a) A =\(\dfrac{x}{x+2}-\dfrac{2x}{x-2}+\dfrac{x^2+12}{x^2-4}\)
A=\(\dfrac{x}{x+2}-\dfrac{2x}{x-2}+\dfrac{x^2+12}{\left(x-2\right)\left(x+2\right)}\) (MTC:(x+2)(x-2)
A=\(\dfrac{x\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}-\dfrac{2x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{x^2+12}{\left(x-2\right)\left(x+2\right)}\)
A=\(\dfrac{x\left(x-2\right)-2x\left(x+2\right)+x^2+12}{\left(x-2\right)\left(x+2\right)}\)
A=\(\dfrac{x^2-2x-2x^2-4x+x^2+12}{\left(x-2\right)\left(x+2\right)}\)
A=\(\dfrac{-6x+12}{\left(x-2\right)\left(x+2\right)}\)
A=\(\dfrac{-6\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\)
A=\(\dfrac{-6}{x+2}\)
b)Để A có kết quả là số nguyên thì \(\dfrac{-6}{x+2}\) phải là số nguyên
=>(x+2)=Ư(-6)={-1;-2;-3;-6;1;2;3;6}
Với x+2=-1 ➜x=-3
x+2=-2 ➜x=-4
x+2=-3 ➜x=-5
x+2=-6 ➜x=-8
x+2=1 ➜x=-1
x+2=2 ➜x=0
x+2=3 ➜x=1
x+2=6 ➜x=3
Vậy x={-3;-4;-5;-8;-1;0;1;3}