a, Để A xác định
$⇔x\neq\pm1,x\neq0$
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}`.
