a. $x = 6 - 2\sqrt{5} = (\sqrt{5})^2 - 2\sqrt{5} + 1 = (\sqrt{5} - 1)^2$
Khi đó:
$A = \dfrac{1 - \sqrt{(\sqrt{5} - 1)^2}}{1 + \sqrt{(\sqrt{5} + 1)^2}} = \dfrac{1 - \sqrt{5} + 1}{1 + \sqrt{5} + 1} = \dfrac{2 - \sqrt{5}}{2 + \sqrt{5}}$
b. ĐKXĐ: $x \geq 0$; $x \neq 25$
$B = (\dfrac{15 - \sqrt{x}}{(\sqrt{x} - 5)(\sqrt{x} + 5)} + \dfrac{2(\sqrt{x} - 5)}{(\sqrt{x} - 5)(\sqrt{x} + 5)}]\dfrac{\sqrt{x} - 5}{\sqrt{x} + 1}$
$B = \dfrac{15 - \sqrt{x} + 2\sqrt{x} - 10}{(\sqrt{x} - 5)(\sqrt{x} + 5)}.\dfrac{\sqrt{x} - 5}{\sqrt{x} + 1}$
$B = \dfrac{\sqrt{x} + 5}{\sqrt{x} + 5}.\dfrac{1}{\sqrt{x} + 1}$
$B = \dfrac{1}{\sqrt{x} + 1}$
