ĐK: $x\ge 0$
$A-B\\=\dfrac{1}{\sqrt x+1}-\dfrac{1-\sqrt x}{1+\sqrt x}\\=\dfrac{1-1+\sqrt x}{\sqrt x+1}\\=\dfrac{\sqrt x}{\sqrt x+1}\\→M=\dfrac{\sqrt x}{\sqrt x+1}=\dfrac{\sqrt x+1-1}{\sqrt x+1}$
Để $M∈\Bbb Z$
$→\dfrac{\sqrt x+1-1}{\sqrt x+1}∈\Bbb Z\\→1-\dfrac{1}{\sqrt x+1}∈\Bbb Z\\→\dfrac{1}{\sqrt x+1}∈\Bbb Z\\→1\vdots \sqrt x+1\\↔\sqrt x+1∈Ư(1)=\{±1\}$
Vì $\sqrt x\ge 0$
$→\sqrt x+1\ge 1\\→\sqrt x+1>0$
mà $\sqrt x+1∈\{±1\}$
$→\sqrt x+1=1\\↔\sqrt x=0\\↔x=0(TM)$
Vậy x=0 thì M có giá trị nguyên
