)的值a-1的绝对值与b-2的绝对值互为相反数,求代数式ab分之1+(a+1)(b+1)分之1+.+(a+2012)(b+2012
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/02 07:36:34
![)的值a-1的绝对值与b-2的绝对值互为相反数,求代数式ab分之1+(a+1)(b+1)分之1+.+(a+2012)(b+2012](/uploads/image/z/1008223-7-3.jpg?t=%EF%BC%89%E7%9A%84%E5%80%BCa-1%E7%9A%84%E7%BB%9D%E5%AF%B9%E5%80%BC%E4%B8%8Eb-2%E7%9A%84%E7%BB%9D%E5%AF%B9%E5%80%BC%E4%BA%92%E4%B8%BA%E7%9B%B8%E5%8F%8D%E6%95%B0%2C%E6%B1%82%E4%BB%A3%E6%95%B0%E5%BC%8Fab%E5%88%86%E4%B9%8B1%2B%EF%BC%88a%2B1%EF%BC%89%EF%BC%88b%2B1%EF%BC%89%E5%88%86%E4%B9%8B1%2B.%2B%EF%BC%88a%2B2012%29%EF%BC%88b%2B2012)
)的值a-1的绝对值与b-2的绝对值互为相反数,求代数式ab分之1+(a+1)(b+1)分之1+.+(a+2012)(b+2012
)的值a-1的绝对值与b-2的绝对值互为相反数,求代数式ab分之1+(a+1)(b+1)分之1+.+(a+2012)(b+2012
)的值a-1的绝对值与b-2的绝对值互为相反数,求代数式ab分之1+(a+1)(b+1)分之1+.+(a+2012)(b+2012
a-1的绝对值与b-2的绝对值互为相反数
∴|a-1|=-|b-2|
∴|a-1|=|b-2|=0
解得:a=1 b=2
∴1/ab+1/(a+1)(b+1)+.+1/(a+2012)(b+2012)
=1/(1*2)+1/(2*3)+……+1/(2013*2014)
=(1-1/2)+(1/2-1/3)+……+(1/2013-1/2014)
=1-1/2014
=2013/2014
有绝对就要大于等于0
所以a-1=0 b-2=0 解得a=1 b=2
1/ab+1/(a+1)(b+1)+......+1/(a+2012)(b+2012)
=1/(1*2)+1/(2*3)+……+1/(2013*2014)
=(1-1/2)+(1/2-1/3)+……+(1/2013-1/2014)
=1-1/20...
全部展开
有绝对就要大于等于0
所以a-1=0 b-2=0 解得a=1 b=2
1/ab+1/(a+1)(b+1)+......+1/(a+2012)(b+2012)
=1/(1*2)+1/(2*3)+……+1/(2013*2014)
=(1-1/2)+(1/2-1/3)+……+(1/2013-1/2014)
=1-1/2014
=2013/2014
收起