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利用规律计算
观察下列各式:
1/2=1/(1*2)=1-1/2
1/6=1/(2*3)=1/2-1/3
1/12=1/(3*4)=1/3-1/4
1.利用上述规律计算:
1/2+1/6+1/12+……+1/[(n-1)n]+1/[(n+1)n]
2.利用上述规律解方程:
1/[(x-4)(x-3)]+1/[(x-3)(x-2)]+1/[(x-2)(x-1)]+1/[(x-1)x]+1/[(x+1)x]=1/(x+1)
人气:256 ℃ 时间:2020-03-27 23:37:26
解答
1、
1/2+1/6+1/12+……+1/[(n-1)n]+1/[(n+1)n]
=1-1/2+1/2-1/3+1/3-1/4+...+1/(n-1)-1/n+1/n-1/(n+1)
=1-1/(n+1)
=n/(n+1)
2、
1/[(x-4)(x-3)]+1/[(x-3)(x-2)]+1/[(x-2)(x-1)]+1/[(x-1)x]+1/[(x+1)x]=1/(x+1)
可化为:
1/(x-4)-1/(x-3)+1/(x-3)-1/(x-2)+1/(x-2)-1/(x-1)+1/(x-1)-1/x+1/x-1/(x+1)=1/(x+1)
所以
1/(x-4)=2/(x+1)
解得x=9
经检验,x=9是原方程的根
所以原方程的根是:x=9
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