac{b(a-b)^2}{2a(a-b)+b(a-b)}$
$=\dfrac{b(a-b)^2}{(2a+b)(a-b)}$
$=\dfrac{b(a-b)}{2a+b}$
$=\dfrac{ab-b^2}{2a+b}$
Câu 2:
a.Ta có:
$\dfrac{2xy-x^2-2y+x}{4x-4x^2}$
$=\dfrac{x(2y-x)-(2y-x)}{4x(1-x)}$
$=\dfrac{(x-1)(2y-x)}{4x(1-x)}$
$=\dfrac{(1-x)(x-2y)}{4x(1-x)}$
$=\dfrac{x-2y}{4x}$
b.Ta có:
$\dfrac{x^2+5x+6}{x^2+6x+9}$
$=\dfrac{x^2+2x+3x+6}{(x+3)^2}$
$=\dfrac{x(x+2)+3(x+2)}{(x+3)^2}$
$=\dfrac{(x+3)(x+2)}{(x+3)^2}$
$=\dfrac{x+2}{x+3}$