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Question 1

Question 2

Решите задачу:

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

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

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

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