Question

|ab-2|和|b-1|为相反数,求1／ab+1/(a+1)(b+1)+1/(a+2)(b+2)...1/(a+2012)(b+2012)的值?

Answer 1

|ab-2|和|b-1|为相反数
所以ab-2=0 b-1=0
解得b=1 a=2

1／ab+1/(a+1)(b+1)+1/(a+2)(b+2)...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

Answer 2

∵|ab-2|和|b-1|为相反数,∴∴

1/ab+1/(a+1)(b+1)+……1/(a+2012)(b+2012)
=1/1*2+1/2*3+1/3*4+……+1/2013*2014
=1-1/2+1/2+1/3-1/3+1/4-1/4+……-1/2013+1/2013-1/2014
=1-1/2014
=2013/2014

Answer 3

|ab-2|和|b-1|为相反数可得：ab-2=b-1=0,所以a=2,b=1
所以1／ab+1/(a+1)(b+1)+1/(a+2)(b+2)...1/(a+2012)(b+2012)=1-1/2+1/2-1/3+1/3-1/4+......+1/2013-1/2014=1-1/2014=2013/2014

Answer 4

绝对值都大于0，所以那两个算式都只有一个情况，那就是同为0，0的相反数是0。则ab=2 b=1 则a=2。代入得：1/2X1+1/3X2+1/4X3+......1/2014X2013= 这个好像涉及到了高中数列求和中的裂项相消法，一直是我的弱项，你自求多福

Answer 5

|ab-2|>=0
|b-1|>=0
|ab-2|和|b-1|为相反数,
则 b=1,a=2
1／ab+1/(a+1)(b+1)+1/(a+2)(b+2)...1/(a+2012)(b+2012)
=1/2+1/(3*2)+1/(4*3)+***+1/(2014*2013)
=1/2+1/2-1/3+1/3-1/4+***+1/2013-1/2014
=1-1/2014
=2013/2014