2道数学题,晚上8点之前给答案赏70,之后赏50!
2个回答
1、已知|a-2|+（b+1)^4=0,求（-a-b)^2004+2^8X（1/a）^9的值
因为|a-2|+（b+1)^4=0
故：a-2=0、b+1=0
故：a=2,b=-1
故：（-a-b)^2004+2^8X（1/a）^9
=（-1)^2004+2^8X（1/2）^9
=3/2
2、若有理数a、b满足|a-1|+（b-3)^2=0,试求
1/ab+1/(a+2)(b+2)+1/(a+4)(b+4)+……+1/(a+100)(b+100)的值.
因为|a-1|+（b-3)^2=0
故：a-1=0,b-3=0
故：a=1,b=3
故：1/ab+1/(a+2)(b+2)+1/(a+4)(b+4)+……+1/(a+100)(b+100)
=1/(1×3) +1/(3×5)+1/(5×7)+……+1/(101×1033)
=1/2×[2/(1×3) +2/(3×5)+2/(5×7)+……+2/(101×1033)]
=1/2×[(3-1)/(1×3) +(5-3)/(3×5)+ (7-5)/(5×7)+……+(103-101)/(101×1033)]
=1/2×[1-1/3+1/3-1/5+1/5-1/7+……+1/101-1/103]
=1/2×(1-1/103)
=51/103
2题似乎少了1/(a+2)(b+2)+
