\(B=\frac{\sqrt{2+\sqrt{3}}}{\sqrt{2}}-\frac{\sqrt{2-\sqrt{3}}}{\sqrt{2}...">
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10 tháng 7 2021

\(B=\frac{\sqrt{4+2\sqrt{3}}}{2}-\frac{\sqrt{4-2\sqrt{3}}}{2}\)

\(B=\frac{\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\left(\sqrt{3}-1\right)^2}}{2}\)

\(B=\frac{\left|\sqrt{3}+1\right|-\left|\sqrt{3}-1\right|}{2}\)

\(B=\frac{\sqrt{3}+1-\sqrt{3}+1}{2}\)

\(B=\frac{2}{2}=1\)

6 tháng 7 2019

\(b,\frac{2+\sqrt{3}}{1-\sqrt{4-2\sqrt{3}}}+\frac{2-\sqrt{3}}{1+\sqrt{4+2\sqrt{3}}}\)

\(=\frac{2+\sqrt{3}}{1-\sqrt{3-2\sqrt{3}+1}}+\frac{2-\sqrt{3}}{1+\sqrt{3+2\sqrt{3}+1}}\)

\(=\frac{2+\sqrt{3}}{1-\sqrt{\left(\sqrt{3}-1\right)^2}}+\frac{2-\sqrt{3}}{1+\sqrt{\left(\sqrt{3}+1\right)^2}}\)

\(=\frac{2+\sqrt{3}}{1-\left(\sqrt{3}-1\right)}+\frac{2-\sqrt{3}}{1+\sqrt{3}+1}\)

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

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

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

\(=14\)

6 tháng 7 2019

\(a,\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=\frac{\sqrt{2}+\sqrt{3}+4+\sqrt{6}+\sqrt{8}}{\sqrt{2}+\sqrt{3}+2}\)

\(=\frac{\sqrt{2}+\sqrt{3}+2}{\sqrt{2}+\sqrt{3}+2}+\frac{\sqrt{2}.\sqrt{2}+\sqrt{2}.\sqrt{3}+\sqrt{2}.2}{\sqrt{2}+\sqrt{3}+2}\)

\(=1+\frac{\sqrt{2}\left(\sqrt{2}+\sqrt{3}+2\right)}{\sqrt{2}+\sqrt{3}+2}\)

\(=1+\sqrt{2}\)

23 tháng 6 2017

a) \(\frac{\sqrt{4-2\sqrt{3}}}{\sqrt{6}-\sqrt{2}}=\frac{\sqrt{3}-1}{\sqrt{2}\left(\sqrt{3}-1\right)}=\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}\)

b) \(\frac{1}{2\sqrt{3}}+\frac{\sqrt{2}}{\sqrt{6}}-\frac{2}{3+\sqrt{3}}=\frac{2\sqrt{3}}{12}+\frac{2\sqrt{3}}{6}-\frac{6-2\sqrt{3}}{6}\)

\(=\frac{2\sqrt{3}}{12}+\frac{4\sqrt{3}}{12}-\frac{12-4\sqrt{3}}{12}=\frac{-12+10\sqrt{3}}{12}=\frac{-6+5\sqrt{3}}{6}\)

8 tháng 7 2018

\(a.\sqrt{\frac{2-\sqrt{3}}{2}}+\frac{1-\sqrt{3}}{2}\)

\(=\sqrt{\frac{2\left(2-\sqrt{3}\right)}{4}}+\frac{1-\sqrt{3}}{2}\)

\(=\frac{\sqrt{4-2\sqrt{3}}}{2}+\frac{1-\sqrt{3}}{2}\)

\(=\frac{\sqrt{\left(\sqrt{3}-1\right)^2}}{2}+\frac{1-\sqrt{3}}{2}\)

\(=\frac{\sqrt{3}-1+1-\sqrt{3}}{2}\) ( Vì \(\sqrt{3}-1>0\))

\(=0\)

b) \(\frac{1}{2+\sqrt{3}}+\frac{\sqrt{2}}{\sqrt{6}}-\frac{2}{3+\sqrt{3}}\)

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

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

\(=\frac{6-3+\sqrt{3}}{3}\)

\(=\frac{3+\sqrt{3}}{3}=\frac{\sqrt{3}+1}{\sqrt{3}}\)

c) \(\frac{3}{2+\sqrt{3}}+\frac{13}{4-\sqrt{3}}+\frac{6}{\sqrt{3}}\)

