如果有理数a,b满足|ab﹣2|+|1-b|=0 求1/ab+1/(a+1)(b+1).1/(a+2003)(b+2003)
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如果有理数a,b满足|ab﹣2|+|1-b|=0 求1/ab+1/(a+1)(b+1).1/(a+2003)(b+2003)
如果有理数a,b满足|ab﹣2|+|1-b|=0 求1/ab+1/(a+1)(b+1).1/(a+2003)(b+2003)
如果有理数a,b满足|ab﹣2|+|1-b|=0 求1/ab+1/(a+1)(b+1).1/(a+2003)(b+2003)
ab﹣2=0 ,1-b=0
则,a=2,b=1
1/ab+1/(a+1)(b+1).1/(a+2003)(b+2003)
=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
解由|ab﹣2|+|1-b|=0
知ab-2=0且1-b=0
即a=2,b=1
又有1/(1+n)(2+n)=1/(n+1)-1/(n+2)
即1/ab=1/1*2=1/1-1/2
1/(a+1)(b+1)=1/(1+1)(2+1)....=1/(2)-1/(3)
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解由|ab﹣2|+|1-b|=0
知ab-2=0且1-b=0
即a=2,b=1
又有1/(1+n)(2+n)=1/(n+1)-1/(n+2)
即1/ab=1/1*2=1/1-1/2
1/(a+1)(b+1)=1/(1+1)(2+1)....=1/(2)-1/(3)
。.....................................
1/(a+2003)(b+2003)=1/(2004)-1/(2005)
上述格式相加得
1/ab+1/(a+1)(b+1)....1/(a+2003)(b+2003)
=(1/1-1/2)+(1/2-1/3)+..........+(1/(2004)-1/(2005))
=1/1-1/2005
=2004/2005
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