Ответы к странице 141

557. Освободитесь от иррациональности в знаменателе дроби:
1) $\frac{\sqrt{2}}{\sqrt{2} + 1}$;
2) $\frac{4}{\sqrt{7} + \sqrt{3}}$;
3) $\frac{15}{\sqrt{15} - \sqrt{12}}$;
4) $\frac{19}{2\sqrt{5} - 1}$;
5) $\frac{1}{\sqrt{a} - \sqrt{b}}$;
6) $\frac{\sqrt{3} + 1}{\sqrt{3} - 1}$.

Решение:

1) $\frac{\sqrt{2}}{\sqrt{2} + 1} = \frac{\sqrt{2}(\sqrt{2} - 1)}{(\sqrt{2} + 1)(\sqrt{2} - 1)} = \frac{2 - \sqrt{2}}{(\sqrt{2})^2 - 1^2} = \frac{2 - \sqrt{2}}{2 - 1} = 2 - \sqrt{2}$

2) $\frac{4}{\sqrt{7} + \sqrt{3}} = \frac{4(\sqrt{7} - \sqrt{3})}{(\sqrt{7} + \sqrt{3})(\sqrt{7} - \sqrt{3})} = \frac{4(\sqrt{7} - \sqrt{3})}{(\sqrt{7})^2 - (\sqrt{3})^2} = \frac{4(\sqrt{7} - \sqrt{3})}{7 - 3} = \frac{4(\sqrt{7} - \sqrt{3})}{4} = \sqrt{7} - \sqrt{3}$

3) $\frac{15}{\sqrt{15} - \sqrt{12}} = \frac{15(\sqrt{15} + \sqrt{12})}{(\sqrt{15} - \sqrt{12})(\sqrt{15} + \sqrt{12})} = \frac{15(\sqrt{15} + \sqrt{12})}{(\sqrt{15})^2 - (\sqrt{12})^2} = \frac{15(\sqrt{15} + \sqrt{12})}{15 - 12} = \frac{15(\sqrt{15} + \sqrt{12})}{3} = 5(\sqrt{15} + \sqrt{12}) = 5(\sqrt{5 * 3} + \sqrt{4 * 3}) = 5(\sqrt{5} * \sqrt{3} + 2\sqrt{3}) = 5\sqrt{3}(\sqrt{5} + 2)$

4) $\frac{19}{2\sqrt{5} - 1} = \frac{19(2\sqrt{5} + 1)}{(2\sqrt{5} - 1)(2\sqrt{5} + 1)} = \frac{19(2\sqrt{5} + 1)}{(2\sqrt{5})^2 - 1^2} = \frac{19(2\sqrt{5} + 1)}{4 * 5 - 1} = \frac{19(2\sqrt{5} + 1)}{20 - 1} = \frac{19(2\sqrt{5} + 1)}{19} = 2\sqrt{5} + 1$

5) $\frac{1}{\sqrt{a} - \sqrt{b}} = \frac{1 * (\sqrt{a} + \sqrt{b})}{(\sqrt{a} - \sqrt{b})(\sqrt{a} + \sqrt{b})} = \frac{\sqrt{a} + \sqrt{b}}{(\sqrt{a})^2 - (\sqrt{b})^2} = \frac{\sqrt{a} + \sqrt{b}}{a - b}$

6) $\frac{\sqrt{3} + 1}{\sqrt{3} - 1} = \frac{(\sqrt{3} + 1)(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = \frac{(\sqrt{3} + 1)^2}{(\sqrt{3})^2 - 1^2} = \frac{(\sqrt{3})^2 + 2 * \sqrt{3} * 1 + 1^2}{3 - 1} = \frac{3 + 2\sqrt{3} + 1}{2} = \frac{4 + 2\sqrt{3}}{2} = \frac{2(2 + \sqrt{3})}{2} = 2 + \sqrt{3}$

558. Освободитесь от иррациональности в знаменателе дроби:
1) $\frac{\sqrt{5}}{\sqrt{5} - 2}$;
2) $\frac{8}{\sqrt{10} - \sqrt{2}}$;
3) $\frac{9}{\sqrt{x} + \sqrt{y}}$;
4) $\frac{2 - \sqrt{2}}{2 + \sqrt{2}}$.

