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25 tháng 9 2018

\(\dfrac{\sqrt{5}-2\sqrt{3}}{\sqrt{5}+\sqrt{3}}-\dfrac{2\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}=\dfrac{\left(\sqrt{5}-2\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}-\dfrac{\left(2\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)}{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}=\dfrac{5-\sqrt{15}-2\sqrt{15}+6}{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}-\dfrac{10+2\sqrt{15}+\sqrt{15}+3}{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}=\dfrac{11-3\sqrt{15}-13-3\sqrt{15}}{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}=\dfrac{-2-6\sqrt{15}}{5-3}=\dfrac{-2\left(1+3\sqrt{15}\right)}{2}=-\left(1+3\sqrt{15}\right)=-1-3\sqrt{15}\)

25 tháng 9 2018

nhân tử vs mẫu vs \(\sqrt{5}-\sqrt{3}\)

và nhân tử vs mẫu vs\(\sqrt{5}-\sqrt{3}\)

31 tháng 5 2017

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

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

b) \(\dfrac{5}{12\left(2\sqrt{5}+3\sqrt{2}\right)}\) - \(\dfrac{5}{12\left(2\sqrt{5}-3\sqrt{2}\right)}\)

= \(\dfrac{5\left(2\sqrt{5}-3\sqrt{2}\right)-5\left(2\sqrt{5}+3\sqrt{2}\right)}{12\left(2\sqrt{5}+3\sqrt{2}\right)\left(2\sqrt{5}-3\sqrt{2}\right)}\)

= \(\dfrac{10\sqrt{5}-15\sqrt{2}-10\sqrt{5}-15\sqrt{2}}{12\left(20-18\right)}\)

= \(\dfrac{-30\sqrt{2}}{24}\) = \(\dfrac{-15\sqrt{2}}{12}\) = \(\dfrac{-5\sqrt{2}}{4}\)

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

= \(\dfrac{25+10\sqrt{5}+5+25-10\sqrt{5}+5}{25-5}\) = \(\dfrac{60}{20}\) = 3

31 tháng 5 2017

d) \(\dfrac{\sqrt{3}}{\sqrt{\sqrt{3+1}}-1}\) - \(\dfrac{\sqrt{3}}{\sqrt{\sqrt{3+1}}+1}\)

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

= \(\dfrac{\sqrt{6}+\sqrt{3}-\sqrt{6}+\sqrt{3}}{2-1}\) = \(2\sqrt{3}\)

a: \(\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{4}}{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}\)

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

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

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

\(=\dfrac{\sqrt{5}+1}{2-\sqrt{5}+1}-\dfrac{\sqrt{5}-1}{3+\sqrt{5}}\)

\(=\dfrac{\left(3+\sqrt{5}\right)\left(\sqrt{5}+1\right)-\left(\sqrt{5}-1\right)\left(3-\sqrt{5}\right)}{4}\)

\(=\dfrac{3\sqrt{5}+3+5+\sqrt{5}-3\sqrt{5}+5+3-\sqrt{5}}{4}\)

\(=4\)

\(=\dfrac{2\sqrt{2}+\sqrt{10}}{2+\sqrt{5}+1}+\dfrac{2\sqrt{2}-\sqrt{10}}{2-\sqrt{5}+1}\)

\(=\dfrac{\sqrt{2}\left(2+\sqrt{5}\right)\left(3-\sqrt{5}\right)+\sqrt{2}\left(2-\sqrt{5}\right)\left(3+\sqrt{5}\right)}{4}\)

\(=\dfrac{\sqrt{2}\left(6-2\sqrt{5}+3\sqrt{5}-5+6+2\sqrt{5}-3\sqrt{5}-5\right)}{4}\)

\(=\sqrt{2}\cdot\dfrac{2}{4}=\dfrac{1}{\sqrt{2}}\)

13 tháng 6 2017

a) \(\sqrt{\dfrac{2-\sqrt{5}}{\sqrt{5}-3}}:\sqrt{\left(\sqrt{5}-3\right)\left(2-\sqrt{5}\right)}\)

\(=\sqrt{\dfrac{2-\sqrt{5}}{\sqrt{5}-3}}:\left(\left(\sqrt{5}-3\right)\cdot\left(2-\sqrt{5}\right)\right)\)

\(=\sqrt{\dfrac{2-\sqrt{5}}{\sqrt{5}-3}:\left(2\sqrt{5}-5-6+3\sqrt{5}\right)}\)

\(=\sqrt{\dfrac{2-\sqrt{5}}{\sqrt{5}-3}:\left(5\sqrt{5}-11\right)}\)

\(=\sqrt{\dfrac{2-\sqrt{5}}{\sqrt{5}-3}\cdot\dfrac{1}{5\sqrt{5}-11}}\)

\(=\sqrt{\dfrac{2-\sqrt{5}}{\left(\sqrt{5}-3\right)\cdot\left(5\sqrt{5}-1\right)}}\)

\(=\sqrt{\dfrac{\left(2-\sqrt{5}\right)\cdot\left(\sqrt{5}+3\right)}{-4\left(5\sqrt{5}-1\right)}}\)

\(=\sqrt{\dfrac{2\sqrt{5}+6-5-3\sqrt{5}}{-4\left(5\sqrt{5}-11\right)}}\)

