化简a(a+1)/1+(a+1)(a+2)/1.+(a+2009)(a+2010)/1

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化简a(a+1)/1+(a+1)(a+2)/1.+(a+2009)(a+2010)/1化简a(a+1)/1+(a+1)(a+2)/1.+(a+2009)(a+2010)/1化简a(a+1)/1+(a+

化简a(a+1)/1+(a+1)(a+2)/1.+(a+2009)(a+2010)/1
化简a(a+1)/1+(a+1)(a+2)/1.+(a+2009)(a+2010)/1

化简a(a+1)/1+(a+1)(a+2)/1.+(a+2009)(a+2010)/1
题目写错了吧,应该是1/[a(a+1)]+1/[(a+1)(a+2)]+...+1/[(a+2009)(a+2010)]
原式=[(a+1)-a]/[a(a+1)]+[(a+2)-(a+1)]/[(a+1)(a+2)]+...+[(a+2010)-(a+2009)]/[(a+2009)(a+2010)]
=[1/a-1/(a+1)]+[1/(a+1)-1/(a+2)]+...+[1/(a+2009)-1/(a+2010)]
=1/a-1/(a+2010)
=2010/[a(a+2010)]
希望能解决您的问题.

a(a+1)/1可化为a/1-(a+1)/1
同理以后的都这样化为两个相减
原式最后就变为a/1-(a+2010)/1
满意请采纳