Решение:

1) $\frac{\sqrt{5}}{\sqrt{5} - 2} = \frac{\sqrt{5}(\sqrt{5} + 2)}{(\sqrt{5} - 2)(\sqrt{5} + 2)} = \frac{\sqrt{5}(\sqrt{5} + 2)}{(\sqrt{5})^2 - 2^2} = \frac{\sqrt{5}(\sqrt{5} + 2)}{5 - 4} = \frac{5 + 2\sqrt{5}}{1} = 5 + 2\sqrt{5}$

2) $\frac{8}{\sqrt{10} - \sqrt{2}} = \frac{8(\sqrt{10} + \sqrt{2})}{(\sqrt{10} - \sqrt{2})(\sqrt{10} + \sqrt{2})} = \frac{8(\sqrt{10} + \sqrt{2})}{(\sqrt{10})^2 - (\sqrt{2})^2} = \frac{8(\sqrt{10} + \sqrt{2})}{10 - 2} = \frac{8(\sqrt{10} + \sqrt{2})}{8} = \sqrt{10} + \sqrt{2}$

3) $\frac{9}{\sqrt{x} + \sqrt{y}} = \frac{9(\sqrt{x} - \sqrt{y})}{(\sqrt{x} + \sqrt{y})(\sqrt{x} - \sqrt{y})} = \frac{9(\sqrt{x} - \sqrt{y})}{(\sqrt{x})^2 - (\sqrt{y})^2} = \frac{9(\sqrt{x} - \sqrt{y})}{x - y}$

4) $\frac{2 - \sqrt{2}}{2 + \sqrt{2}} = \frac{(2 - \sqrt{2})(2 - \sqrt{2})}{(2 + \sqrt{2})(2 - \sqrt{2})} = \frac{(2 - \sqrt{2})^2}{2^2 - (\sqrt{2})^2} = \frac{2^2 - 2 * 2\sqrt{2} + (\sqrt{2})^2}{4 - 2} = \frac{4 - 4\sqrt{2} + 2}{2} = \frac{6 - 4\sqrt{2}}{2} = \frac{2(3 - 2\sqrt{2})}{2} = 3 - 2\sqrt{2}$

559. Докажите равенство:
1) $\frac{1}{5 - 2\sqrt{6}} + \frac{1}{5 + 2\sqrt{6}} = 10$;
2) $\frac{2}{3\sqrt{2} + 4} - \frac{2}{3\sqrt{2} - 4} = -8$;
3) $\frac{\sqrt{2} + 1}{\sqrt{2} - 1} - \frac{\sqrt{2} - 1}{\sqrt{2} + 1} = 4\sqrt{2}$.

Решение:

1) $\frac{1}{5 - 2\sqrt{6}} + \frac{1}{5 + 2\sqrt{6}} = \frac{5 + 2\sqrt{6} + 5 - 2\sqrt{6}}{(5 - 2\sqrt{6})(5 + 2\sqrt{6})} = \frac{10}{5^2 - (2\sqrt{6})^2)} = \frac{10}{25 - 4 * 6} = \frac{10}{25 - 24} = \frac{10}{1} = 10$

2) $\frac{2}{3\sqrt{2} + 4} - \frac{2}{3\sqrt{2} - 4} = \frac{2(3\sqrt{2} - 4) - 2(3\sqrt{2} + 4)}{(3\sqrt{2} + 4)(3\sqrt{2} - 4)} = \frac{6\sqrt{2} - 8 - 6\sqrt{2} - 8}{(3\sqrt{2})^2 - 4^2} = \frac{-16}{9 * 2 - 16} = \frac{-16}{18 - 16} = \frac{-16}{2} = -8$

3) $\frac{\sqrt{2} + 1}{\sqrt{2} - 1} - \frac{\sqrt{2} - 1}{\sqrt{2} + 1} = \frac{(\sqrt{2} + 1)^2 - (\sqrt{2} - 1)^2}{(\sqrt{2} - 1)(\sqrt{2} + 1)} = \frac{2 + 2\sqrt{2} + 1 - (2 - 2\sqrt{2} + 1)}{(\sqrt{2})^2 - 1^2} = \frac{3 + 2\sqrt{2} - 2 + 2\sqrt{2} - 1}{2 - 1} = \frac{4\sqrt{2}}{1} = 4\sqrt{2}$

