Giải thích các bước giải:
a.Ta có:
$\dfrac{1}{5x+3}-\dfrac{1}{5x-3}$
$=\dfrac{(5x-3)-(5x+3)}{(5x-3)(5x+3)}$
$=\dfrac{-6}{(5x-3)(5x+3)}$
b.Ta có:
$\dfrac{x^2}{x-1}-\dfrac{1}{x-1}$
$=\dfrac{x^2-1}{x-1}$
$=\dfrac{(x-1)(x+1)}{x-1}$
$=x+1$
c.Ta có:
$\dfrac{x^2+7x}{x^2-9}-\dfrac{x^2+6x+9}{x^2-9}$
$=\dfrac{x^2+7x-(x^2+6x+9)}{x^2-9}$
$=\dfrac{x^2+7x-x^2-6x-9}{x^2-9}$
$=\dfrac{x-9}{x^2-9}$
d.Ta có:
$\dfrac{x^2-9}{x+7}:\dfrac{x^2+3x}{2x^2+14x}$
$=\dfrac{x^2-9}{x+7}:\dfrac{x(x+3)}{x(2x+14)}$
$=\dfrac{x^2-9}{x+7}:\dfrac{x+3}{2x+14}$
$=\dfrac{(x-3)(x+3)}{x+7}\cdot \dfrac{2x+14}{x+3}$
$=\dfrac{(x-3)(x+3)}{x+7}\cdot \dfrac{2(x+7)}{x+3}$
$=2(x-3)$
