Решите 2,3,4 задачи срочно со всеми действами даю 20 баллов

Ответ

Проверено экспертом

Автор - Universalka

$2)sqrt{(frac{13}{12})^{-2}+(2,6)^{-2}}+sqrt{(-1frac{3}{7})^{-2}}+sqrt{(-1frac{2}{3})^{-2}}=sqrt{(frac{12}{13})^{2}+(frac{5}{13})^{2}}+sqrt{(frac{7}{10})^{2}}+sqrt{(frac{3}{5})^{2}}=sqrt{frac{144}{169}+frac{25}{169}}+frac{7}{10}+frac{3}{5}=sqrt{frac{169}{169}}+frac{13}{10}=1+1,3=2,3\\Otvet:boxed{2,3}$

$3)frac{39}{5-2sqrt{3}}+frac{10}{3sqrt{3}-4sqrt{2}}+frac{4}{3-2sqrt{2}}=frac{39*(5+2sqrt{3})}{(5-2sqrt{3})(5+2sqrt{3})}+frac{10*(3sqrt{3}+4sqrt{2})}{(3sqrt{3}-4sqrt{2})(3sqrt{3}+4sqrt{2})}+frac{4*(3+2sqrt{2})}{(3-2sqrt{2})(3+2sqrt{2})}=frac{39*(5+2sqrt{3})}{25-12}+frac{10*(3sqrt{3}+4sqrt{2})}{27-32}+frac{4*(3+2sqrt{2})}{9-8}=frac{39*(5+2sqrt{3}}{13}-frac{10*(3sqrt{3}+4sqrt{2}}{5}+4*(3+2sqrt{2})=3(5+2sqrt{3})-2(3sqrt{3}+4sqrt{2})+4(3+2sqrt{2})$ $=15+6sqrt{3}-6sqrt{3}-8sqrt{2}+12+8sqrt{2}=27\\Otvet:boxed{27}$

$4)sqrt{22-sqrt{160}}*(sqrt{10}+1)=sqrt{22-4sqrt{10}} *(sqrt{10}+1)=sqrt{2*(11-2sqrt{10})}*(sqrt{10}+1)=sqrt{2*(10-2sqrt{10}+1)}*(sqrt{10}+1)=sqrt{2}*sqrt{(sqrt{10})^{2}-2sqrt{10}+1^{2})}*(sqrt{10}+1)=sqrt{2}*sqrt{(sqrt{10}-1)^{2}}*(sqrt{10}+1)=sqrt{2}*|sqrt{10}-1|*(sqrt{10}+1)=sqrt{2}*(sqrt{10}-1)(sqrt{10}+1)=sqrt{2}*[(sqrt{10})^{2}-1^{2}]=9sqrt{2}\\Otvet:boxed{9sqrt{2} }$