Giúp tôi giải bài 8 nhé

`a)` 

`A = (4)/(x+1) - 1` với `x \ne -1`

Thay `x = 1` `(\text{TMĐK})` vào `A`, ta có: 

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

Vậy `A = 1` khi `x = 1`

`b)` 

`A = (3-x)/(x+1)`; `B = ((3-x)(3+x))/((x+1)^2)` với `x \ne \pm 3; x \ne -1`

`P = (A)/(B)`

`= (3-x)/(x+1) : ((3-x)(3+x))/((x+1)^2)`

`= (3-x)/(x+1) . ((x+1)^2)/((3-x)(3+x))`

`= (x+1)/(x+3)`

Vậy `P = (x+1)/(x+3)` với `x \ne \pm 3; x \ne -1`

`c)`

`P = (x+1)/(x+3)` với `x \ne \pm 3; x \ne -1`

`P = 1 + (-2)/(x+3)`

Để `P \in \mathbb{Z}` thì `1 + (-2)/(x+3) \in \mathbb{Z}`

Mà `1 \in \mathbb{Z}`, `x \in \mathbb{Z}`

`-> (-2)/(x+3) \in \mathbb{Z}`

`-> x + 3 \in Ư(-2) = {\pm 1; \pm 2}`

Lập bảng

$$\begin{array}{|c|c|c|c|c|}\hline x+3&-2&-1&1&2\\\hline x&-5&-4&-2&-1\\\hline \text{KTĐK}&\text{TM}&\text{TM}&\text{TM}&\text{KTM}\\\hline P&2&3&-1&\text{/}\\\hline \text{KTĐK}&\text{TM}&\text{TM}&\text{TM}&\text{KTM}\\\hline\end{array}$$

Vậy $x \in \{-5; -4; -2\}$ thì $P \in \mathbb{Z}$

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