Giải thích các bước giải:
$d.\dfrac{x+2y}{xy}+\dfrac{x-2y}{xy}=\dfrac{x+2y+x-2y}{xy}=\dfrac{2x}{xy}=\dfrac{2}{y}$
$g.\dfrac{3}{2x^2+2x}+\dfrac{2x-1}{x^2-1}-\dfrac{2}{x}$
$=\dfrac{3}{2x(x+1)}+\dfrac{2x-1}{(x-1)(x+1)}-\dfrac{4(x+1)}{2x(x+1)}$
$=\dfrac{3}{2x(x+1)}-\dfrac{4(x+1)}{2x(x+1)}+\dfrac{(2x(2x-1)}{2x(x-1)(x+1)}$
$=\dfrac{3-4(x+1)}{2x(x+1)}+\dfrac{2x(2x-1)}{2x(x-1)(x+1)}$
$=\dfrac{-4x-1}{2x(x+1)}+\dfrac{2x(2x-1)}{2x(x-1)(x+1)}$
$=\dfrac{(-4x-1)(x-1)}{2x(x+1)(x-1)}+\dfrac{2x(2x-1)}{2x(x-1)(x+1)}$
$=\dfrac{(-4x-1)(x-1)+2x(2x-1)}{2x(x+1)(x-1)}$
$=\dfrac{-4x^2+3x+1+4x^2-2x}{2x(x+1)(x-1)}$
$=\dfrac{x+1}{2x(x+1)(x-1)}$
$=\dfrac{x+1}{2x(x+1)}$
