Question

Решите контрльная работа

15

оба варианта или только 1

Answer 1

Вариант 1.

A1.

$a) \sqrt{1600} = 40$

$b) \sqrt{6\frac{1}{4} } = \sqrt{\frac{25}{4} } = \frac{5}{2} = 2.5$

А2.

$a) \sqrt{0.36\times 81 } = \sqrt{\frac{9}{25}\times 81} = \sqrt{\frac{9}{25}} \times \sqrt{81 } = \frac{3}{5} \times 9 = \frac{27}{5} = 5.4$

$b) \frac{\sqrt{4500} }{\sqrt{500} } = \sqrt{\frac{4500}{500} } = \sqrt{9} } = 3$

$c) \sqrt{(-31)^{2} } = |-31| = 31$

A3.

$2\sqrt{a} + 6\sqrt{a} - 7\sqrt{a} = (2+6-7)\sqrt{a} = \sqrt{a}$

A4.

$a) \frac{b}{\sqrt{7} } = \frac{b\times \sqrt{7} }{\sqrt{7} \times \sqrt{7} } = \frac{b \sqrt{7} }{7}$

$b) \frac{5}{\sqrt{13} + \sqrt{3} } = \frac{5 \times (\sqrt{13} - \sqrt{3}) }{(\sqrt{13} + \sqrt{3}) \times (\sqrt{13} - \sqrt{3}) } = \frac{5 \times (\sqrt{13} - \sqrt{3}) }{13 - 3} = \frac{5 \times (\sqrt{13} - \sqrt{3}) }{10} = \frac{\sqrt{13} - \sqrt{3} }{2}$

В1.

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

Вариант 2.

А1.

$a) \sqrt{900} = 30\\$

$b) \sqrt{3 \frac{1}{16} } = \sqrt{\frac{49}{16} } = \frac{7}{4} = 1.75$

А2.

$a) \sqrt{225 \times 0.09 } = \sqrt{225 \times \frac{9}{100} } = \sqrt{9 \times \frac{9}{4} } = \sqrt{9} \times \sqrt{\frac{9}{4} } = 3 \times \frac{3}{2} = \frac{9}{2} = 4.5\\$

$b) \frac{\sqrt{108} }{\sqrt{3} } = \sqrt{\frac{108}{3} } = \sqrt{36} = 6$

$c) \sqrt{6^{4}} = |6^{2}| = 6^{2} = 36$

А3.

$\sqrt{49c} - \sqrt{16c} + \sqrt{25c} = \sqrt{49} \times \sqrt{c} - \sqrt{16} \times \sqrt{c} + \sqrt{25} \times \sqrt{c} = 7\sqrt{c} - 4\sqrt{c} + 5\sqrt{c} = (7 - 4 + 5)\sqrt{c} = 8\sqrt{c}$

А4.

$a) \frac{1}{\sqrt{3} } = \frac{1 \times \sqrt{3} }{\sqrt{3} \times \sqrt{3} } = \frac{\sqrt{3}}{3}$

$b) \frac{6}{\sqrt{5} + 1} = \frac{6 \times (\sqrt{5} - 1) }{(\sqrt{5} + 1) \times (\sqrt{5} - 1) } = \frac{6(\sqrt{5} - 1) }{5 - 1} = \frac{6(\sqrt{5} - 1) }{4} = \frac{3(\sqrt{5} - 1) }{2}$

В1.

$\frac{\sqrt{10} + \sqrt{6} }{\sqrt{10} - \sqrt{6} } - \frac{\sqrt{10} - \sqrt{6} }{\sqrt{10} + \sqrt{6} } = \frac{(\sqrt{10} + \sqrt{6}) \times (\sqrt{10} + \sqrt{6}) - (\sqrt{10} - \sqrt{6}) \times (\sqrt{10} - \sqrt{6}) }{(\sqrt{10} - \sqrt{6}) \times (\sqrt{10} + \sqrt{6}) } = \frac{(\sqrt{10} + \sqrt{6})^{2} - (\sqrt{10} - \sqrt{6})^{2} }{10 - 6} = \frac{10 + 2\sqrt{60} + 6 - 10 + 2\sqrt{60} - 6}{4} = \frac{4\sqrt{60}}{4} = \sqrt{60} = \sqrt{4 \times 15} = 2\sqrt{15}$