560. Докажите, что значением выражения является рациональное число:
1) $\frac{6}{3 + 2\sqrt{3}} + \frac{6}{3 - 2\sqrt{3}}$;
2) $\frac{\sqrt{11} + \sqrt{6}}{\sqrt{11} - \sqrt{6}} + \frac{\sqrt{11} - \sqrt{6}}{\sqrt{11} + \sqrt{6}}$.

Решение:

1) $\frac{6}{3 + 2\sqrt{3}} + \frac{6}{3 - 2\sqrt{3}} = \frac{6(3 - 2\sqrt{3}) + 6(3 + 2\sqrt{3})}{(3 + 2\sqrt{3})(3 - 2\sqrt{3})} = \frac{18 - 12\sqrt{3} + 18 + 12\sqrt{3}}{3^2 - (2\sqrt{3})^2} = \frac{36}{9 - 4 * 3} = \frac{36}{9 - 12} = \frac{36}{-3} = -12$ − рациональное число

2) $\frac{\sqrt{11} + \sqrt{6}}{\sqrt{11} - \sqrt{6}} + \frac{\sqrt{11} - \sqrt{6}}{\sqrt{11} + \sqrt{6}} = \frac{(\sqrt{11} + \sqrt{6})^2 + (\sqrt{11} - \sqrt{6})^2}{(\sqrt{11} - \sqrt{6})(\sqrt{11} + \sqrt{6})} = \frac{(\sqrt{11})^2 + 2 * \sqrt{11} * \sqrt{6} + (\sqrt{6})^2 + (\sqrt{11})^2 - 2 * \sqrt{11} * \sqrt{6} + (\sqrt{6})^2}{(\sqrt{11})^2 - (\sqrt{6})^2} = \frac{11 + 2\sqrt{66} + 6 + 11 - 2\sqrt{66} + 6}{11 - 6} = \frac{34}{5} = 6\frac{4}{5}$ − рациональное число

561. Упростите выражение:
1) $\frac{a}{\sqrt{a} - 2} - \frac{4\sqrt{a} - 4}{\sqrt{a} - 2}$;
2) $\frac{\sqrt{m} + 1}{\sqrt{m} - 2} - \frac{\sqrt{m} + 3}{\sqrt{m}}$;
3) $\frac{\sqrt{y} + 4}{\sqrt{xy} + y} - \frac{\sqrt{x} - 4}{x + \sqrt{xy}}$;
4) $\frac{\sqrt{a}}{\sqrt{a} + 4} - \frac{a}{a - 16}$;
5) $\frac{a}{\sqrt{ab} - b} + \frac{\sqrt{b}}{\sqrt{b} - \sqrt{a}}$;
6) $\frac{a + \sqrt{a}}{\sqrt{b}} * \frac{b}{2\sqrt{a} + 2}$;
7) $\frac{\sqrt{c} - 5}{\sqrt{c}} : \frac{с - 25}{3с}$;
8) $(\sqrt{a} - \frac{a}{\sqrt{a} + 1}) : \frac{\sqrt{a}}{a - 1}$;
9) $(\frac{\sqrt{a} + \sqrt{b}}{\sqrt{b}} + \frac{\sqrt{b}}{\sqrt{a} - \sqrt{b}}) : \frac{\sqrt{a}}{\sqrt{b}}$;
10) $(\frac{\sqrt{x} - 3}{\sqrt{x} + 3} + \frac{12\sqrt{x}}{x - 9}) : \frac{\sqrt{x} + 3}{x - 3\sqrt{x}}$.

