• 所有问题

    • 1/ab+1/（a+1)(b+1)+1/(a+2)(b+2)+......1/(a+2006)(b+2006)r如何简便

    问题是没有值

    举报该问题
    推荐答案   2010-09-26
    1/ab=(b-a)/ab *1/(b-a)=(1/a-1/b)*1/(b-a)
    1/(a+1)(b+1)= [(b+1)-(a+1)]/(a+1)(b+1) *1/(b-a)=[1/(a+1)-1/(b+1)]*1/(b-a)
    ....
    1/(a+2006)(b+2006)=[(b+2006)-(a+2006)]/(a+2006)(b+2006) *1/(b-a)=[1/(a+2006)-1/(b+2006)]*1/(b-a)

    原式=1/(b-a)* {(1/a-1/b)+[1/(a+1)-1/(b+1)]+....+[1/(a+2006)-1/(b+2006)]}
    =1/(b-a)*{[1/a+1/(a+1)+...+1/(a+2006)]-[1/b+1/(b+1)+....+1/(b+2006)]}

    只能做到这一步,如果还要化简,当知道a,b的差值后,可以消去很多项.
    温馨提示：答案为网友推荐，仅供参考
    相似回答
    ...ab+1/(a+1)(b+1)+1/(a+2)(b+2)+...+1/(a+2003)(b+2003)的值_百度知 ...答：所以两个都等于0 所以ab-2=0,a-1=0 ab=2,a=1 b=2/a=2 原式=1/1*2+1/2*3+……+1/2004*2005 =1-1/2+1/2-1/3+……+1/2004-1/2005 =1-1/2005 =2004/2005
    1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+...1/(a+2006)(b+2006)r如何简便答：1/ab=(b-a)/ab *1/(b-a)=(1/a-1/b)*1/(b-a)1/(a+1)(b+1)= [(b+1)-(a+1)]/(a+1)(b+1) *1/(b-a)=[1/(a+1)-1/(b+1)]*1/(b-a)...1/(a+2006)(b+2006)=[(b+2006)-(a+2006)]/(a+2006)(b+2006) *1/(b-a)=[1/(a+2006)-1/(b+200...
    ...1/ab+1/(a+1)(b+1)+…+1/(a+9)(b+9),其中(a-1)^2+|ab-2|=2 我要...答：必须有a-1=0,a=1 和ab-2=0,ab=2,b=2 代入原式得:1/2+1/(2*3)+1/(3*4)+...+1/(10*11)=1-1/2+1/2-1/3+1/3-1/4+...+1/10-1/11 =1-1/11 =10/11 原式值为10/11
    1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+、、+1/(a+2010)(b+2010)的值(|ab-2...答：所以 |a-2|+(b-3)^2=0 从而 a-2=0,b-3=0 所以 b=3,a=2 1/ab+1/（a+1）（b+1）+1/（a+2）（b+2）+、、+1/（a+2010）（b+2010）=1/2*3+1/3*4+...+1/2012*2013 =1/2-1/3+1/3-1/4+...+1/2012-1/2013 =1/2-1/2013 =2011/4026 ...
    1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+1/(a+3)(b+3)+...+1/(a+208)(b+208...答：把a=2,b=1代入得 原式=1/(2*1)+1/(3*2)+1/(4*3)+.+1/(210*209)=(2-1)/(2*1)+(3-2)/(3*2)+.+(210*209)=1/1-1/2+1/2-1/3+.+1/209-1/210 =1-1/210 =209/210
    ...ab+1/(a+1)(b+1)+1/(a+2)(b+2)+……1/(a+2004)(b+2004)答：因为|ab-2|和|a-1|互为相反数，所以二者都为0，推出a=1,b=2 所以原式=1/(1*2)+1/(2*3)+...+1/(2005*2006)=1-1/2+1/2-1/3+...+1/2005-1/2006 (约去中间项)=1-1/2006 =2005/2006 这一类题目都是这么用拆项法解。
    大家正在搜
    ab=0,r(a)+r(b)≤nr(ab)≤r(a)r(ab)=r(b)p(ab)和p(a∩b)的区别p(ab)=p(a逆b逆)a+b>=2√abab178847a+b=abab+a非b非