a/ ĐKXĐ: $x\ge 0,x\ne 1$
$A=\left(\dfrac{1}{\sqrt x -1}+\dfrac{1}{\sqrt x +1}\right)\left(\dfrac{x-1}{\sqrt x -1}-2\right)\\=\left(\dfrac{\sqrt x +1}{(\sqrt x- 1)(\sqrt x +1)}+\dfrac{\sqrt x -1}{(\sqrt x -1)(\sqrt x +1)}\right)\left(\dfrac{(\sqrt x -1)(\sqrt x +1)}{\sqrt x -1}-2\right)\\=\dfrac{\sqrt x+1+\sqrt x-1}{(\sqrt x -1)(\sqrt x +1)}(\sqrt x +1-2)\\=\dfrac{2\sqrt x}{(\sqrt x -1)(\sqrt x +1)}.(\sqrt x -1)\\=\dfrac{2\sqrt x}{\sqrt x +1}$
b/ $A∈\Bbb Z\\→\dfrac{2\sqrt x}{\sqrt x +1}∈\Bbb Z\\→\dfrac{2\sqrt x +2-2}{\sqrt x +1}∈\Bbb Z\\→\dfrac{2(\sqrt x +1)-2}{\sqrt x +1}∈\Bbb Z\\→2-\dfrac{2}{\sqrt x +1}∈\Bbb Z\\→\dfrac{2}{\sqrt x +1}∈\Bbb Z\\→2\vdots \sqrt x +1\\→\sqrt x +1∈Ư(2)=\{±1;±2\}$
mà $\sqrt x +1>0$
$→\sqrt x +1∈\{1;2\}\\↔\sqrt x ∈\{0;1\}\\↔x∈\{0;1\}$
mà $x\ne 1$
$→x=0$
Vậy $x=0$
