谁是数学天才,解方程.

x提出后

1/(1×2)+1/(2×3)+1/(3×4)…………… +1/(2007×2008)
＝(1-1/2)+(1/2-1/3)+……+(1/2007-1/2008)
=1-1/2008
=2007/2008

所以x=2008

x[1/(1*2)+1/(2*3)+1/(3*4)……+1/(2007*2008)]=2007
x[(1-1/2)+（1/2-1/3)+(1/3-1/4)……+（1/2007-1/2008)]=2007
中括号内各项可以抵消 如下
x[1-1/2008]=2007
故x=2008

首先你要知道1/(1×2)=1/x-1/2 1/(2×3)=1/2-1/3

x/(1×2)+x/(2×3)+x/(3×4)…………… +x/(2007×2008)
=x(1/1-1/2+1/2-1/3+1/3-1/4+……+1/2007-1/2008)
=x-x/2008

x-x/2008=2007
x=2008