Question

Пожалуйста, помогите! Номер 136

Answer 1

а)

$2\sqrt{20}-3\sqrt{45}+3\sqrt{80}-\sqrt{125}=\\\\=2\sqrt{4*5}-3\sqrt{9*5}+3\sqrt{16*5}-\sqrt{25*5}=\\\\=2*2\sqrt{5}-3*3\sqrt{5}+3*4\sqrt{5}-5\sqrt{5}=\\\\=4\sqrt{5}-9\sqrt{5}+12\sqrt{5}-5\sqrt{5}=2\sqrt{5}$

б)

$10\sqrt{\frac{2}{5}}-0,5\sqrt{160}+3\sqrt{1\frac{1}{9}}=\\\\=10\sqrt{\frac{2*5}{5*5}}-0,5\sqrt{16*10}+3\sqrt{\frac{10}{9}}=\\\\=10*\frac{1}{5}*\sqrt{10}-0,5*4*\sqrt{10}+3*\frac{1}{3}*\sqrt{10}=\\\\=2\sqrt{10}-2\sqrt{10}+\sqrt{10}= \sqrt{10}$

в)

$(\sqrt{15}-\sqrt{6})^2+6\sqrt{10}=\\\\=(\sqrt{15})^2-2*\sqrt{15}*\sqrt{6}+(\sqrt{6})^2+6\sqrt{10}=\\\\=15-2\sqrt{90}+6+6\sqrt{10}=\\\\=21-2\sqrt{9*10}+6\sqrt{10}=\\ \\ =21-6\sqrt{10}+6\sqrt{10}=21$

г)

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