Question

一道数学啊，求高手帮写过程和为什么，关于分式加减的

有一题计算：1/x(x+1)+1/(x+1)(x+2)+1/(x+2)(x+3)+...+1/(x+2008)(x+2009)

高手帮忙啊

Answer 1

解：∵ 1/x(x+1)=1/x-1/(x+1)
1/(x+1)(x+2)=1/(x+1)-1/(x+2)
1/(x+2)(x+3)=1/(x+2)-1/(x+3)
...
1/(x+2008)(x+2009)=1/(x+2008)-1/(x+2009)
∴ 1/x(x+1)+1/(x+1)(x+2)+1/(x+2)(x+3)+...+1/(x+2008)(x+2009)
=1/x-1/(x+1)+1/(x+1)-1/(x+2)+1/(x+2)-1/(x+3)+...+1/(x+2008)-1/(x+2009)
=1/x-1/(x+2009)
=(x+2008)/(x(x+2009))

Answer 2

切
等于0

Answer 3

原式=1/x-1/(x+1)+1/(x+1)-1/(x+2)+...+1/(x+2008)-1/(x+2009)
=1/x-1/(x+2009)

Answer 4

1/x(x+1)+1/(x+1)(x+2)+1/(x+2)(x+3)+...+1/(x+2008)(x+2009)
=1/x-1/(x+1)+1/(x+1)-1/(x+2)+。。。。。。+1/(x+2008)-1/(x+2009)
=1/x-1/(x+2009)
=2008/x(x+2009)