Đáp án:

 

 

Bài 2:

a. 

`x\ne +-1`

b.

\(P=(\dfrac{x+1}{x-1}-\dfrac{4x}{x^2-1})÷\dfrac{x-1}{2}\\P=(\dfrac{x+1}{x-1}-\dfrac{4x}{(x-1)(x+1)})\times \dfrac{2}{x-1}\\P= \dfrac{(x+1)^2-4x}{(x-1)(x+1)}\times \dfrac{2}{x-1}\\P= \dfrac{x^2 +2x+1-4x}{(x-1)(x+1)}\times \dfrac{2}{x-1}\\P= \dfrac{x^2-2x+1}{(x-1)(x+1)}\times \dfrac{2}{x-1}\\P=\dfrac{(x-1)^2}{(x-1)(x+1)}\times \dfrac{2}{x-1}\\P = \dfrac{2}{x+1} \)

c.

Để `P \in \mathbb{Z}` thì `x+1 \vdots 2`

Hay `x+1 \in Ư(2)={+-1,+-2}`

Ta có bảng:

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

Kết hợp điều kiện, ta thấy `x=1` không thỏa mãn

Vậy `x\in {-2,-3,0}` thì `P \in \mathbb{Z}`

Bài 3:

a.

`x\ne +-1, \quad x \ne 0`

b.

\(P=(\dfrac{2+x}{x^2 +2x+1}-\dfrac{x-2}{x^2-1})\times \dfrac{1-x^2}{x}\\P= (\dfrac{2+x}{(x+1)^2}-\dfrac{x-2}{(x-1)(x+1)}\times \dfrac{(1-x)(1+x)}{x}\\P= \dfrac{(x-1)(2+x)-(x+1)(x-2)}{(x-1)(x+1)^2}\times \dfrac{-(x-1)(x+1)}{x}\\P=\dfrac{2x+x^2 -2-x-(x^2-2x+x-2)}{(x-1)(x+1)}\times \dfrac{-(x-1)}{x}\\P =\dfrac{2x}{(x-1)(x+1)}\times \dfrac{-(x-1)}{x}\\P = -\dfrac{2}{x+1}\)

 c.

Để `P \in \mathbb{Z}` thì `x+1 \vdots -2`

Hay `x+1 \in Ư(-2)={-2,-1,1,2}`

Ta có bảng:

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

Kết hợp điều kiện, ta thấy `x=1` không thỏa mãn

Vậy `x\in {-3,-2,0}` thì \(P\in \mathbb{Z}\)