Решение:

1) $\frac{a}{\sqrt{a} - 2} - \frac{4\sqrt{a} - 4}{\sqrt{a} - 2} = \frac{a - (4\sqrt{a} - 4)}{\sqrt{a} - 2} = \frac{a - 4\sqrt{a} + 4}{\sqrt{a} - 2} = \frac{(\sqrt{a})^2 - 2 * 2\sqrt{a} + 2^2}{\sqrt{a} - 2} = \frac{(\sqrt{a} - 2)^2}{\sqrt{a} - 2} = \sqrt{a} - 2$

2) $\frac{\sqrt{m} + 1}{\sqrt{m} - 2} - \frac{\sqrt{m} + 3}{\sqrt{m}} = \frac{\sqrt{m}(\sqrt{m} + 1) - (\sqrt{m} + 3)(\sqrt{m} - 2)}{\sqrt{m}(\sqrt{m} - 2)} = \frac{m + \sqrt{m} - (m + 3\sqrt{m} - 2\sqrt{m} - 6)}{\sqrt{m}(\sqrt{m} - 2)} = \frac{m + \sqrt{m} - m - 3\sqrt{m} + 2\sqrt{m} + 6}{\sqrt{m}(\sqrt{m} - 2)} = \frac{6}{m - 2\sqrt{m}}$

3) $\frac{\sqrt{y} + 4}{\sqrt{xy} + y} - \frac{\sqrt{x} - 4}{x + \sqrt{xy}} = \frac{\sqrt{y} + 4}{\sqrt{x} * \sqrt{y} + (\sqrt{y})^2} - \frac{\sqrt{x} - 4}{(\sqrt{x})^2 + \sqrt{x} * \sqrt{y}} = \frac{\sqrt{y} + 4}{\sqrt{y}(\sqrt{x} + \sqrt{y})} - \frac{\sqrt{x} - 4}{\sqrt{x}(\sqrt{x} + \sqrt{y})} = \frac{\sqrt{x}(\sqrt{y} + 4) - \sqrt{y}(\sqrt{x} - 4)}{\sqrt{xy}(\sqrt{x} + \sqrt{y})} = \frac{\sqrt{xy} + 4\sqrt{x} - \sqrt{xy} + 4\sqrt{y}}{\sqrt{xy}(\sqrt{x} + \sqrt{y})} = \frac{4\sqrt{x} + 4\sqrt{y}}{\sqrt{xy}(\sqrt{x} + \sqrt{y})} = \frac{4(\sqrt{x} + \sqrt{y})}{\sqrt{xy}(\sqrt{x} + \sqrt{y})} = \frac{4}{\sqrt{xy}}$

4) $\frac{\sqrt{a}}{\sqrt{a} + 4} - \frac{a}{a - 16} = \frac{\sqrt{a}}{\sqrt{a} + 4} - \frac{a}{(\sqrt{a})^2 - 4^2} = \frac{\sqrt{a}}{\sqrt{a} + 4} - \frac{a}{(\sqrt{a} - 4)(\sqrt{a} + 4)} = \frac{\sqrt{a}(\sqrt{a} - 4) - a}{(\sqrt{a} - 4)(\sqrt{a} + 4)} = \frac{a - 4\sqrt{a} - a}{a - 16} = \frac{-4\sqrt{a}}{a - 16} = -\frac{4\sqrt{a}}{a - 16}$

5) $\frac{a}{\sqrt{ab} - b} + \frac{\sqrt{b}}{\sqrt{b} - \sqrt{a}} = \frac{a}{\sqrt{b} * \sqrt{a} - (\sqrt{b})^2} + \frac{\sqrt{b}}{\sqrt{b} - \sqrt{a}} = \frac{a}{\sqrt{b}(\sqrt{a} - \sqrt{b})} + \frac{\sqrt{b}}{\sqrt{b} - \sqrt{a}} = \frac{a}{\sqrt{b}(\sqrt{a} - \sqrt{b})} - \frac{\sqrt{b}}{\sqrt{a} - \sqrt{b}} = \frac{a - (\sqrt{b})^2}{\sqrt{b}(\sqrt{a} - \sqrt{b})} = \frac{(\sqrt{a})^2 - (\sqrt{b})^2}{\sqrt{b}(\sqrt{a} - \sqrt{b})} = \frac{(\sqrt{a} - \sqrt{b})(\sqrt{a} + \sqrt{b})}{\sqrt{b}(\sqrt{a} - \sqrt{b})} = \frac{\sqrt{a} + \sqrt{b}}{\sqrt{b}}$

