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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

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

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

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

\(\Rightarrow A=0\)

29 tháng 6 2017

Ta có : \(\left(\frac{1}{49}-\frac{1}{2^2}\right)\left(\frac{1}{49}-\frac{1}{3^2}\right)\left(\frac{1}{49}-\frac{1}{4^2}\right).......\left(\frac{1}{49}-\frac{1}{40^2}\right)\)

\(=\left(\frac{1}{49}-\frac{1}{2^2}\right)\left(\frac{1}{49}-\frac{1}{3^2}\right)......\left(\frac{1}{49}-\frac{1}{7^2}\right)......\left(\frac{1}{49}-\frac{1}{40^2}\right)\)

\(=\left(\frac{1}{49}-\frac{1}{2^2}\right)\left(\frac{1}{49}-\frac{1}{3^2}\right)......0......\left(\frac{1}{49}-\frac{1}{40^2}\right)\)

\(=0\)