Алгебра: Решите 2,3,4 со всеми действами 20

Решите 2,3,4 со всеми действами 20

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Ответ:

Глеб0707

06.10.2020 08:32

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

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

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