Đáp án + Giải thích các bước giải:

`e)` Điều kiện : `x ne -4`

`5/(x+4)=>x+4inƯ(5)={pm1;pm5}`

Lập bảng :

$\begin{array}{|c|c|c||c|}\hline \text{x+4}&\text{1}&\text{-1}&\text{5}&\text{-5}\\\hline \text{x}&\text{-3}&\text{-5}&\text{1}&\text{-9}\\\hline \end{array}$ `(tm)`

Vậy `x in {-3;-5;1;-9}`

`g)` Điều kiện : ` x ne 3`

`(-2)/(x-3)=>x-3inƯ(-2)={pm1;pm2}`

Lập bảng :

$\begin{array}{|c|c|c||c|}\hline \text{x-3}&\text{1}&\text{-1}&\text{2}&\text{-2}\\\hline \text{x}&\text{4}&\text{2}&\text{5}&\text{1}\\\hline \end{array}$ `(tm)`

Vậy `x in {4;2;5;1}`

`h)` Điều kiện : `x ne 5`

`(x+7)/(x-5)=(x-5+12)/(x-5)=1+12/(x-5)` $\\$ `x - 5 in Ư(12)={pm1;pm2;pm3;pm4;pm6;pm12}`

Lập bảng :

$\begin{array}{|c|c|c||c|}\hline \text{x-5}&\text{1}&\text{-1}&\text{2}&\text{-2}&\text{3}&\text{-3}&\text{4}&\text{-4}&\text{6}&\text{-6}&\text{12}&\text{-12}\\\hline \text{x}&\text{6}&\text{4}&\text{7}&\text{3}&\text{8}&\text{2}&\text{9}&\text{1}&\text{11}&\text{-1}&\text{17}&\text{-7}\\\hline \end{array}$ `(tm)`

Vậy `x in {6;4;7;3;8;2;9;1;11;-1;17;-7}`

`i)` Điều kiện : `x ne -1`

`(2x - 1)/(x + 1)=[2(x+1)-3]/(x+1)=2-3/(x+1)`

`=>(x+1)inƯ(-3)={pm1;pm3}`

Lập bảng :

$\begin{array}{|c|c|c||c|}\hline \text{x+1}&\text{1}&\text{-1}&\text{3}&\text{-3}\\\hline \text{x}&\text{0}&\text{-2}&\text{2}&\text{-4}\\\hline \end{array}$ `(tm)` 

Vậy `x in {0;-2;2;-4}`

`k)` Điều kiện : `x ne -1/2`

`(4x - 5)/(2x + 1)=[2(2x+1)-7]/(2x+1)=2-7/(2x+1)`

`=> 2x+1inƯ(-7)={pm1;pm7}`

Lập bảng :

$\begin{array}{|c|c|c||c|}\hline \text{2x+1}&\text{1}&\text{-1}&\text{7}&\text{-7}\\\hline \text{x}&\text{0}&\text{-1}&\text{3}&\text{-4}\\\hline \end{array}$ `(tm)` 

Vậy `x in {0;-1;3;-4}`

`l)` Điều kiện : `x ne -1/2`

`(x - 1)/(2x +1)=[4(x-1)]/[2(2x+1)]=(4x-4)/(4x+2)=(4x+2-6)/(4x+2)=1-6/(4x+2)` $\\$ `4x + 2 in Ư(-6)={pm1;pm2;pm3;pm6}`

Lập bảng :$\begin{array}{|c|c|c||c|}\hline \text{4x+2}&\text{1}&\text{-1}&\text{2}&\text{-2}&\text{3}&\text{-3}&\text{6}&\text{-6}\\\hline \text{x}&\text{-1/4}&\text{-3/4}&\text{0}&\text{-1}&\text{1/4}&\text{-5/4}&\text{1}&\text{-2}\\\hline \end{array}$ 

Vậy `x in {0;-1;1;-2}`