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2 tháng 4 2017

\(A=3+\frac{1}{1+\frac{1}{1+\frac{1}{\frac{4}{3}}}}=3+\frac{1}{1+\frac{1}{1+\frac{3}{4}}}\)

\(=3+\frac{1}{1+\frac{1}{\frac{7}{4}}}=3+\frac{1}{1+\frac{4}{7}}=3+\frac{1}{\frac{11}{4}}=3+\frac{4}{11}=\frac{37}{11}\)

\(B=-5+\frac{1}{1-\frac{1}{2+\frac{1}{\frac{3}{4}}}}=-5+\frac{1}{1-\frac{1}{2+\frac{4}{3}}}\)

\(=-5+\frac{1}{1-\frac{1}{\frac{10}{3}}}=-5+\frac{1}{1-\frac{3}{10}}=-5+\frac{1}{\frac{7}{10}}=-5+\frac{10}{7}=\frac{-25}{7}\)

25 tháng 3 2018

\(A=\frac{\left(1+2+...+100\right)\left(\frac{1}{2}^2-...-\frac{1}{5}\right)\left(2,4.42-21.4,8\right)}{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}}\)

=> \(A=\frac{\left(1+2+...+100\right)\left(\frac{1}{2}-...-\frac{1}{5}\right).0}{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}}\)=     0

18 tháng 9 2019

Thêm đk \(a,b,c\ne0\)

Ta có: \(\frac{ab}{a+b}=\frac{1}{3}\Rightarrow\frac{a+b}{ab}=3\)

\(\frac{bc}{b+c}=\frac{1}{4}\Rightarrow\frac{bc}{b+c}=4\)

\(\frac{ca}{c+a}=\frac{1}{5}\Rightarrow\frac{c+a}{ca}=5\)

\(\Rightarrow\frac{a+b}{ab}+\frac{b+c}{bc}+\frac{c+a}{ca}=12\)

\(\Leftrightarrow\frac{1}{b}+\frac{1}{a}+\frac{1}{c}+\frac{1}{b}+\frac{1}{a}+\frac{1}{c}=12\)

\(\Leftrightarrow2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=12\)

\(\Leftrightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=6\)

22 tháng 11 2017

giup minh voi cac ban

8 tháng 8 2019

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

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

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

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

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

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

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

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

\(=1+1=2\)

8 tháng 8 2019

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

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

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

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

\(A=\frac{\left(3-\sqrt{3}\right)\left(2+\sqrt{3}\right)+\left(2-\sqrt{3}\right)\left(3+\sqrt{3}\right)}{6}\)

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

\(A=\frac{6}{6}=1\)

13 tháng 6 2020

(1-1/3)x(1-1/5)x(1-1/7)x(1-1/9)x(1-1/2)x(1-1/4)x(1-1/6)x(1-1/8)x(1-1/10)

=2/3x4/5x6/7x8/9x1/2x3/4x5/6x7/8x9/10

=2x4x6x8x1x3x5x7x9 /3x5x7x9x2x4x6x8x10

=1/10

29 tháng 7 2019

B = \(\frac{-2}{3}+\frac{3}{4}-\frac{-1}{6}+\frac{-2}{5}=\frac{-240+270+60-144}{360}=\frac{-54}{360}=-0,15\)