Question

ПОМОГИТЕ СРОЧНО!!!! ВСЕ БАЛЛЫ ОТДАМ. ​

Answer 1

$6.1) \frac{c-36}{\sqrt{c}-6} = \frac{(c-36)(\sqrt{c}+6)}{(\sqrt{c}-6)(\sqrt{c}+6)} = \frac{(c-36)(\sqrt{c}+6)}{c-36} = \sqrt{c} + 6\\\\\\6.2) \frac{7+3\sqrt{7}}{\sqrt{7}} = \frac{(7+3\sqrt{7})\sqrt{7}}{7} = \frac{7\sqrt{7}+3\cdot7}{7} = \frac{7\sqrt{7}+21}{7} = \sqrt{7} + 3\\\\\\6.3) \frac{b-4}{b+4\sqrt{b}+4} = \frac{(\sqrt{b}-2)(\sqrt{b}+2)}{(\sqrt{b}+2)(\sqrt{b}+2)} = \frac{\sqrt{b}-2}{\sqrt{b}+2}$

$7.1) \frac{12}{7\sqrt{3}} = \frac{12\cdot \sqrt{3}}{7\sqrt{3} \cdot \sqrt{3}} = \frac{12\sqrt{3}}{21} = \frac{4\sqrt{3}}{7}\\\\\\7.2) \frac{18}{\sqrt{7}+1} = \frac{18 \cdot (\sqrt{7}-1)}{(\sqrt{7}+1)(\sqrt{7}-1)} = \frac{18\sqrt{7} - 18}{7-1} = \frac{18\sqrt{7}-18}{6} = 3\sqrt{7} - 3 = 3(\sqrt{7} - 1)$

$8.1) \sqrt{7y^2} = -y\sqrt{7}\\\\8.2) \sqrt{32a^8} = 4a^4\sqrt{2}\\\\8.3) \sqrt{-b^{15}} =(b^7\sqrt{-b})\\\\ 8.4)\sqrt{-x^{14}y^3} = -x^7y\sqrt{y}$

$9) \sqrt{(5-\sqrt{12})^2} + \sqrt{(3-\sqrt{12})^2} = 5-\sqrt{12} + 3-\sqrt{12} = 8 - 2\sqrt{12}$