2/[(x+1)(x+3)]+2/[(x+3)(x+5)]+``````+2/[(x+2007)(x+2009)]
问题描述：
2/[(x+1)(x+3)]+2/[(x+3)(x+5)]+``````+2/[(x+2007)(x+2009)]
答
[2/(x+1)(x+3)]+[2/(x+3)(x+5)]+···+[2/(x+2007)(x+2009)] =[1/(x+1)]-[1/(x+3)]+1/[(x+3)]-[1/(x+5)]·+··+[1/(x+2007)]-[1/(x+2009)] =[1/(x+1)]-[1/(x+2009)] =(x+2009-x-1)/(x+1)(x+2009) =2008/(x+1)(x+2008))
答
此题为典型的拆分法! 适用条件：分母为两个因式的乘积、二者的差为一常数,且前后项的分母的因式有一项相同. 如3/[（x+3）（x+5）]=[1/（x+3）+1/（x+5）]*1.5 因此此题可以拆成1/(x+1)-1/(x+3)+1/(x+3)-1/(x+5)+...+1/(x+2005)-1/(x+2007) 再前后抵消得1/(x+1)-1/(x+2007) 即得2006/[(x+1)(x+2007)] 希望采纳!谢谢