`a)`
`A= ((2+x)/(2-x) - (4x^2)/(x^2 -4) - (2-x)/(2+x)) : (x^2 - 3x)/(2x^2- x^3) (ĐKXĐ : x \ne +-2 ; x \ne 0 ; x \ne 3)`
`=(( -x-2)/(x-2) - (4x^2)/(x^2 -4) - (2-x)/(2+x)) : (x. (x-3))/(x^2 . (2-x))`
` = (((-x-2). (x+2))/(x^2 -4) - (4x^2)/(x^2-4) - ((2-x)(x-2))/(x^2 -4)) . (x^2 . (2-x))/(x.(x-3))`
` = ((-x-2).(x+2) - 4x^2 - (2-x).(x-2)) / (x^2-4) . (x^2 . (2-x))/(x.(x-3))`
` = (-x^2 - 2x - 2x- 4 - 4x^2 + x^2 - 4x + 4)/((x-2).(x+2)) . (x^2 . (2-x))/(x.(x-3))`
` = (-4x^2 - 8x )/(x-2)(x+2) . (x^2 . (2-x))/(x.(x-3))`
`= (-4x . (x+2))/(x-2)(x+2) . (x^2 . (2-x))/(x.(x-3))`
`= (-4x)/(x-2) . (x^2 . (2-x))/(x.(x-3))`
` = (4x)/(2-x) . (x^2 . (2-x))/(x.(x-3))`
` = (4x^2)/(x-3)`
``
`b)`
`|x-5| = 2`
`=> x - 5 = 2` hoặc `x - 5 =-2`
`+) x - 5 =2 => x = 7 (TMĐKXĐ)`
`+) x - 5 = -2 => x = 3 (KTMĐKXĐ)`
Với `x = 7 (TMĐKXĐ)` thì ta có :
`A = (4 . 7^2)/(7-3)`
` = 196/4`
`= 49`
``
`c)`
Để `A \vdots 4` thì `(4x^2)/(x-3) \vdots 4` và `x \in ZZ; x \ne +-2 ; x \ne 0 ; x \ne 3`
`<=> 4 . (x^2)/(x-3) \vdots 4`
`<=> (x^2)/(x-3) \in ZZ (do\ 4 vdots 4)`
`<=> x^2 \vdots x - 3`
`<=> x. (x-3) + 3 (x - 3) + 9 \vdots x -3`
`<=> 9 \vdots x-3`
`<=> x - 3 \in Ư(9)`
`<=> x - 3 \in{ 1 ; -1 ; 3 ; -3 ; 9 ;-9}`
`<=> x \in{ 4 ; 2 ; 6 ; 0 ; 12 ; -6}`
Mà `x \ne +-2 ; x \ne 0 ; x \ne 3` nên `x \in { 4 ; 6 ; 12 ; -6 }`
Vậy với `x \in { 4 ; 6 ; 12 ; -6 }` thì `A \vdots 4`
