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Trả lời:
\(\frac{2\sqrt{5}-5\sqrt{2}}{\sqrt{2}-\sqrt{5}}+\frac{6}{2-\sqrt{10}}+\sqrt{67+12\sqrt{7}}\)
\(=\frac{\sqrt{2}.\sqrt{5}.\left(\sqrt{2}-\sqrt{5}\right)}{\sqrt{2}-\sqrt{5}}-\frac{6}{\sqrt{10}-2}+\sqrt{63+12\sqrt{7}+4}\)
\(=\sqrt{2}.\sqrt{5}-\frac{6.\left(\sqrt{10}+2\right)}{10-4}+\sqrt{\left(3\sqrt{7}+2\right)^2}\)
\(=\sqrt{10}-\sqrt{10}-2+3\sqrt{7}+2\)
\(=3\sqrt{7}\)
Trả lời:
\(\frac{2}{\sqrt{5}+\sqrt{3}}-\sqrt{\frac{2}{4-\sqrt{15}}}+6\sqrt{\frac{1}{3}}\)
\(=\frac{2.\left(\sqrt{5}-\sqrt{3}\right)}{5-3}-\sqrt{\frac{2\times2}{2\times\left(4-\sqrt{15}\right)}}+6\times\frac{1}{\sqrt{3}}\)
\(=\frac{2.\left(\sqrt{5}-\sqrt{3}\right)}{2}-\sqrt{\frac{4}{8-2\sqrt{15}}}+6\times\frac{\sqrt{3}}{3}\)
\(=\sqrt{5}-\sqrt{3}-\sqrt{\frac{4}{5-2\sqrt{15}+3}}+2\sqrt{3}\)
\(=\sqrt{5}-\sqrt{3}-\sqrt{\frac{4}{\left(\sqrt{5}-\sqrt{3}\right)^2}}+2\sqrt{3}\)
\(=\sqrt{5}-\sqrt{3}-\frac{2}{\sqrt{5}-\sqrt{3}}+2\sqrt{3}\)
\(=\sqrt{5}+\sqrt{3}-\frac{2}{\sqrt{5}-\sqrt{3}}\)
\(=\frac{\left(\sqrt{5}-\sqrt{3}\right).\left(\sqrt{5}+3\right)-2}{\sqrt{5}-\sqrt{3}}\)
\(=\frac{5-3-2}{\sqrt{5}-\sqrt{3}}\)
\(=0\)
Học tốt
\(\sqrt{9-4\sqrt{5}}-\sqrt{14+6\sqrt{5}}=\sqrt{5-4\sqrt{5}+4}-\sqrt{9+6\sqrt{5}+5}\)
\(=\sqrt{\left(\sqrt{5}-2\right)^2}-\sqrt{\left(3+\sqrt{5}\right)^2}=\left(\sqrt{5}-2\right)-\left(3+\sqrt{5}\right)=-5\)
Trả lời:
\(\sqrt{9-4\sqrt{5}}-\sqrt{14+6\sqrt{5}}\)
\(=\sqrt{5-4\sqrt{5}+4}-\sqrt{9+6\sqrt{5}+5}\)
\(=\sqrt{\left(\sqrt{5}-2\right)^2}-\sqrt{\left(3+\sqrt{5}\right)^2}\)
\(=\sqrt{5}-2-3-\sqrt{5}\)
\(=-5\)
1. Trục căn thức ở mẫu:
\(A=\frac{1}{1+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{9}}+\frac{1}{\sqrt{9}+\sqrt{13}}+....+\frac{1}{\sqrt{2001}+\sqrt{2005}}+\frac{1}{\sqrt{2005}+\sqrt{2009}}\)
=\(\frac{\sqrt{5}-1}{4}+\frac{\sqrt{9}-\sqrt{5}}{4}+\frac{\sqrt{13}-\sqrt{9}}{4}+....+\frac{\sqrt{2005}-\sqrt{2001}}{4}+\frac{\sqrt{2009}-\sqrt{2005}}{4}\)
\(=\frac{\sqrt{2009}-1}{4}\)
2/ \(x=\sqrt[3]{3+2\sqrt{2}}+\sqrt[3]{3-2\sqrt{2}}\)
=> \(x^3=\left(\sqrt[3]{3+2\sqrt{2}}+\sqrt[3]{3-2\sqrt{2}}\right)^3\)
\(=3+2\sqrt{2}+3-2\sqrt{2}+3\left(\sqrt[3]{3+2\sqrt{2}}+\sqrt[3]{3-2\sqrt{2}}\right).\sqrt[3]{3+2\sqrt{2}}.\sqrt[3]{3-2\sqrt{2}}\)
\(=6+3x\)
=> \(x^3-3x=6\)
=> \(B=x^3-3x+2000=6+2000=2006\)
\(A=\frac{1-\sqrt{5}}{1-5}+\frac{\sqrt{5}-\sqrt{9}}{5-9}+\frac{\sqrt{9}-\sqrt{13}}{9-13}+...+\frac{\sqrt{2001}-\sqrt{2005}}{2001-2005}\)
\(A=\frac{1-\sqrt{5}+\sqrt{5}-\sqrt{9}+\sqrt{9}-\sqrt{13}+...+\sqrt{2001}-\sqrt{2005}}{-4}\)
\(A=\frac{1-\sqrt{2005}}{-4}=\frac{\sqrt{2005}-1}{4}\)
\(\sqrt{\left(5+2\sqrt{6}\right).\left(5-2\sqrt{6}\right)}\\ =\sqrt{5^2-\left(2\sqrt{6}\right)^2}\\ =\sqrt{25-24}\\ =\sqrt{1}=1\)