Question

Решите задания. 8-класс.

Answer 1

$10)\ \ \sqrt{5\dfrac{1}{7}\cdot 3\dfrac{4}{7}}=\sqrt{\dfrac{36}{7}\cdot \dfrac{25}{7}}=\dfrac{6\cdot 5}{7}=\dfrac{30}{7}=4\dfrac{2}{7}\\\\\\(\sqrt{32}+\sqrt8)^2=32+2\sqrt{32\cdot 8}+8=40+2\sqrt{256}=40+2\cdot 16=72$

$11.3)\ \ \dfrac{2}{\sqrt{11}+3}+\dfrac{7}{\sqrt{11}-2}=\dfrac{2\sqrt{11}-4+7\sqrt{11}+21}{11-2\sqrt{11}+3\sqrt{11}-6}=\dfrac{9\sqrt{11}+17}{5+\sqrt{11}}=\\\\\\=\dfrac{(8\sqrt{11}+17)(5-\sqrt{11})}{5-11}=\dfrac{40\sqrt{11}-88+85-17\sqrt{11}}{-6}=\\\\\\=\dfrac{23\sqrt{11}-3}{-6}=\dfrac{3-23\sqrt{11}}{6}$

$11.2)\ \ 15\sqrt{1,2}+\dfrac{1}{3}\sqrt{270}-2\sqrt{30}=\sqrt{270}+\dffrac{1}{3}\sqrt{270}-2\sqrt{30}=\\\\=14\sqrt{270}-2\sqrt{30}=14\cdot 3\sqrt{30}-2\sqrt{30}=40\sqrt{30}$

$11.4)\ \ \dfrac{4}{3-\sqrt5}+\dfrac{1}{2-\sqrt5}+\dfrac{3\sqrt5}{4}=\dfrac{8-4\sqrt5+3-\sqrt5}{6-3\sqrt5-2\sqrt5+5}+\dfrac{3\sqrt5}{4}=\\\\\\=\dfrac{11-5\sqrt5}{11-5\sqrt5}+\dfrac{3\sqrt5}{4}=1+\dfrac{3\sqrt5}{4}=\dfrac{4+3\sqrt5}{4}$