=\(\frac{1}{2}x\left(\frac{2}{5x7}+\frac{2}{7x9}+\frac{2}{9x11}+...+\frac{2}{2015x2017}\right)\)
=\(\frac{1}{2}x\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{2015}-\frac{1}{2017}\right)\)
=\(\frac{1}{2}x\left(\frac{1}{5}-\frac{1}{2017}\right)\)
=\(\frac{1}{2}x\frac{2012}{10085}\)
=\(\frac{1006}{10085}\)
ta có : công thức 1/n.(n+k) = 1/k.(1/n-1/k)
=> 1/5.7+1/7.9.........+1/2015.2017
= 1/2.(1/5-1/7)+1/2.(1/7-1/9)...............+1/2.(1/2015-1/2017)
= 1/2.(1/5-1/7+1/7-1/9+.........+1/2015-1/2017)
=1/2.(1/5-1/2017)
=1/2.2012/10085
=1006/10085
( có thể là kết quả sai nhưng cách làm là đug rùi đấy ) * ^_^ *
Đặt \(D=\) \(\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+...+\frac{1}{2015.2017}\)
\(\Leftrightarrow D=2.\left(\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+...+\frac{1}{2015.2017}\right)\)
\(\Leftrightarrow D=\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{2015.2017}\)
\(\Leftrightarrow D=\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{2015}-\frac{1}{2017}\)
\(\Leftrightarrow D=\frac{1}{5}-\frac{1}{2017}\)
\(\Leftrightarrow D=\frac{2012}{10085}\)
A = \(\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+...+\frac{1}{2015.2017}\)
A.2 = \(\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+...+\frac{2}{2015.2017}\)
A.2 = \(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}...+\frac{1}{2015}-\frac{1}{2017}\)
A.2 = \(\frac{1}{5}-\frac{1}{2017}=\frac{2012}{10085}\)
A = \(\frac{2012}{10085}:2=\frac{1006}{10085}\)
*Cách khác:
A = \(\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+...+\frac{1}{2015.2017}\)
A = \(\frac{1}{2}\). \(\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+...+\frac{2}{2015.2017}\)
A = \(\frac{1}{2}\). (\(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}...+\frac{1}{2015}-\frac{1}{2017}\))
A = \(\frac{1}{2}\).(\(\frac{1}{5}-\frac{1}{2017}\))
A = \(\frac{1}{2}.\frac{2012}{10085}=\frac{1006}{10085}\)
Mình không biết là đúng hay sai nữa! Có thể là đúng rùi đấy * ^ _ ^ *.
Có lẽ thế hihi!