a) K = ($\frac{\sqrt{a}}{\sqrt{a} - 1}$ - $\frac{1}{a - \sqrt{a}}$):($\frac{1}{\sqrt{a} + 1}$ + $\frac{2}{a - 1}$) Đkxđ: x > 0; x $\neq$ 1
= $\frac{\sqrt{a}.\sqrt{a} - 1}{\sqrt{a}.(\sqrt{a} - 1)}$ : $\frac{1.(\sqrt{a} - 1) + 2}{(\sqrt{a} - 1)(\sqrt{a} + 1)}$
= $\frac{a - 1}{\sqrt{a}.(\sqrt{a} - 1)}$ . $\frac{(\sqrt{a} - 1)(\sqrt{a} + 1)}{\sqrt{a} - 1 + 2}$
= $\frac{(\sqrt{a} - 1)(\sqrt{a} + 1)}{\sqrt{a}.(\sqrt{a} - 1)}$ . $\frac{(\sqrt{a} + 1)(\sqrt{a} - 1)}{\sqrt{a} + 1}$
= $\frac{\sqrt{a} + 1}{\sqrt{a}}$ . ($\sqrt{a}$ - 1)
= $\frac{(\sqrt{a} + 1)(\sqrt{a} - 1)}{\sqrt{a}}$
= $\frac{a - 1}{\sqrt{a}}$
b) Có: a = 3 + 2$\sqrt{2}$
= 2 + 2$\sqrt{2}$ + 1
= ($\sqrt{2}$ + 1)² (Thoả mãn)
Thay a = ($\sqrt{2}$ + 1)² vào K ta được:
K = $\frac{(\sqrt{2} + 1)² - 1}{\sqrt{(\sqrt{2} + 1)²} }$
= $\frac{(\sqrt{2} + 1 - 1)(\sqrt{2} + 1 + 1)}{|\sqrt{2} + 1|}$
= $\frac{\sqrt{2}.(\sqrt{2} + 2)}{\sqrt{2} + 1}$
= $\frac{2 + 2\sqrt{2}}{\sqrt{2} + 1}$
= $\frac{(2 + 2\sqrt{2})(\sqrt{2} - 1)}{(\sqrt{2} + 1)(\sqrt{2} - 1)}$
= $\frac{2\sqrt{2} - 2 + 4 - 2\sqrt{2}}{2 - 1}$
= 2
Vậy K = 2 khi x = 3 + 2$\sqrt{2}$
