a,
ĐK: $x\ge 0; x\ne 4$
b,
$A=\dfrac{\sqrt{x}+1}{\sqrt{x}-2}+\dfrac{2\sqrt{x}}{\sqrt{x}+2}-\dfrac{5\sqrt{x}+2}{(\sqrt{x}-2)(\sqrt{x}+2)}$
$=\dfrac{ (\sqrt{x}+1)(\sqrt{x}+2)+2\sqrt{x}(\sqrt{x}-2)-5\sqrt{x}-2}{(\sqrt{x}-2)(\sqrt{x}+2)}$
$=\dfrac{x+3\sqrt{x}+2+2x-4\sqrt{x}-5\sqrt{x}-2}{(\sqrt{x}-2)(\sqrt{x}+2)}$
$=\dfrac{3x-6\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}+2)}$
$=\dfrac{3\sqrt{x}(\sqrt{x}-2)}{(\sqrt{x}-2)(\sqrt{x}+2)}$
$=\dfrac{3\sqrt{x}}{\sqrt{x}+2}$
c,
Để $A=2$:
$\dfrac{3\sqrt{x}}{\sqrt{x}+2}=2$
$\to 3\sqrt{x}=2(\sqrt{x}+2)$
$\to \sqrt{x}=4$
$\to x=16$ (TM)
Vậy $x=16$