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+200
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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)
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
再问: 姐姐请讲解一下。三克油 ^ 3 ^
再答: |ab﹣2|≥0 |1-b|≥0 所以, ab﹣2=0 ,1-b=0 解得,a=2,b=1 把a=2,b=1代入.则 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
再问: 可是这有没有简便算法?难道你们都是一个一个算的( ⊙ o ⊙ )!(>^ω^
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
再问: 姐姐请讲解一下。三克油 ^ 3 ^
再答: |ab﹣2|≥0 |1-b|≥0 所以, ab﹣2=0 ,1-b=0 解得,a=2,b=1 把a=2,b=1代入.则 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
再问: 可是这有没有简便算法?难道你们都是一个一个算的( ⊙ o ⊙ )!(>^ω^
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