Đáp án:
Giải thích các bước giải:
$a) (x+\dfrac{1}{x^2}-2x-8)(4-\dfrac{x}{x^2}+x)$
$= x(4-\dfrac{x}{x^2}+x)+\dfrac{1}{x^2}(4-\dfrac{x}{x^2}+x)-2x(4-\dfrac{x}{x^2}+x)-8(4-\dfrac{x}{x^2}+x)$
$=4x-x.\dfrac{x}{x^2}+x^2+\dfrac{1}{x^2}.4-\dfrac{1}{x^2}.\dfrac{x}{x^2}+\dfrac{1}{x^2}.x-2x.4+2x.\dfrac{x}{x^2}-2x.x-8.4+8.\dfrac{x}{x^2}-8.x$
$=4x-1+x^2+\dfrac{4}{x^2}-\dfrac{1}{x^3}+\dfrac{1}{x}-8x+2-2x^2-32+\dfrac{8}{x}-8x$
$=-\dfrac{1}{x^3}-12x-x^2+\dfrac{4}{x^2}+\dfrac{9}{x^3}-33$
$a) \dfrac{x+1}{x^2-2x-8} . \dfrac{4-x}{x^2+x}$
$= \dfrac{(x+1)(4-x)}{(x^2-2x-8)(x^2+x)}$
$= \dfrac{(x+1)(4-x)}{x(x^2-2x-8)(x+1)}$
$= \dfrac{4-x}{x(x^2-2x-8)}$
$= \dfrac{4-x}{x(x^2+2x-4x-8}$
$= \dfrac{4-x}{x[x(x+2)-4(x+2)]}$
$= \dfrac{4-x}{x(x+2)(x-4)}$
$= -\dfrac{4-x}{x(x+2)(4-x)}$
$= -\dfrac{1}{x(x+2)}$
$= -\dfrac{1}{x^2+2x}$
$\text { Do không biết đề nào đúng nên làm cả 2 }$
