Giải thích các bước giải:
a.Ta có:
$\dfrac{20x^2-45}{(2x+3)^2}$
$=\dfrac{5(4x^2-9)}{(2x+3)^2}$
$=\dfrac{5((2x)^2-3^2)}{(2x+3)^2}$
$=\dfrac{5(2x-3)(2x+3)}{(2x+3)^2}$
$=\dfrac{5(2x-3)}{2x+3}$
b.Ta có:
$\dfrac{8xy(3x-1)^3}{12x^3(1-3x)}$
$=-\dfrac{8xy(3x-1)^3}{12x^3(3x-1)}$
$=-\dfrac{2y(3x-1)^2}{3x^2}$
c.Ta có:
$\dfrac{9-(x+5)^2}{x^2+4x+4}$
$=\dfrac{3^2-(x+5)^2}{(x+2)^2}$
$=\dfrac{(3-(x+5))(3+(x+5))}{(x+2)^2}$
$=\dfrac{(-x-2)(x+8)}{(x+2)^2}$
$=-\dfrac{(x+2)(x+8)}{(x+2)^2}$
$=-\dfrac{x+8}{x+2}$
