算1/1+根号2+1/根号2+根号3+1/根号3+根号4……+1/根号99+根号100
来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/05/23 17:55:07
算1/1+根号2+1/根号2+根号3+1/根号3+根号4……+1/根号99+根号100
求,10点以前回答
求,10点以前回答
1/1+根号2+1/根号2+根号3+1/根号3+根号4……+1/根号99+根号100
=(√2-1)/[(√2+1)(√2-1)]+(√3-√2)/[(√3+√2)(√3-√2)]+(√4-√3)/[(√4+√3)(√4-√3)]+……+(√100-√99)/[(√100+√99)(√100-√99)]
=√2-1+√3-√2+√4-√3+……+√100-√99
=√100-1
=10-1
=9
再问: 为何1/1+根号2+1/根号2+根号3+1/根号3+根号4……+1/根号99+根号100 =(√2-1)/[(√2+1)(√2-1)]+(√3-√2)/[(√3+√2)(√3-√2)]+(√4-√3)/[(√4+√3)(√4-√3)]+……+(√100-√99)/[(√100+√99)(√100-√99)]
再答: 1/(√2+1) =(√2-1)/[(√2+1)(√2-1)] =(√2-1)/(2-1) =√2-1 1/(√3-√2) =(√3-√2)/[(√3+√2)(√3-√2)] =(√3-√2)/(3-2) =√3-√2 …… 1/(√100+√99) =(√100-√99)/[(√100+√99)(√100-√99)] =(√100-√99)/(100-99) =√100-√99
=(√2-1)/[(√2+1)(√2-1)]+(√3-√2)/[(√3+√2)(√3-√2)]+(√4-√3)/[(√4+√3)(√4-√3)]+……+(√100-√99)/[(√100+√99)(√100-√99)]
=√2-1+√3-√2+√4-√3+……+√100-√99
=√100-1
=10-1
=9
再问: 为何1/1+根号2+1/根号2+根号3+1/根号3+根号4……+1/根号99+根号100 =(√2-1)/[(√2+1)(√2-1)]+(√3-√2)/[(√3+√2)(√3-√2)]+(√4-√3)/[(√4+√3)(√4-√3)]+……+(√100-√99)/[(√100+√99)(√100-√99)]
再答: 1/(√2+1) =(√2-1)/[(√2+1)(√2-1)] =(√2-1)/(2-1) =√2-1 1/(√3-√2) =(√3-√2)/[(√3+√2)(√3-√2)] =(√3-√2)/(3-2) =√3-√2 …… 1/(√100+√99) =(√100-√99)/[(√100+√99)(√100-√99)] =(√100-√99)/(100-99) =√100-√99
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