答:
|ab-2|+|b-1|=0
因为绝对值具有非负性质
同时为0时和为0
所以:
ab-2=0
b-1=0
解得: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/(2×3)+1/(3×4)+.+1/(2013×2014)
=1-1/2+1/2-1/3+1/3-1/4+.+1/2013-1/2014
=1-1/2014
=2013/2014
答:
|ab-2|+|b-1|=0
因为绝对值具有非负性质
同时为0时和为0
所以:
ab-2=0
b-1=0
解得: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/(2×3)+1/(3×4)+.+1/(2013×2014)
=1-1/2+1/2-1/3+1/3-1/4+.+1/2013-1/2014
=1-1/2014
=2013/2014