Làm nhanh hộ mk ak mk cần gấp

`a)``A=(3x-2)/(x)-(x-7)/(x-5)-(10)/(x^2-5x)(xne0;xne5)`

`=(3x-2)/(x)-(x-7)/(x-5)-(10)/(x(x-5))`

`=((3x-2)(x-5)-x(x-7)-10)/(x(x-5))`

`=(3x^2-15x-2x+10-x^2+7x-10)/(x(x-5))`

`=(2x^2-10x)/(x(x-5))`

`=(2x(x-5))/(x(x-5))`

`=2`

Vậy `A=2` khi `xne0;xne5`

`b)``B=A.(x+1)/(x-1)=2.(x+1)/(x-1)=(2(x+1))/(x-1)=(2x+2)/(x-1)=2+4/(x-1)`

Để `B in ZZ` thì `4/(x-1) in ZZ`

`<=>4 vdots x-1`

`<=>x-1 in Ư(4)={+-1;+-2;+-4}`

Lập bảng :

$\begin{array}{|c|c|c|}\hline\text{x- 1}&\text{-4}&\text{-2}&\text{-1}&\text{1}&\text{2}&\text{4}\\\hline\text{x}&\text{-3}&\text{-1}&\text{0}&\text{2}&\text{3}&\text{5}\\\hline\text{ĐCĐK}&\text{TM}&\text{TM}&\text{KTM}&\text{TM }&\text{TM}&\text{KTM}\\\hline\end{array}$

Vậy `x in {-3;-1;2;3}` thì `Bin ZZ`