1/1*3 + 1/3*5 + 1/5*7 + ... + 1/2007*2009
= 1/2(2/1*3 + 2/3*5 + 2/5*7 + ... + 2/2007*2009)
= 1/2(1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + ... + 1/2007 - 1/2009)
= 1/2( 1- 1/2009)
= 1/2 * 2008/2009
= 1009/2009
#)Giải :
Gọi A = 1/1.3 + 1/3.5 + 1/5.7 + ... + 1/2007.2009
A = 1/2 . ( 1/1 - 1/3 + 1/3 - 1/5 + ... + 1/2007 - 1/2009
A = 1/2 . ( 1/1 - 1/2009 )
A = 1/2 . 2008/2009
A = 1004/2009
#)Chúc bn học tốt :D
\(\frac{1}{1.3}\)+\(\frac{1}{3.5}\)+...+\(\frac{1}{2007.2009}\)=(\(\frac{2}{1.3}\)+\(\frac{2}{3.5}\)+\(\frac{2}{2007.2009}\)) (*)
Ta có:\(\frac{2}{n.\left(n+2\right)}\)=\(\frac{\left(n+2\right)-n}{n.\left(n+2\right)}\)=\(\frac{n+2}{n.\left(n+2\right)}\)-\(\frac{n}{n.\left(n+2\right)}\)=\(\frac{1}{n}\)-\(\frac{1}{n+2}\)
Áp dụng vào (*),ta được:\(\frac{2}{1.3}\)+\(\frac{2}{3.5}\)+\(\frac{2}{2007.2009}\)=\(\frac{1}{1}\)-\(\frac{1}{3}\)+\(\frac{1}{3}\)-\(\frac{1}{5}\)+...+\(\frac{1}{2007}\)-\(\frac{1}{2009}\)=1-\(\frac{1}{2009}\)=\(\frac{2008}{2009}\)
Vậy:\(\frac{1}{1.3}\)+\(\frac{1}{3.5}\)+...+\(\frac{1}{2007.2009}\)=\(\frac{2008}{2009}\)
dễ thế mà ko làm dc con lạy mấy má luôn.
của mk làm đúng lun đó :V, các bn cứ ra mấy cái đáp án linh tinh j :v
Gọi A=1/1.3+1/3.5+....+1/2007.2009
2A=2/1.3+2/3.5+....+2/2007.2009
2A=1/1-1/3+1/3-1/5+.......+1/2007-1/2009
2A=1-1/2009
2A=2009/2009-1/2009
2A=2008/2009
A=2008/2009:2
A=1004/2009·
