Question

已知|a+2|+(b+1)^2=0,求1/ab+1/(a-2)(b-2)+…+/1/(a-2009)(b-2009)的值？

Answer 1

那么：由相反数定义,A-2的绝对值+（b+1）^2=0.因为一个数的绝对值和平方都是大于等于零的,当他们相加为零时,则两者皆为零.A=2,b=-1.b^a是1.a^3+b^15=7,6,|a+2|+(b+1)^2=0
有
a=-2,b=-1
b-a=1
1/ab=1/a-1/b=1/a-1/(a+1)
1/(a-2)(b-2)=1/(a-2)-1/(b-2)=1/(a-2)-1/(a-1);
1/(a-3)(b-3)=1/(a-3)-1/(b-3)=1/(a-3)-1/(a-2);
...
1/(a-2009)(b-...,2,