Предмет: Алгебра
Автор: Толенген
Вопрос задан 7 лет назад

Помогите с решением!

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Ответы

Ответ дал: UluanaV

$1) (2sqrt{175}-3sqrt{28}+2sqrt{63})^2-60sqrt[3]{1000}=\ \=(2sqrt{25*7}-3sqrt{4*7}+2sqrt{9*7})^2-60sqrt[3]{10^3}=\ \=(2sqrt{25}sqrt{7} -3sqrt{4}sqrt{7}+2sqrt{9}sqrt{7})^2-60*10=\ \=(2*5sqrt{7}-3*2sqrt{7}+2*3sqrt{7})^2-600=\ \=(10sqrt{7}-6sqrt{7}+6sqrt{7})^2-600=\ \=(10sqrt{7})^2-600=10^2*(sqrt{7})^2-600=\ \=100*7-600=700-600=100\ \100=100$

$2) frac{1}{3}(2sqrt{150}+3sqrt{24}-5sqrt{54})^2+15sqrt[4]{625}=\ \=frac{1}{3}(2sqrt{25*6}+3sqrt{4*6}-5sqrt{9*6})^2+15sqrt[4]{5^4}=\ \=frac{1}{3}(2sqrt{25}sqrt{6}+3sqrt{4}sqrt{6}-5sqrt{9}sqrt{6})^2+15*5=\ \=frac{1}{3}(2*5sqrt{6}+3*2sqrt{6}-5*3sqrt{6})^2+75=\ \=frac{1}{3}(10sqrt{6}+6sqrt{6}-15sqrt{6})^2+75=\ \=frac{1}{3}(sqrt{6})^2+75=frac{1}{3}*6+75=2+75=77\ \77=77$

$3) (sqrt[6]{5+2sqrt{6}}+sqrt[3]{sqrt{3}+sqrt{2}})sqrt[3]{sqrt{2}-sqrt{3}}=\ \=sqrt[6]{5+2sqrt{6}}*sqrt[3]{sqrt{2}-sqrt{3}}+sqrt[3]{sqrt{3}+sqrt{2}}*sqrt[3]{sqrt{2}-sqrt{3}}=\ \=sqrt[6]{(5+2sqrt{6})(sqrt{2}-sqrt{3})^2}+sqrt[3]{(sqrt{2}+sqrt{3})(sqrt{2}-sqrt{3})}=\ \=sqrt[6]{(5+2sqrt{6})(2-2sqrt{2}sqrt{3}+3)}+sqrt[3]{(sqrt{2})^2-(sqrt{3})^2}=\ \=sqrt[6]{(5+2sqrt{6})(2-2sqrt{6}+3)}+sqrt[3]{2-3}=\ \=sqrt[6]{5*2-5*2sqrt{6}+5*3+2sqrt{6}*2-2sqrt{6}*2sqrt{6}+2sqrt{6}*3}+sqrt[3]{2-3}=\ \=sqrt[6]{10-10sqrt{6}+15+4sqrt{6}-4*6+6sqrt{6}}+sqrt[3]{-1}=\ \=sqrt[6]{25-24}-1=sqrt[6]{1}-1=1-1=0\ \0=0$

$4) sqrt{20,25}+sqrt[3]{24}-sqrt[4]{0,1296}-frac{2}{5}sqrt[3]{375}+frac{1}{3}sqrt[5]{7frac{19}{32}}=\ \=sqrt{frac{2025}{100} }+sqrt[3]{8*3}-sqrt[4]{frac{1296}{10000}}-frac{2}{5}sqrt[3]{125*3}+frac{1}{3}sqrt[5]{frac{243}{32}}=\ \=sqrt{frac{25*81}{100} }+sqrt[3]{2^3*3}-sqrt[4]{frac{16*81}{10000}}-frac{2}{5}sqrt[3]{5^3*3}+frac{1}{3}sqrt[5]{frac{3^5}{2^5}}=\ \=sqrt{frac{5^2*9^2}{10^2} }+2sqrt[3]{3}-sqrt[4]{frac{2^4*3^4}{10^4}}-frac{2}{5}*5sqrt[3]{3}+frac{1}{3}*frac{3}{5}=\ \=frac{5*9}{10}+2sqrt[3]{3}-frac{2*3}{10}-2sqrt[3]{3}+frac{1}{5}=\ \=frac{45}{10}-frac{6}{10}+frac{1^{(2}}{5}=\ \=frac{45-6+2}{10}=frac{44}{10}=4,4\ \4,4=4,4$