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 

-2.6314...

Bài làm:

Ta có: \(\frac{\sqrt{7}-\sqrt{5}}{\sqrt{7}+\sqrt{5}}=\frac{\left(\sqrt{7}-\sqrt{5}\right)^2}{\left(\sqrt{7}+\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)}=\frac{7-2.\sqrt{7}.\sqrt{5}-5}{7-5}\)

\(=\frac{2-2\sqrt{35}}{2}=1-\sqrt{35}\)

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: 

\(\sqrt{9-4\sqrt{5}}-\sqrt{14+6\sqrt{5}}\)

\(=\sqrt{5-4\sqrt{5}+4}-\sqrt{9+6\sqrt{5}+5}\)

\(=\sqrt{\left(\sqrt{5}-2\right)^2}-\sqrt{\left(3+\sqrt{5}\right)^2}\)

\(=\sqrt{5}-2-3-\sqrt{5}\)

\(=-5\)

Trả lời:

\(\frac{2\sqrt{5}-5\sqrt{2}}{\sqrt{2}-\sqrt{5}}+\frac{6}{2-\sqrt{10}}+\sqrt{67+12\sqrt{7}}\)

\(=\frac{\sqrt{2}.\sqrt{5}.\left(\sqrt{2}-\sqrt{5}\right)}{\sqrt{2}-\sqrt{5}}-\frac{6}{\sqrt{10}-2}+\sqrt{63+12\sqrt{7}+4}\)

\(=\sqrt{2}.\sqrt{5}-\frac{6.\left(\sqrt{10}+2\right)}{10-4}+\sqrt{\left(3\sqrt{7}+2\right)^2}\)

\(=\sqrt{10}-\sqrt{10}-2+3\sqrt{7}+2\)

\(=3\sqrt{7}\)

Bài làm:

\(\frac{\sqrt{5}+2}{\sqrt{5}-2}=\frac{\left(\sqrt{5}+2\right)^2}{\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)}=\frac{5+2.2.\sqrt{5}+4}{5-4}\)

\(=9+4\sqrt{5}\)

Học tốt!!!!

\(\frac{\sqrt{5}+2}{\sqrt{5}-2}=\frac{5+2\sqrt{5}+4+2\sqrt{5}}{\sqrt{5}^2-2^2}\)

\(=\frac{9+4\sqrt{5}}{5-4}=9+4\sqrt{5}\)

@Học tốt@

Trả lời 

\(\frac{\sqrt{5}+2}{\sqrt{5}-2}=\frac{\left(\sqrt{5}+2\right).\left(\sqrt{5}+2\right)}{\left(\sqrt{5}-2\right).\left(\sqrt{5}+2\right)}\) 

                    \(=\frac{5+4\sqrt{5}+4}{5-4}\)

                    \(=\frac{9+4\sqrt{5}}{1}\)

                    \(=9+4\sqrt{5}\)

Bài 1:

Ta có: \(\sqrt{16x-32}+\sqrt{25x-50}=18+\sqrt{9x-18}\)

\(\Leftrightarrow\sqrt{16\left(x-2\right)}+\sqrt{25\left(x-2\right)}=18+\sqrt{9\left(x-2\right)}\)

\(\Leftrightarrow4\sqrt{x-2}+5\sqrt{x-2}=18+3\sqrt{x-2}\)

\(\Leftrightarrow6\sqrt{x-2}=18\)

\(\Leftrightarrow\sqrt{x-2}=3\)

\(\Leftrightarrow\left(\sqrt{x-2}\right)^2=3^2\)

\(\Leftrightarrow x-2=9\)

\(\Leftrightarrow x=11\)

Vậy tập nghiệm của PT \(S=\left\{11\right\}\)

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/ \(x=\sqrt[3]{3+2\sqrt{2}}+\sqrt[3]{3-2\sqrt{2}}\)

=> \(x^3=\left(\sqrt[3]{3+2\sqrt{2}}+\sqrt[3]{3-2\sqrt{2}}\right)^3\)

\(=3+2\sqrt{2}+3-2\sqrt{2}+3\left(\sqrt[3]{3+2\sqrt{2}}+\sqrt[3]{3-2\sqrt{2}}\right).\sqrt[3]{3+2\sqrt{2}}.\sqrt[3]{3-2\sqrt{2}}\)

\(=6+3x\)

=> \(x^3-3x=6\)

=> \(B=x^3-3x+2000=6+2000=2006\)

\(A=\frac{1-\sqrt{5}}{1-5}+\frac{\sqrt{5}-\sqrt{9}}{5-9}+\frac{\sqrt{9}-\sqrt{13}}{9-13}+...+\frac{\sqrt{2001}-\sqrt{2005}}{2001-2005}\)

\(A=\frac{1-\sqrt{5}+\sqrt{5}-\sqrt{9}+\sqrt{9}-\sqrt{13}+...+\sqrt{2001}-\sqrt{2005}}{-4}\)

\(A=\frac{1-\sqrt{2005}}{-4}=\frac{\sqrt{2005}-1}{4}\)