Question

已知丨ab-2丨+(a-1)的平方=0,求ab分之1+（a+1）（b+1）分之1+（a+2）（b+2）分之1+•••+

已知丨ab-2丨+(a-1)的平方=0,求ab分之1+（a+1）（b+1）分之1+（a+2）（b+2）分之1+•••+（a+2008）（b+2008）分之1

Answer 1

ab-2=0
a-1=0
a=1
b=2
ab分之1+（a+1）（b+1）分之1+（a+2）（b+2）分之1+•••+（a+2008）（b+2008）分之1=1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+•••+1/(a+2008)(b+2008)=1/(2*3)+1/(3*4)+1/(4*5)+1/(5*6)+……+1/(2009*2010)=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+……+1/2009-1/2010=1-1/2010=2009/2010

Answer 2

丨ab-2丨=0
ab=2
(a-1)的平方=0
a=1,b=2
ab分之1+（a+1）（b+1）分之1+（a+2）（b+2）分之1+•••+（a+2008）（b+2008）分之1
=1-1/2+1/2-1/3+......+1/2009-1/2010
=1-1/2010
=2009/2010

Answer 3

a-1=0,ab-2=0,解得a=1,b=2
ab分之1+（a+1）（b+1）分之1+（a+2）（b+2）分之1+•••+（a+2008）（b+2008）分之1
=1/1-1/2+1/2-1/3+•••+1/2009-1/2010
=1-1/2010
=2009/2010

Answer 4

丨ab-2丨+(a-1)的平方=0,则ab=2; a=1,故b=2.
ab分之1+（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.