a, Để A xác định

$⇔\begin{cases}x-1\neq0\\x\neq0\\x+1\neq0\end{cases}$

$⇔\begin{cases}x\neq1\\x\neq0\\x\neq-1\end{cases}$

$⇒\begin{cases}x\neq\pm1\\x\neq0\end{cases}$

Vậy để A xác định thì $x\neq\pm1,x\neq0$.

b, $A=(\dfrac{x}{x-1}-\dfrac{x+1}{x}):(\dfrac{x}{x+1}-\dfrac{x-1}{x})$      $(x\neq\pm1,x\neq0)$

$A=(\dfrac{x^2}{x(x-1)}-\dfrac{(x-1)(x+1)}{x(x-1)}):(\dfrac{x^2}{x(x+1)}-\dfrac{(x-1)(x+1)}{x(x+1)})$

$A=\dfrac{x^2-x^2+1}{x(x-1)}:\dfrac{x^2-x^2+1}{x(x+1)}$

$A=\dfrac{1}{x(x-1)}\cdot\dfrac{x(x+1)}{1}$

$A=\dfrac{x+1}{x-1}$

Vậy $A=\dfrac{x+1}{x-1}$ với $x\neq\pm1,x\neq0$

c, $A=\dfrac{x+1}{x-1}=\dfrac{x-1+2}{x-1}=1+\dfrac{2}{x-1}$

Để A nguyên

$\dfrac{2}{x-1}$ nguyên

`⇒x-1∈Ư(2)={±1;±2}`.

· $x-1=1⇒x=2(tm)$

· $x-1=-1⇒x=0(ktm)$

· $x-1=2⇒x=3(tm)$

· $x-1=-2⇒x=-1(ktm)$

Vậy để A nguyên thì `x\in{2;3}`.

Đáp án:

$\begin{array}{l}
A = \left( {\frac{x}{{x - 1}} - \frac{{x + 1}}{x}} \right):\left( {\frac{x}{{x + 1}} - \frac{{x - 1}}{x}} \right)\\
a)Đkxđ:\left\{ \begin{array}{l}
x - 1 \ne 0\\
x \ne 0\\
x + 1 \ne 0
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
x \ne 1\\
x \ne 0\\
x \ne  - 1
\end{array} \right.\\
b)A = \left( {\frac{x}{{x - 1}} - \frac{{x + 1}}{x}} \right):\left( {\frac{x}{{x + 1}} - \frac{{x - 1}}{x}} \right)\\
 = \frac{{{x^2} - \left( {x - 1} \right)\left( {x + 1} \right)}}{{x\left( {x - 1} \right)}}:\frac{{{x^2} - \left( {x - 1} \right)\left( {x + 1} \right)}}{{x\left( {x + 1} \right)}}\\
 = \frac{{{x^2} - {x^2} + 1}}{{x\left( {x - 1} \right)}}.\frac{{x\left( {x + 1} \right)}}{{{x^2} - {x^2} + 1}}\\
 = \frac{{x + 1}}{{x - 1}}\\
c)Đkxđ:x \ne  - 1;x \ne 0;x \ne 1\\
A = \frac{{x + 1}}{{x - 1}} = \frac{{x - 1 + 2}}{{x - 1}} = 1 + \frac{2}{{x - 1}}\\
A \in Z \Rightarrow \frac{2}{{x - 1}} \in Z\\
 \Rightarrow 2 \vdots \left( {x - 1} \right)\\
 \Rightarrow \left( {x - 1} \right) \in Ư\left( 2 \right) = {\rm{\{ }} - 2; - 1;1;2\} \\
 \Rightarrow x \in {\rm{\{ }} - 1;0;2;3\} \\
Mà:\,x \ne  - 1;x \ne 1;x \ne 0\\
 \Rightarrow x \in {\rm{\{ }}0;2;3\} 
\end{array}$