Help me plssss mn ơi!!!!!!!!!!

$a)$ Thay $x=-5$ vào biểu thức $B$, có:

$\dfrac{2x-2}{5}$

$=\dfrac{2.(-5)-2}{5}$

$=\dfrac{(-10)-2}{5}$

$=\dfrac{-12}{5}$

Vậy khi $x=5$ thì biểu thức $B=\dfrac{-12}{5}$

$b)A=\dfrac{2x}{x-3}-\dfrac{1}{x+3}+\dfrac{2x^2-12}{9-x^2}$

$⇒ \dfrac{2x}{x-3}-\dfrac{1}{x+3}-\dfrac{2x^2-12}{x^2-9}$

$⇒ \dfrac{2x(x+3)}{(x-3)(x+3)}-\dfrac{1(x-3)}{(x+3)(x-3)}-\dfrac{2x^2-12}{(x-3)(x+3)}$

$⇒ \dfrac{2x^2+6x}{(x-3)(x+3)}-\dfrac{x-3}{(x-3)(x+3)}-\dfrac{2x^2-12}{(x-3)(x+3)}$

$⇒ \dfrac{2x^2+6x-(x-3)-(2x^2-12)}{(x-3)(x+3)}$

$⇒ \dfrac{2x^2+6x-x+3-2x^2+12}{(x-3)(x+3)}$

$⇒ \dfrac{5x+15}{(x-3)(x+3)}$

$⇒ \dfrac{5(x+3)}{(x-3)(x+3)}$

$⇒ \dfrac{5}{x-3}$

$c)M=A.B$

$⇒ M=\dfrac{5}{x-3}.\dfrac{2x-2}{5}$

$⇒ M=\dfrac{5.(2x-2)}{(x-3).5}$

$⇒ M=\dfrac{2x-2}{x-3}$

$⇒ M=\dfrac{2x-6+4}{x-3}$

$⇒ M=\dfrac{2(x-3)+4}{x-3}$

$⇒ M=2+\dfrac{4}{x-3}$

Để $A∈Z$

$⇔ x-3 ∈ Ư(4)=$ {$±1;±2;±4$}

Vậy để $M=A.B$ là số nguyên thì $x∈${$±1;±2;±4$}

$#mie1208$