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\(\sqrt{8+\sqrt{8}+\sqrt{20}+\sqrt{40}}=\sqrt{8+2\sqrt{2}+2\sqrt{5}+2\sqrt{10}}\)
\(=\sqrt{2\left(4+\sqrt{2}+\sqrt{5}+\sqrt{10}\right)}\)
a/ \(\sqrt{2}+\sqrt{6}\)
b/ Sửa đề:
\(\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}.\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{3}}}}=1\)
c/ \(1+\sqrt{2}+\sqrt{5}\)
\(A=\sqrt{8+\sqrt{8}+\sqrt{20}+\sqrt{40}}\)
\(=\sqrt{\sqrt{5}^5+\sqrt{2}^2+1^2+2\sqrt{2}.1+2\sqrt{2}.\sqrt{5}}\)
\(=\sqrt{\left(\sqrt{5}+\sqrt{2}+1\right)^2}\)
\(=\sqrt{5}+\sqrt{2}+1\)
a) \(2\sqrt{40\sqrt{12}}-2\sqrt{\sqrt{75}}-3\sqrt{5\sqrt{48}}\)
\(=2\sqrt{40.2\sqrt{3}}-2\sqrt{5\sqrt{3}}-3\sqrt{5.4\sqrt{3}}\)
\(=\left(2\sqrt{80}-2\sqrt{5}-3\sqrt{20}\right).\sqrt{\sqrt{3}}\)
\(=\left(8\sqrt{5}-2\sqrt{5}-6\sqrt{5}\right).\sqrt{\sqrt{3}}=0\)
b) \(2\sqrt{8\sqrt{3}}-2\sqrt{5\sqrt{3}}-3\sqrt{20\sqrt{3}}\)
\(=\left(4\sqrt{2}-2\sqrt{5}-6\sqrt{5}\right).\sqrt{\sqrt{3}}\)
\(=\left(4\sqrt{2}-8\sqrt{5}\right).\sqrt{\sqrt{3}}\)
\(=\sqrt{\sqrt{3}}\left(\sqrt{2}-2\sqrt{5}\right)\)
a: \(=2\sqrt{\sqrt{3}}\cdot4\sqrt{5}-2\cdot\sqrt{\sqrt{3}}\cdot\sqrt{5}-3\cdot\sqrt{\sqrt{3}}\cdot2\sqrt{5}\)
\(=2\sqrt{\sqrt{3}}\left(4\sqrt{5}-\sqrt{5}-3\sqrt{5}\right)=0\)
b: \(=2\cdot2\sqrt{2}\cdot\sqrt{\sqrt{3}}-2\cdot\sqrt{5}\cdot\sqrt{\sqrt{3}}-3\cdot2\sqrt{5}\cdot\sqrt{\sqrt{3}}\)
\(=2\sqrt{\sqrt{3}}\left(2\sqrt{2}-\sqrt{5}-3\sqrt{5}\right)=2\sqrt{\sqrt{3}}\cdot\left(2\sqrt{2}-4\sqrt{5}\right)\)
\(\sqrt{8+2\sqrt{2}+2\sqrt{5}+2\sqrt{10}}-\sqrt{7+2\sqrt{10}}\)
\(=\sqrt{\sqrt{2}^2+\sqrt{5}^2+1^2+2\sqrt{2}+2\sqrt{5}+2\sqrt{2}.\sqrt{5}}-\sqrt{\sqrt{5}^2+2\sqrt{10}+\sqrt{2}^2}\)
\(=\sqrt{\left(\sqrt{5}+\sqrt{2}+1\right)^2}-\sqrt{\left(\sqrt{5}+\sqrt{2}\right)^2}\)
\(=\sqrt{5}+\sqrt{2}+1-\sqrt{5}-\sqrt{2}\)
\(=1\)
những bài thế này thì suy luận kiểu gì