1) $$\frac{a - 64}{\sqrt{a} - 8} = \frac{(\sqrt{a} - 8)(\sqrt{a} + 8)}{\sqrt{a} - 8} = \sqrt{a} + 8$$ при условии $$a \ge 0$$ и $$a
e 64$$
2) $$\frac{\sqrt{11} - 11}{\sqrt{11}} = \frac{\sqrt{11} - (\sqrt{11})^2}{\sqrt{11}} = \frac{\sqrt{11}(1 - \sqrt{11})}{\sqrt{11}} = 1 - \sqrt{11}$$
3) $$\frac{a - 5}{a + 2\sqrt{5a} + 5} = \frac{(\sqrt{a} - \sqrt{5})(\sqrt{a} + \sqrt{5})}{(\sqrt{a} + \sqrt{5})^2} = \frac{\sqrt{a} - \sqrt{5}}{\sqrt{a} + \sqrt{5}}$$ при условии $$a \ge 0$$
Ответ: 1) $$\sqrt{a} + 8$$; 2) $$1 - \sqrt{11}$$; 3) $$\frac{\sqrt{a} - \sqrt{5}}{\sqrt{a} + \sqrt{5}}$$.