\(=\sqrt{\dfrac{-\sqrt{5}+1}{-4\left(5\sqrt{5}-11\right)}}\)

\(=\sqrt{-\dfrac{\left(-\sqrt{5}+1\right)\cdot\left(5\sqrt{5}+11\right)}{16}}\)

\(=\sqrt{-\dfrac{-25-11\sqrt{5}+5\sqrt{5}+11}{16}}\)

\(=\sqrt{-\dfrac{-14-6\sqrt{5}}{16}}\)

\(=\sqrt{-\dfrac{2\left(-7-3\sqrt{5}\right)}{16}}\)

\(=\sqrt{-\dfrac{-7-3\sqrt{5}}{8}}\)

\(=\dfrac{\sqrt{-\left(-7-3\sqrt{5}\right)}}{\sqrt{8}}\)

\(=\dfrac{\sqrt{7+3\sqrt{5}}}{2\sqrt{2}}\)

\(=\dfrac{\sqrt{\left(7+3\sqrt{5}\right)\cdot2}}{4}\)

\(=\dfrac{\sqrt{14+6\sqrt{5}}}{4}\)

\(=\dfrac{\sqrt{\left(3+\sqrt{5}\right)^2}}{4}\)

\(=\dfrac{3+\sqrt{5}}{4}\)

b) \(\dfrac{2+3\sqrt{5}}{\sqrt{5}-2}-\dfrac{\sqrt{5}+1}{\sqrt{5}+2}\)

\(=\left(2+3\sqrt{5}\right)\cdot\left(\sqrt{5}+2\right)-\left(\sqrt{5}+1\right)\cdot\left(\sqrt{5}-2\right)\)

\(=2\sqrt{5}+4+15+6\sqrt{5}-\left(5-2\sqrt{5}+\sqrt{5}-2\right)\)

\(=2\sqrt{5}+4+15+6\sqrt{5}-\left(3-\sqrt{5}\right)\)

\(=2\sqrt{5}+4+15+6\sqrt{5}-3+\sqrt{5}\)

\(=9\sqrt{5}+16\)

c) \(\dfrac{1+\sqrt{2}}{\sqrt{4-2\sqrt{3}}}:\dfrac{\sqrt{3}+1}{\sqrt{2}-1}\)

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

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

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

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

\(=\dfrac{2-1}{2}\)

\(=\dfrac{1}{2}\)

13 tháng 6 2017

a) \(\sqrt{\dfrac{2-\sqrt{5}}{\sqrt{5}-3}}:\sqrt{\left(\sqrt{5}-3\right)\left(2-\sqrt{5}\right)}\)= \(\dfrac{\sqrt{2-\sqrt{5}}}{\sqrt{\sqrt{5}-3}}.\dfrac{1}{\sqrt{\sqrt{5}-3}\sqrt{2-\sqrt{5}}}\)

= \(\dfrac{1}{\sqrt{\sqrt{5}-3}}.\dfrac{1}{\sqrt{\sqrt{5}-3}}\) = \(\dfrac{1}{\sqrt{\sqrt{5}-3}^2}\) = \(\dfrac{1}{3-\sqrt{5}}\)

b) \(\dfrac{2+3\sqrt{5}}{\sqrt{5}-2}-\dfrac{\sqrt{5}+1}{\sqrt{5}+2}\) = \(\dfrac{\left(2+3\sqrt{5}\right)\left(\sqrt{5}+2\right)-\left(\sqrt{5}+1\right)\left(\sqrt{5}-2\right)}{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}\)

= \(\dfrac{2\sqrt{5}+4+15+6\sqrt{5}-\left(5-2\sqrt{5}+\sqrt{5}-2\right)}{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}\)

= \(\dfrac{8\sqrt{5}+19-5+2\sqrt{5}-\sqrt{5}+2}{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}\) = \(\dfrac{9\sqrt{5}+16}{5-4}\) = \(9\sqrt{5}+16\)

c) \(\dfrac{1+\sqrt{2}}{\sqrt{4-2\sqrt{3}}}:\dfrac{\sqrt{3}+1}{\sqrt{2}-1}\) = \(\dfrac{1+\sqrt{2}}{\sqrt{\left(\sqrt{3}-1\right)^2}}:\dfrac{\sqrt{3}+1}{\sqrt{2}-1}\)

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

= \(\dfrac{1}{2}\)

30 tháng 10 2017

\(\frac{1}{(n+1)\sqrt{n}+n\sqrt{n+1} }=\frac{1}{\sqrt{n(n+1)}(\sqrt{n+1)+\sqrt{n}) } } =\frac{\sqrt{n+1}-\sqrt{n} }{\sqrt{n(n+1)} } =\frac{1}{\sqrt{n} }-\frac{1}{\sqrt{n+1} } \)

=>K=1-\( \frac{1}{5} \)=\(\frac{4}{5} \)

19 tháng 1 2018

nhan them \(\sqrt{2}\)

vao mau voi tu la dc

21 tháng 1 2018

ko hiểu

b: Ta có: \(\dfrac{1}{2+\sqrt{3}}+\dfrac{\sqrt{2}}{\sqrt{6}}-\dfrac{2}{3+\sqrt{3}}\)

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

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