Question

已知丨ab-2丨与丨b-1丨互为相反数，试求代数式ab分之一+（a+1)(b+1)分之1+（a+2)(b+2)+.....+

(a+2008)(b+2008)的值

Answer 1

丨ab-2丨与丨b-1丨互为相反数
则ab-2=0，b-1=0,b=1，a=2
所以ab分之一+（a+1)(b+1)分之1+（a+2)(b+2)分之1+.....+(a+2008)(b+2008)分之1
=1/1*2+1/2*3+1/3*4+……+1/2009*2010
=1-1/2+1/2-1/3+1/3-1/4+……1/2009-1/2010
=1-1/2010
=2009/2010

祝你开心

Answer 2

即|ab-2|+|b-1|=0
所以ab-2=0,b-1=0
b=1,a=2/b=2

所以原式=1/1*2+1/2*3+……+1/2009*2010
=1-1/2+1/2-1/3+……+1/2009-1/2010
=1-1/2010
=2009/2010

Answer 3

原式=1/1*2+1/2*3+……+1/2009*2010
=1-1/2+1/2-1/3+……+1/2009-1/2010
=1-1/2010
=2009/2010