6) $\frac{a + \sqrt{a}}{\sqrt{b}} * \frac{b}{2\sqrt{a} + 2} = \frac{(\sqrt{a})^2 + \sqrt{a}}{\sqrt{b}} * \frac{(\sqrt{b})^2}{2(\sqrt{a} + 1)} = \frac{\sqrt{a}(\sqrt{a} + 1)}{1} * \frac{\sqrt{b}}{2(\sqrt{a} + 1)} = \frac{\sqrt{a}}{1} * \frac{\sqrt{b}}{2} = \frac{\sqrt{ab}}{2}$

7) $\frac{\sqrt{c} - 5}{\sqrt{c}} : \frac{с - 25}{3с} = \frac{\sqrt{c} - 5}{\sqrt{c}} * \frac{3с}{с - 25} = \frac{\sqrt{c} - 5}{\sqrt{c}} * \frac{3 * (\sqrt{c})^2}{(\sqrt{c})^2 - 5^2} = \frac{\sqrt{c} - 5}{1} * \frac{3\sqrt{c}}{(\sqrt{c} - 5)(\sqrt{c} + 5)} = \frac{3\sqrt{c}}{\sqrt{c} + 5}$

8) $(\sqrt{a} - \frac{a}{\sqrt{a} + 1}) : \frac{\sqrt{a}}{a - 1} = \frac{\sqrt{a}(\sqrt{a} + 1) - a}{\sqrt{a} + 1} * \frac{(\sqrt{a})^2 - 1^2}{\sqrt{a}} = \frac{a + \sqrt{a} - a}{\sqrt{a} + 1} * \frac{(\sqrt{a} - 1)(\sqrt{a} + 1)}{\sqrt{a}} = \frac{\sqrt{a}}{1} * \frac{\sqrt{a} - 1}{\sqrt{a}} = \sqrt{a} - 1$

9) $(\frac{\sqrt{a} + \sqrt{b}}{\sqrt{b}} + \frac{\sqrt{b}}{\sqrt{a} - \sqrt{b}}) : \frac{\sqrt{a}}{\sqrt{b}} = \frac{(\sqrt{a} - \sqrt{b})(\sqrt{a} + \sqrt{b}) + (\sqrt{b})^2}{\sqrt{b}(\sqrt{a} - \sqrt{b})} * \frac{\sqrt{b}}{\sqrt{a}} = \frac{a - b + b}{\sqrt{a} - \sqrt{b}} * \frac{1}{\sqrt{a}} = \frac{a}{\sqrt{a} - \sqrt{b}} * \frac{1}{\sqrt{a}} = \frac{(\sqrt{a})^2}{\sqrt{a} - \sqrt{b}} * \frac{1}{\sqrt{a}} = \frac{\sqrt{a}}{\sqrt{a} - \sqrt{b}}$

10) $(\frac{\sqrt{x} - 3}{\sqrt{x} + 3} + \frac{12\sqrt{x}}{x - 9}) : \frac{\sqrt{x} + 3}{x - 3\sqrt{x}} = (\frac{\sqrt{x} - 3}{\sqrt{x} + 3} + \frac{12\sqrt{x}}{(\sqrt{x})^2 - 3^2}) * \frac{x - 3\sqrt{x}}{\sqrt{x} + 3} = \frac{(\sqrt{x} - 3)^2 + 12\sqrt{x}}{(\sqrt{x} - 3)(\sqrt{x} + 3)} * \frac{(\sqrt{x})^2 - 3\sqrt{x}}{\sqrt{x} + 3} = \frac{(\sqrt{x})^2 - 2 * 3\sqrt{x} + 3^2 + 12\sqrt{x}}{(\sqrt{x} - 3)(\sqrt{x} + 3)} * \frac{\sqrt{x}(\sqrt{x} - 3)}{\sqrt{x} + 3} = \frac{x - 6\sqrt{x} + 9 + 12\sqrt{x}}{\sqrt{x} + 3} * \frac{\sqrt{x}}{\sqrt{x} + 3} = \frac{x + 6\sqrt{x} + 9}{\sqrt{x} + 3} * \frac{\sqrt{x}}{\sqrt{x} + 3} = \frac{(\sqrt{x})^2 + 2 * 3\sqrt{x} + 3^2}{\sqrt{x} + 3} * \frac{\sqrt{x}}{\sqrt{x} + 3} = \frac{(\sqrt{x} + 3)^2}{\sqrt{x} + 3} * \frac{\sqrt{x}}{\sqrt{x} + 3} = \sqrt{x}$

