Question

Answer 1

a, đk : x khác 2 ; -2 ; -3 

\(A=\left(\dfrac{x+x+2}{\left(x-2\right)\left(x+2\right)}\right).\dfrac{x^2-4}{x+3}=\dfrac{2x+2}{x+3}\)

Vậy ta có đpcm 

b, \(A=\dfrac{2x+2}{x+3}=\dfrac{2\left(x+3\right)-4}{x+3}=2-\dfrac{4}{x+3}\Rightarrow x+3\inƯ\left(-4\right)=\left\{\pm1;\pm2;\pm4\right\}\)

x + 3	1	-1	2	-2	4	-4
x	-2	-4	-1	-5	1	-7

 

Answer 2

a.A=\(\left(\dfrac{x}{x^2-4}+\dfrac{1\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\right).\dfrac{x^2-4}{x+3}\)

   A=\(\dfrac{x+x+2}{x^2-4}.\dfrac{x^2-4}{x+3}\)

A=\(\dfrac{2x+2}{x+3}\) ( đfcm)

 

Answer 3