`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`

Đáp án:

 

Giải thích các bước giải:

 `a;A=[3x-2]/x -[x-7]/[x-5]-[10]/[x^2 -5x](đkxđ:x\ne 0 ;x\ne 5)`

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

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

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

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

`=2`

`b;`

`B=A.[x+1]/[x-1]`

`=[2x+2]/[x-1]`

`=[2(x-1)+4]/[x-1]`

`=2+4/[x-1]`

`B in ZZ`

`<=>4/[x-1] in ZZ`

`<=>x-1 in Ư (4)`

`<=>x in {+-1;+-2;+-4}`

`<=>x in {0(l);2(tm);-1(tm);3(tm);-3(tm);5(l)}`

Vậy`B in ZZ <=>x in {-1;3;-3;2}`