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Đáp án `+` Giải thích các bước giải:

`a,`

`A = ( (4x)/(x^2 - 4) + (2x-4)/(x+2) ) . (x+2)/(2x) +  2/(2-x)` `(x \ne 0; x\ne +-2)`

`=[ (4x)/( (x-2)(x+2) ) + ( 2(x-2)^2 )/( (x-2)(x+2) ) ] . (x+2)/(2x) + 2/(2-x)`

`=( 4x + 2(x^2 - 4x + 4) )/( (x-2)(x+2) ) . (x+2)/(2x) + 2/(2-x)`

`=( 4x + 2x^2 - 8x + 8)/( (x-2)(x+2) ) . (x+2)/(2x) + 2/(2-x)`

`=(2x^2 - 4x + 8)/( (x-2)(x+2) ) . (x+2)/(2x) + 2/(2-x)`

`=( 2(x^2  - 2x + 4) )/( (x-2)(x+2) ) . (x+2)/(2x) + 2/(2-x)`

`=(x^2 - 2x + 4)/( x(x-2) ) + 2/(2-x)`

`=(x^2 - 2x + 4)/( x(x-2) ) - (2x)/(x(x-2) )`

`=(x^2 - 2x + 4 - 2x)/( x(x-2) )`

`=(x^2 -4x+4)/( x(x-2))`

`=(x-2)^2/(x(x-2))`

`=(x-2)/x`

`b`

Tại `x=4` (TMĐKXĐ) ta có:

`(4-2)/4 = 2/4=1/2`

`c,`

`A = (x-2)/x=x/x-2/x=1 - 2/x`

Để `A in ZZ` thì `1 - 2/x in ZZ`

Hay `2/x in ZZ`

`=>x in Ư(2)`

`=> x in {-2;-1;1;2}`

Mà `x \ne +-2`

`=> x in {+-1}`

Vậy `...`