562. Упростите выражение:
1) $\frac{\sqrt{a} - 3}{\sqrt{a} + 1} - \frac{\sqrt{a} - 4}{\sqrt{a}}$;
2) $\frac{\sqrt{a} + 1}{a - \sqrt{ab}} - \frac{\sqrt{b} + 1}{\sqrt{ab} - b}$;
3) $\frac{\sqrt{x}}{y - 2\sqrt{y}} : \frac{\sqrt{x}}{3\sqrt{y} - 6}$;
4) $\frac{\sqrt{m}}{\sqrt{m} - \sqrt{n}} : (\frac{\sqrt{m} + \sqrt{n}}{\sqrt{n}} + \frac{\sqrt{n}}{\sqrt{m} - \sqrt{n}})$;
5) $(\frac{\sqrt{x} + 1}{\sqrt{x} - 1} - \frac{4\sqrt{x}}{x - 1}) * \frac{x + \sqrt{x}}{\sqrt{x} - 1}$;
6) $\frac{a - 64}{\sqrt{a} + 3} * \frac{1}{a + 8\sqrt{a}} - \frac{\sqrt{a} + 8}{a - 3\sqrt{a}}$.

Решение:

1) $\frac{\sqrt{a} - 3}{\sqrt{a} + 1} - \frac{\sqrt{a} - 4}{\sqrt{a}} = \frac{\sqrt{a}(\sqrt{a} - 3) - (\sqrt{a} + 1)(\sqrt{a} - 4)}{\sqrt{a}(\sqrt{a} + 1)} = \frac{a - 3\sqrt{a} - (a + \sqrt{a} - 4\sqrt{a} - 4)}{\sqrt{a}(\sqrt{a} + 1)} = \frac{a - 3\sqrt{a} - a - \sqrt{a} + 4\sqrt{a} + 4}{\sqrt{a}(\sqrt{a} + 1)} = \frac{4}{a + \sqrt{a}}$

2) $\frac{\sqrt{a} + 1}{a - \sqrt{ab}} - \frac{\sqrt{b} + 1}{\sqrt{ab} - b} = \frac{\sqrt{a} + 1}{\sqrt{a}(\sqrt{a} - \sqrt{b})} - \frac{\sqrt{b} + 1}{\sqrt{b}(\sqrt{a} - \sqrt{b})} = \frac{\sqrt{b}(\sqrt{a} + 1) - \sqrt{a}(\sqrt{b} + 1)}{\sqrt{ab}(\sqrt{a} - \sqrt{b})} = \frac{\sqrt{ab} + \sqrt{b} - \sqrt{ab} - \sqrt{a}}{\sqrt{ab}(\sqrt{a} - \sqrt{b})} = \frac{\sqrt{b} - \sqrt{a}}{\sqrt{ab}(\sqrt{a} - \sqrt{b})} = -\frac{\sqrt{a} - \sqrt{b}}{\sqrt{ab}(\sqrt{a} - \sqrt{b})} = -\frac{1}{\sqrt{ab}} = -\sqrt{\frac{1}{ab}}$

3) $\frac{\sqrt{x}}{y - 2\sqrt{y}} : \frac{\sqrt{x}}{3\sqrt{y} - 6} = \frac{\sqrt{x}}{\sqrt{y}(\sqrt{y} - 2)} : \frac{\sqrt{x}}{3(\sqrt{y} - 2)} = \frac{\sqrt{x}}{\sqrt{y}(\sqrt{y} - 2)} * \frac{3(\sqrt{y} - 2)}{\sqrt{x}} = \frac{1}{\sqrt{y}} * \frac{3}{1} = \frac{3}{\sqrt{y}}$

