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7 tháng 9 2016

hề gặp bạn oy

 

7 tháng 9 2016

ò

bài kho quá à

17 tháng 2 2017

\(A=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}}{\frac{2010}{1}+\frac{2009}{2}+...+\frac{1}{2010}}\)

\(\Rightarrow A=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}}{\left(1+1+1+...+1\right)+\frac{2009}{2}+\frac{2008}{3}+...+\frac{1}{2010}}\)

\(\Rightarrow A=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}}{1+\left(1+\frac{2009}{2}\right)+\left(1+\frac{2008}{3}\right)+...+\left(1+\frac{1}{2010}\right)}\)

\(\Rightarrow A=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}}{\frac{2011}{2}+\frac{2011}{3}+...+\frac{2011}{2010}+\frac{2011}{2011}}\)

\(\Rightarrow A=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}}{2011.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}\right)}\)

\(\Rightarrow A=\frac{1}{2011}\)

17 tháng 2 2017

\(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2011}}{\frac{2010}{1}+\frac{2009}{2}+\frac{2008}{3}+...+\frac{1}{2010}}\)

\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2011}}{\left(\frac{2009}{2}+1\right)+\left(\frac{2008}{3}+1\right)+....+\left(\frac{1}{2010}+1\right)}\)

\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2011}}{\frac{2011}{2}+\frac{2011}{3}+....+\frac{2011}{2010}}\)

\(=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}}{2011\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}\right)}\)

\(=\frac{1}{2011}\)

17 tháng 8 2019

\(a;\)

\(=0-1\)

\(=-1\)

17 tháng 8 2019

\(b;\)

\(=0-4\)

\(=-4\)

7 tháng 10 2019

\(A=\left(\frac{1}{125}-\frac{1}{1^3}\right).\left(\frac{1}{125}-\frac{1}{2^3}\right)...\left(\frac{1}{125}-\frac{1}{100^3}\right)\)

\(A=\left(\frac{1}{125}-\frac{1}{1^3}\right).\left(\frac{1}{125}-\frac{1}{2^3}\right)...\left(\frac{1}{125}-\frac{1}{5^3}\right)...\left(\frac{1}{125}-\frac{1}{100^3}\right)\)

\(A=\left(\frac{1}{125}-\frac{1}{1^3}\right).\left(\frac{1}{125}-\frac{1}{2^3}\right)...0...\left(\frac{1}{125}-\frac{1}{100^3}\right)\)

\(\Rightarrow A=0\)

18 tháng 10 2020

\(M=\left(\frac{0,4-\frac{2}{9}+\frac{2}{11}}{1,4-\frac{7}{9}+\frac{7}{11}}-\frac{\frac{1}{3}-0,25+\frac{1}{5}}{1\frac{1}{6}-0,875+0,7}\right):\frac{2014}{2015}\)

\(=\left(\frac{\frac{2}{5}-\frac{2}{9}+\frac{2}{11}}{\frac{7}{5}-\frac{7}{9}+\frac{7}{11}}-\frac{\frac{1}{3}-\frac{1}{4}+\frac{1}{5}}{\frac{7}{6}-\frac{7}{8}+\frac{7}{10}}\right):\frac{2014}{2015}\)

\(=\left[\frac{2.\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}{7.\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}-\frac{\frac{1}{3}-\frac{1}{4}+\frac{1}{5}}{\frac{7}{2}.\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{5}\right)}\right].\frac{2015}{2014}\)

\(=\left(\frac{2}{7}-\frac{1}{\frac{7}{2}}\right).\frac{2015}{2014}=\left(\frac{2}{7}-\frac{2}{7}\right).\frac{2015}{2014}=0\)

18 tháng 10 2020

Ta có M = \(\left(\frac{0,4-\frac{2}{9}+\frac{2}{11}}{1,4-\frac{7}{9}+\frac{7}{11}}-\frac{\frac{1}{3}-0,25+\frac{1}{5}}{1\frac{1}{6}-0,875+0,7}\right):\frac{2014}{2015}\)

\(\left(\frac{2\left(0,2-\frac{1}{9}+\frac{1}{11}\right)}{7\left(0,2-\frac{1}{9}+\frac{1}{11}\right)}-\frac{\frac{1}{3}-0,25+\frac{1}{5}}{\frac{7}{2}\left(\frac{1}{3}-0,25+\frac{1}{5}\right)}\right):\frac{2014}{2015}\)

\(\left(\frac{2}{7}-\frac{1}{\frac{7}{2}}\right):\frac{2014}{2015}=\left(\frac{2}{7}-\frac{2}{7}\right):\left(\frac{2014}{2015}\right)=0\)