c) Ta có
$\dfrac{2x-1}{x} + \dfrac{2x+5}{4x-3} + \dfrac{2x^2 + x + 3}{3x-4x^2} = \dfrac{(2x-1)(4x-3)}{x(4x-3)} + \dfrac{(2x+5)x}{x(4x-3)} - \dfrac{2x^2 + x +3 }{x(4x-3)}$
$= \dfrac{8x^2 -10x + 3 + 2x^2 + 5x - 2x^2 - x - 3}{x(4x-3)}$
$= \dfrac{8x^2 -6x}{x(4x-3)}$
$= \dfrac{2x(4x-3)}{x(4x-3)} = 2$
d) Ta có
$\dfrac{x}{x-1} - \dfrac{x+1}{x} + \dfrac{x-2}{x^2-x} = \dfrac{x^2}{x(x-1)} - \dfrac{(x+1)(x-1)}{x(x-1)} + \dfrac{x-2}{x(x-1)}$
$= \dfrac{x^2 - (x^2-1) + x-2}{x(x-1)}$
$= \dfrac{x-1}{x(x-1)} = \dfrac{1}{x}$