b) Mẫu thức chung là $(x-3)(x+3) = x^2-9$
Ta có
$\dfrac{x}{x+3} = \dfrac{x(x-3)}{(x-3)(x+3)} = \dfrac{x^2-3x}{x^2-9}$
$\dfrac{1}{3-x} = \dfrac{-1}{x-3} = \dfrac{-(x+3)}{(x-3)(x+3)} = \dfrac{-x-3}{x^2-9}$
c) Mẫu thức chung là $xy(x-y)(x+y) = xy(x^2-y^2)$
Khi đó
$\dfrac{1}{x^2+xy} = \dfrac{1}{x(x+y)} = \dfrac{y(x-y)}{xy(x-y)(x+y)} = \dfrac{xy-y^2}{xy(x^2-y^2)}$
$\dfrac{1}{xy-y^2} = \dfrac{1}{y(x-y)} = \dfrac{x(x+y)}{xy(x-y)(x+y)} = \dfrac{x^2 + xy}{xy(x^2-y^2)}$
$\dfrac{2}{y^2-x^2} = \dfrac{-2}{x^2-y^2} = \dfrac{-2xy}{xy(x^2-y^2)}$.
d) Mẫu thức chung là $x^3 + 1 = (x+1)(x^2-x+1)$
Khi đó
$\dfrac{2}{x+1} = \dfrac{2(x^2-x+1)}{(x+1)(x^2-x+1)} = \dfrac{2x^2-2x+2}{x^3+1}$
$\dfrac{3}{x^2-x+1} = \dfrac{3(x+1)}{(x+1)(x^2-x+1)} = \dfrac{3x+3}{x^3+1}$