\(=\frac{2\left(2-\sqrt{3}\right)}{1}+\frac{13\left(1+\sqrt{3}\right)}{13}+2\sqrt{3}\)

\(=4-2\sqrt{3}+1-\sqrt{3}+2\sqrt{3}\)

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

8 tháng 7 2018

ban mai thanh xuân ơi cầu c sai

24 tháng 5 2017

\(C=\sqrt{2-\sqrt{3}}.\left(\sqrt{6}+\sqrt{2}\right)\)

\(=\sqrt{\frac{4-2\sqrt{3}}{2}}.\left[\sqrt{2}.\left(\sqrt{3}+\sqrt{1}\right)\right]\)

\(=\sqrt{\frac{\left(\sqrt{3}-1\right)^2}{2}}.\sqrt{2}.\left(\sqrt{3}+1\right)\)

\(=\frac{\sqrt{3}-1}{\sqrt{2}}.\sqrt{2}\left(\sqrt{3}+1\right)=\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)=3-1=2\)

24 tháng 5 2017

\(D=\frac{8+2\sqrt{2}}{3-\sqrt{2}}-\frac{2+3\sqrt{2}}{\sqrt{2}}+\frac{\sqrt{2}}{1-\sqrt{2}}\)

\(=\frac{\left(8+2\sqrt{2}\right).\left(3+\sqrt{2}\right)}{9-2}-\frac{\sqrt{2}.\left(2+3\sqrt{2}\right)}{2}+\frac{\sqrt{2}.\left(1+\sqrt{2}\right)}{1-2}\)

\(=\frac{24+14\sqrt{2}+4}{7}-\frac{2\sqrt{2}+6}{2}-\frac{\sqrt{2}+2}{1}\)

\(=\frac{28+14\sqrt{2}}{7}-\sqrt{2}-3-\sqrt{2}-2\)

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

\(=-1\)

30 tháng 10 2019

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

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

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

\(=6+3x\)

=> \(x^3-3x=6\)

=> \(B=x^3-3x+2000=6+2000=2006\)

30 tháng 10 2019

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

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

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

30 tháng 8 2018

a) \(\frac{\sqrt{6}+\sqrt{14}}{2\sqrt{3}+\sqrt{28}}=\frac{\sqrt{2}\left(\sqrt{3}+\sqrt{7}\right)}{2\left(\sqrt{3}+\sqrt{7}\right)}=\frac{1}{\sqrt{2}}\)

b) \(\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\left(\sqrt{4}+\sqrt{6}+\sqrt{8}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)\left(1+\sqrt{2}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=1+\sqrt{2}\)

15 tháng 7 2019

a ) \(A=\frac{1}{\sqrt{5}+\sqrt{3}}-\frac{1}{\sqrt{5}-\sqrt{3}}\)

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

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

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

\(=-\sqrt{3}\)

15 tháng 7 2019

c ) \(\frac{1}{2+\sqrt{3}}+\frac{\sqrt{2}}{\sqrt{6}}-\frac{2}{3+\sqrt{3}}\)

\(=\frac{1}{2+\sqrt{3}}+\frac{1}{\sqrt{3}}-\frac{2}{\sqrt{3}\left(\sqrt{3}+1\right)}\)

\(=\frac{\sqrt{3}\left(\sqrt{3}+1\right)+\left(2+\sqrt{3}\right)\left(\sqrt{3}+1\right)-2\left(2+\sqrt{3}\right)}{\sqrt{3}\left(\sqrt{3}+1\right)\left(2+\sqrt{3}\right)}\)

\(=\frac{2\sqrt{3}+4}{\sqrt{3}\left(\sqrt{3}+1\right)\left(2+\sqrt{3}\right)}\)

\(=\frac{2\left(\sqrt{3}+2\right)}{\sqrt{3}\left(\sqrt{3}+1\right)\left(2+\sqrt{3}\right)}\)

\(=\frac{2.\sqrt{3}\left(\sqrt{3}-1\right)}{3\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}\)

\(=\frac{2\sqrt{3}\left(\sqrt{3}-1\right)}{3.\left(3-1\right)}\)

\(=\frac{\sqrt{3}\left(\sqrt{3}-1\right)}{3}\)

\(=\frac{3-\sqrt{3}}{3}\)

\(=1-\frac{\sqrt{3}}{3}\)