Đáp án:
$a)\dfrac{A}{x(x-y)^2}\\ b)\dfrac{A}{(5-x^2)(4x-3)}$
Giải thích các bước giải:
$a)=\dfrac{y-x}{-x}\\ =\dfrac{x-y}{x}\\ =\dfrac{(x-y)(x-y)^2}{x(x-y)^2}\\ =\dfrac{(x-y)^3}{x(x-y)^2}\\ =\dfrac{x^3-3x^2y+3xy^2-y^3}{x(x-y)^2}\\ =\dfrac{A}{x(x-y)^2}\\ b)\dfrac{3-4x}{x^2-5}\\ =\dfrac{4x-3}{5-x^2}\\ =\dfrac{(4x-3)(4x+3)}{(5-x^2)(4x-3)}\\ =\dfrac{16x^2-9}{(5-x^2)(4x-3)}\\ =\dfrac{A}{(5-x^2)(4x-3)}$