4) $\frac{\sqrt{m}}{\sqrt{m} - \sqrt{n}} : (\frac{\sqrt{m} + \sqrt{n}}{\sqrt{n}} + \frac{\sqrt{n}}{\sqrt{m} - \sqrt{n}}) = \frac{\sqrt{m}}{\sqrt{m} - \sqrt{n}} : \frac{(\sqrt{m} - \sqrt{n})(\sqrt{m} + \sqrt{n}) + (\sqrt{n})^2}{\sqrt{n}(\sqrt{m} - \sqrt{n})} = \frac{\sqrt{m}}{\sqrt{m} - \sqrt{n}} : \frac{m - n + n}{\sqrt{n}(\sqrt{m} - \sqrt{n})} = \frac{\sqrt{m}}{\sqrt{m} - \sqrt{n}} : \frac{m}{\sqrt{n}(\sqrt{m} - \sqrt{n})} = \frac{\sqrt{m}}{\sqrt{m} - \sqrt{n}} * \frac{\sqrt{n}(\sqrt{m} - \sqrt{n})}{m} = \frac{\sqrt{n}}{\sqrt{m}} = \sqrt{\frac{n}{m}}$

5) $(\frac{\sqrt{x} + 1}{\sqrt{x} - 1} - \frac{4\sqrt{x}}{x - 1}) * \frac{x + \sqrt{x}}{\sqrt{x} - 1} = (\frac{\sqrt{x} + 1}{\sqrt{x} - 1} - \frac{4\sqrt{x}}{(\sqrt{x} - 1)(\sqrt{x} + 1)}) * \frac{\sqrt{x}(\sqrt{x} + 1)}{\sqrt{x} - 1} = \frac{(\sqrt{x} + 1)^2 - 4\sqrt{x}}{(\sqrt{x} - 1)(\sqrt{x} + 1)} * \frac{\sqrt{x}(\sqrt{x} + 1)}{\sqrt{x} - 1} = \frac{x + 2\sqrt{x} + 1 - 4\sqrt{x}}{\sqrt{x} - 1} * \frac{\sqrt{x}}{\sqrt{x} - 1} = \frac{x - 2\sqrt{x} + 1}{\sqrt{x} - 1} * \frac{\sqrt{x}}{\sqrt{x} - 1} = \frac{\sqrt{x}(\sqrt{x} - 1)^2}{(\sqrt{x} - 1)^2} = \sqrt{x}$

6) $\frac{a - 64}{\sqrt{a} + 3} * \frac{1}{a + 8\sqrt{a}} - \frac{\sqrt{a} + 8}{a - 3\sqrt{a}} = \frac{(\sqrt{a} - 8)(\sqrt{a} + 8)}{\sqrt{a} + 3} * \frac{1}{\sqrt{a}(\sqrt{a} + 8)} - \frac{\sqrt{a} + 8}{\sqrt{a}(\sqrt{a} - 3)} = \frac{\sqrt{a} - 8}{\sqrt{a} + 3} * \frac{1}{\sqrt{a}} - \frac{\sqrt{a} + 8}{\sqrt{a}(\sqrt{a} - 3)} = \frac{\sqrt{a} - 8}{\sqrt{a}(\sqrt{a} + 3)} - \frac{\sqrt{a} + 8}{\sqrt{a}(\sqrt{a} - 3)} = \frac{(\sqrt{a} - 8)(\sqrt{a} - 3) - (\sqrt{a} + 8)(\sqrt{a} + 3)}{\sqrt{a}(\sqrt{a} - 3)(\sqrt{a} + 3)} = \frac{a - 8\sqrt{a} - 3\sqrt{a} + 24 - (a + 8\sqrt{a} + 3\sqrt{a} + 24)}{\sqrt{a}(\sqrt{a} - 3)(\sqrt{a} + 3)} = \frac{a - 8\sqrt{a} - 3\sqrt{a} + 24 - a - 8\sqrt{a} - 3\sqrt{a} - 24}{\sqrt{a}(\sqrt{a} - 3)(\sqrt{a} + 3)} = \frac{-22\sqrt{a}}{\sqrt{a}(a - 9)} = -\frac{22}{a - 9}$