Resposta :
Olá Aline,
boa noite!
[tex]\frac{\sqrt{20-\sqrt[5]{35-\sqrt[3]{25+\sqrt[5]{\sqrt{1024}}}}}}{\sqrt{576}-\sqrt{392}+\sqrt{450}-\sqrt{441}}=\\\\\\\frac{\sqrt{20-\sqrt[5]{35-\sqrt[3]{25+\sqrt[5\times2]{2^{10}}}}}}{24-\sqrt{2\times14^2}+\sqrt{2\times15^2}-21}=\\\\\\\frac{\sqrt{20-\sqrt[5]{35-\sqrt[3]{25+2}}}}{24-14\sqrt{2}+15\sqrt{2}-21}=\\\\\\\frac{\sqrt{20-\sqrt[5]{35-\sqrt[3]{3^3}}}}{3+\sqrt{2}}=[/tex]
[tex]\frac{\sqrt{20-\sqrt[5]{35-3}}}{3+\sqrt{2}}=\\\\\\\frac{\sqrt{20-\sqrt[5]{2^5}}}{3+\sqrt{2}}=\\\\\\\frac{\sqrt{20-2}}{3+\sqrt{2}}=\\\\\\\frac{\sqrt{2\times3^2}}{3+\sqrt{2}}=\\\\\\\frac{3\sqrt{2}}{3+\sqrt{2}}\times\frac{3-\sqrt{2}}{3-\sqrt{2}}=[/tex]
[tex]\frac{3\sqrt{2}\left(3-\sqrt{2}\right)}{9-2}=\\\\\\\boxed{\frac{9\sqrt{2}-6}{7}}[/tex]
boa noite!
[tex]\frac{\sqrt{20-\sqrt[5]{35-\sqrt[3]{25+\sqrt[5]{\sqrt{1024}}}}}}{\sqrt{576}-\sqrt{392}+\sqrt{450}-\sqrt{441}}=\\\\\\\frac{\sqrt{20-\sqrt[5]{35-\sqrt[3]{25+\sqrt[5\times2]{2^{10}}}}}}{24-\sqrt{2\times14^2}+\sqrt{2\times15^2}-21}=\\\\\\\frac{\sqrt{20-\sqrt[5]{35-\sqrt[3]{25+2}}}}{24-14\sqrt{2}+15\sqrt{2}-21}=\\\\\\\frac{\sqrt{20-\sqrt[5]{35-\sqrt[3]{3^3}}}}{3+\sqrt{2}}=[/tex]
[tex]\frac{\sqrt{20-\sqrt[5]{35-3}}}{3+\sqrt{2}}=\\\\\\\frac{\sqrt{20-\sqrt[5]{2^5}}}{3+\sqrt{2}}=\\\\\\\frac{\sqrt{20-2}}{3+\sqrt{2}}=\\\\\\\frac{\sqrt{2\times3^2}}{3+\sqrt{2}}=\\\\\\\frac{3\sqrt{2}}{3+\sqrt{2}}\times\frac{3-\sqrt{2}}{3-\sqrt{2}}=[/tex]
[tex]\frac{3\sqrt{2}\left(3-\sqrt{2}\right)}{9-2}=\\\\\\\boxed{\frac{9\sqrt{2}-6}{7}}[/tex]