a)  Ta có:   \(A=\dfrac{x}{x+2}-\dfrac{2x}{x-2}+\dfrac{x^2+12}{x^2-4}\left(x\ne\pm2\right)\)

\(A=\dfrac{x\left(x-2\right)-2x\left(x+2\right)+x^2+12}{\left(x-2\right)\left(x+2\right)}\)

\(A=\dfrac{x^2-2x-2x^2-4x+x^2+12}{\left(x-2\right)\left(x+2\right)}\)

\(A=\dfrac{-6\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\)

\(A=\dfrac{-6}{x+2}\)

b) Để A có giá trị nguyên thì \(x+2\inƯ\left(6\right)\)

Mà \(Ư\left(6\right)=\left\{\pm1;\pm2;\pm3;\pm6\right\}\)

Từ đó, ta có:

\(x+1=1\Leftrightarrow x=0\) ( nhận )

\(x+1=-1\Leftrightarrow x=-2\)  ( loại )

\(x+1=2\Rightarrow x=1\) ( nhận )

\(x+1=-2\Rightarrow x=-3\) ( nhận )

\(x+1=3\Rightarrow x=2\) ( loại )

\(x+1=-3\Rightarrow x=-4\) ( nhận )

\(x+1=6\Rightarrow x=5\) ( nhận )

\(x+1=-6\Rightarrow x=-7\) ( nhận )

Vậy để A nhận giá trị nguyên thì \(x\in\left\{-7;-4;-3;0;1;5\right\}\)

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

a, \(\dfrac{x}{x+2}\) + \(\dfrac{2x}{x-2}\) -\(\dfrac{3x^2-4}{x^2-4}\)

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

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

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

= \(\dfrac{2x-4}{\left(x+2\right)\left(x-2\right)}=\dfrac{2\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{2}{x+2}\)

Có vài bước mình làm tắc á nha :>

\(a,A=\dfrac{x^2-3x+2+x^2+3x+2-x^2+2x-4}{\left(x+2\right)\left(x-2\right)}=\dfrac{x^2+2x}{\left(x+2\right)\left(x-2\right)}\\ A=\dfrac{x\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{x}{x-2}\\ b,A=\dfrac{x-2+2}{x-2}=1+\dfrac{2}{x-2}\in Z\\ \Rightarrow x-2\inƯ\left(2\right)=\left\{-2;-1;1;2\right\}\\ \Rightarrow x\in\left\{0;1;3;4\right\}\)

\(a,A=\dfrac{x^2-6x+9-x^2+9}{x\left(x-3\right)}\cdot\dfrac{x}{2\left(x-1\right)}\\ A=\dfrac{-6x+18}{2\left(x-3\right)\left(x-1\right)}=\dfrac{-6\left(x-3\right)}{2\left(x-3\right)\left(x-1\right)}=\dfrac{-3}{x-1}\\ b,A\in Z\Leftrightarrow x-1\inƯ\left(-3\right)=\left\{-3;-1;1;3\right\}\\ \Leftrightarrow x\in\left\{-2;0;2;4\right\}\)

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

a: \(A=\dfrac{x+2+x^2-2x+x-2}{\left(x-2\right)\left(x+2\right)}=\dfrac{x^2-2x}{\left(x-2\right)\left(x+2\right)}=\dfrac{x}{x+2}\)

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

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