\(\sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}}\)

\(=\sqrt{\left(1+\sqrt{5}\right)^2}+\sqrt{\left(1-\sqrt{5}\right)^2}\)'

\(=|1+\sqrt{5}|+|1-\sqrt{5}|\)

\(=1+\sqrt{5}+\sqrt{5}-1\)

\(=2\sqrt{5}\)

\(\sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}}\)

= \(\sqrt{5}+1+\sqrt{5}-1\)

= \(2\sqrt{5}\)

\(A=\sqrt{14+6\sqrt{5}}+\sqrt{14-6\sqrt{5}}\)

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

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

\(A=3+\sqrt{5}+3-\sqrt{5}=6\)

b) \(B=\sqrt{7-4\sqrt{3}}-\sqrt{7+4\sqrt{3}}\)

\(B=\sqrt{3-4\sqrt{3}+4}-\sqrt{3+4\sqrt{3}+4}\)

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

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

Câu a tách 14 thành 5+9 . Có hằng đẳng thức

Câu b tương tự tách 7 thành 4+ 3 nhé

\(A=\sqrt{6-2\sqrt{5}}-\sqrt{6+2\sqrt{5}}\)

\(A=\sqrt{\left(\sqrt{5}-1\right)^2}-\sqrt{\left(\sqrt{5}+1\right)^2}\)

\(A=\sqrt{5}-1-\sqrt{5}-1\)

\(A=-2\)

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

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

\(B=\sqrt{5}+2-\sqrt{5}+2\)

\(B=4\)

Sửa đề :

\(C=\sqrt{14-6\sqrt{5}}-\sqrt{14+6\sqrt{5}}\)

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

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

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

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

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

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

\(=6+3x\)

=> \(x^3-3x=6\)

=> \(B=x^3-3x+2000=6+2000=2006\)

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

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

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

\(a.\sqrt{2}\cdot\sqrt{7-\sqrt{6}}\)

\(=\sqrt{2\cdot\left(7-\sqrt{6}\right)}\)

\(=\sqrt{14-2\cdot\sqrt{6}}\)

b) \(\sqrt{5-2\cdot\sqrt{6}}-\sqrt{5+2\cdot\sqrt{6}}\)

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

\(=\sqrt{3}-\sqrt{2}-\sqrt{3}+\sqrt{2}\)( VÌ \(\sqrt{3}>\sqrt{2}>0\))

\(=0\)

Mình làm hơi tắt nha !!

\(\sqrt{3}+\sqrt{11+6\sqrt{2}}-\sqrt{5+2\sqrt{6}}\)

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

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

\(=3\)

\(\sqrt{3}+\sqrt{11+6\sqrt{2}}-\sqrt{5+2\sqrt{6}}\)

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

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

\(=3\)

(Tíck cho mìk vs nhé!)

\(\frac{\sqrt{9-4\sqrt{5}}}{2-\sqrt{5}}\)

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

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

= \(\frac{\sqrt{5}-2}{2-\sqrt{5}}\)

= -1

Chúc bạn làm bài tốt :)

\(A=\sqrt{6-2\sqrt{5}}-\sqrt{6+2\sqrt{5}}\)

\(A=\sqrt{5-2\sqrt{5}+1}-\sqrt{5+2\sqrt{5}+1}\)

\(A=\sqrt{\left(\sqrt{5}-1\right)^2}-\sqrt{\left(\sqrt{5}+1\right)^2}\)

\(A=\sqrt{5}-1-\sqrt{5}-1\)

\(A=-2\)

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

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

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

\(B=\sqrt{5}+2-\sqrt{5}+2\)

\(B=4\)

Học tốt 

Bài làm:

a) \(A=\sqrt{6-2\sqrt{5}}-\sqrt{6+2\sqrt{5}}\)

\(A=\sqrt{5-2\sqrt{5}+1}-\sqrt{5+2\sqrt{5}+1}\)

\(A=\sqrt{\left(\sqrt{5}-1\right)^2}-\sqrt{\left(\sqrt{5}+1\right)^2}\)

\(A=\sqrt{5}-1-\sqrt{5}-1=-2\)

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

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

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

\(B=2+\sqrt{5}-\sqrt{5}+2\)

\(B=4\)

\(\frac{6-\sqrt{6}}{\sqrt{6}-1}+\frac{6+\sqrt{6}}{\sqrt{6}}\)\(=\frac{\sqrt{6}\left(\sqrt{6}-1\right)}{\sqrt{6}-1}+\frac{6}{\sqrt{6}}+\frac{\sqrt{6}}{\sqrt{6}}\)\(=\sqrt{6}+\frac{6}{\sqrt{6}}+1\)\(=\sqrt{6}\left(1+\frac{\sqrt{6}}{\sqrt{6}}\right)+1\)\(=\sqrt{6}\left(1+1\right)+1\)\(=\sqrt{6}.2+1\)
\(\frac{\sqrt{10}-\sqrt{2}}{\sqrt{5}-1}+\frac{2-\sqrt{2}}{\sqrt{2}-1}\)\(=\frac{\sqrt{2}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}+\frac{\sqrt{2}\left(\sqrt{2}-1\right)}{\sqrt{2}-1}\)\(=\sqrt{2}+\sqrt{2}=2\sqrt{2}\)
\(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)\(=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{20-2.3\sqrt{20}+9}}}\)\(=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(\sqrt{20}-3\right)^2}}}\)\(=\sqrt{\sqrt{5}-\sqrt{3-I\sqrt{20}-3I}}\)\(=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{20}+3}}\)\(=\sqrt{\sqrt{5}-\sqrt{5-2\sqrt{5}+1}}\)\(=\sqrt{\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}}\)\(=\sqrt{\sqrt{5}-I\sqrt{5}-1I}\)\(=\sqrt{\sqrt{5}-\sqrt{5}+1}\)\(=\sqrt{1}=1\)