Question

若|b-1|+(a-2)的平方=0，试求1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+

若|b-1|+(a-2)的平方=0，试求 1/ab + 1/(a+1)(b+1) + 1/(a+2)(b+2)+...+1/(a+2004)(b+2004) (-3)的40次方乘(-1/3)的41次方= 化简(-2)的100次方+(-2)的101次方=

Answer 1

由
|b-1|+(a-2)的平方=0
得
B-1=0，则B=1，
A-2=0，A=2
则
1/ab
+
1/(a+1)(b+1)
+
1/(a+2)(b+2)+...+1/(a+2004)(b+2004)=1/2+1/(2+3)+...+1/(2005+2006)
=(1-1/2)+(1/2-1/3)+...+(1/2005-1/2006)
=1-1/2+1/2-1/3+...+1/2005-1/2006
=1-1/2006
=2005/2006
(-3)的40次方乘(-1/3)的41次方
=(-3)^40*(-1/3)^40*(-1/3)
=(-3*-1/3)^40*(-1/3)
=1^40*(-1/3)
=-1/3
(-2)的100次方+(-2)的101次方也是一样的，自己去做咯~