9) $\dfrac{-x^2 - 2}{x^3 - 1} + \dfrac{1}{x^2 + x + 1} + \dfrac{1}{x - 1}$
Ta có: $x^3 - 1 = (x - 1)(x^2 + x + 1)$
$\dfrac{-x^2 - 2}{(x - 1)(x^2 + x + 1)} + \dfrac{1}{x^2 + x + 1} + \dfrac{1}{x - 1}$
$= \dfrac{-x^2 - 2}{(x - 1)(x^2 + x + 1)} + \dfrac{x - 1}{(x - 1)(x^2 + x + 1)} + \dfrac{x^2 + x + 1}{(x - 1)(x^2 + x + 1)}$
$= \dfrac{(-x^2 - 2) + (x - 1) + (x^2 + x + 1)}{(x - 1)(x^2 + x + 1)}$
$= \dfrac{-x^2 - 2 + x - 1 + x^2 + x + 1}{(x - 1)(x^2 + x + 1)}$
$= \dfrac{2x - 2}{(x - 1)(x^2 + x + 1)}$
$= \dfrac{2(x - 1)}{(x - 1)(x^2 + x + 1)}$
$= \dfrac{2}{x^2 + x + 1}$
Vậy $\dfrac{-x^2 - 2}{x^3 - 1} + \dfrac{1}{x^2 + x + 1} + \dfrac{1}{x - 1} = \dfrac{2}{x^2 + x + 1}$
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10) $\dfrac{1}{x^2 + x + 1} + \dfrac{1}{1 - x^3} + \dfrac{2x}{1 - x^3}$
Ta có: $1 - x^3 = -(x^3 - 1) = -(x - 1)(x^2 + x + 1)$
$\dfrac{1}{x^2 + x + 1} + \dfrac{1}{1 - x^3} + \dfrac{2x}{1 - x^3}$
$= \dfrac{1 - x^3}{(x^2 + x + 1)(1 - x^3)} + \dfrac{1}{1 - x^3} + \dfrac{2x}{1 - x^3}$
$= \dfrac{1 - x^3}{(x^2 + x + 1)(1 - x^3)} + \dfrac{1 + 2x}{1 - x^3}$
$= \dfrac{1 - x^3}{(x^2 + x + 1)(1 - x^3)} + \dfrac{(1 + 2x)(x^2 + x + 1)}{(1 - x^3)(x^2 + x + 1)}$
$= \dfrac{1 - x^3 + (1 + 2x)(x^2 + x + 1)}{(1 - x^3)(x^2 + x + 1)}$
$= \dfrac{1 - x^3 + (x^2 + x + 1 + 2x^3 + 2x^2 + 2x)}{(1 - x^3)(x^2 + x + 1)}$
$= \dfrac{1 - x^3 + 3x^2 + 3x + 1 + 2x^3}{(1 - x^3)(x^2 + x + 1)}$
$= \dfrac{x^3 + 3x^2 + 3x + 2}{(1 - x^3)(x^2 + x + 1)}$
$= \dfrac{(x + 1)^3 + 1}{(1 - x^3)(x^2 + x + 1)}$
Vậy $\dfrac{1}{x^2 + x + 1} + \dfrac{1}{1 - x^3} + \dfrac{2x}{1 - x^3} = \dfrac{x^3 + 3x^2 + 3x + 2}{(1 - x^3)(x^2 + x + 1)}$