a/ ĐKXĐ: $x\ge 0,x\ne 1$
$A\,=\dfrac{\sqrt x-1}{x-1}+\dfrac{\sqrt x}{\sqrt x+1}\\\quad =\dfrac{\sqrt x-1}{(\sqrt x-1)(\sqrt x+1)}+\dfrac{\sqrt x(\sqrt x-1)}{(\sqrt x-1)(\sqrt x+1)}\\\quad =\dfrac{\sqrt x-1+x-\sqrt x}{(\sqrt x-1)(\sqrt x+1)}\\\quad =\dfrac{x-1}{x-1}\\\quad =1$
Vậy $A=1$ với $x\ge 0,x\ne 1$
b/ Vì $A=1$ với $x\ge 0,x\ne 1$
$→x=\dfrac{1}{4}(TM)$ thì $A=1$
Vậy $A=1$ với $x=\dfrac{1}{4}$