58问答库 > 1⼀（x-1）（x-2）+1⼀（x-2）（x-3）+1⼀（x-3)(x-4)+......1⼀（x-2008)(x-2009)求解

1⼀（x-1）（x-2）+1⼀（x-2）（x-3）+1⼀（x-3)(x-4)+......1⼀（x-2008)(x-2009)求解

麻烦给我详细过程，还有这样做的原因

2024-11-01 23:55:55

推荐回答（4个）

回答（1）：

1/(x-1)(x-2)=[(x-1)-(x-2)]/(x-1)(x-2)=(x-1)/(x-1)(x-2)-(x-2)/(x-1)(x-2)=1/(x-2)-1/(x-1)
其余类同
∴1/（x-1）（x-2）+1/（x-2）（x-3）+1/（x-3)(x-4)+......1/（x-2008)(x-2009)
=1/(x-2)-1/(x-1)+1/(x-3)-1/(x-2)+1/(x-4)-1/(x-3)+……+1/(x-2009)-1/(2008)
观察1、4项，3、6项、5、8项……可相互抵消
=1/(x-2009)-1/(x-1)
=[(x-1)-(x-2009)]/(x-1)(x-2009)
=2008/(x²-2010x+2009)

回答（2）：

1/[（x-1）（x-2）]+1/[（x-2）（x-3）]+1/[（x-3)(x-4)]+......1/[（x-2008)(x-2009)]
=-[1/(x-1)-1/(x-2)]-[1/(x-2)-1/(x-3)]-[1/(x-3)-1/(x-4)]-……-[1/(x-2008)-1/(x-2009)]
=-1/(x-1)+1/(x-2)-1/(x-2)+1/(x-3)-1/(x-3)+1/(x-4)-……-1/(x-2008)+1/(x-2009)
=-1/(x-1)+1/(x-2009)
=-(x-2009)/[(x-1)(x-2009)]+(x-1)/[(x-1)(x-2009)]
=[-(x-2009)+(x-1)]/[(x-1)(x-2009)]
=2008/[(x-1)(x-2009)]

∵1/n-1/(n+1)
=1/[n(n-1)]

回答（3）：

1/（x-1）（x-2）+1/（x-2）（x-3）+1/（x-3)(x-4)+......1/（x-2008)(x-2009)
=1/(x-2)-1/(x-1)+1/(x-3)-1/(x-2)+1/(x-4)-1/(x-3)+……+1/(x-2009)-1/(x-2008)
=1/(x-2009)-1/(x-1)=2008/(x-1)(x-2009)

回答（4）：

原式=[1/(x-2)-1/(x-1)]+[1/(x-3)-1/(x-2)]+...........+[1/(x-2009)-1/(x-2008)]
=1/(x-2009)-1/(x-1)=2008/[(x-1)(x-2009)