第30題
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第30題第30題第30題第30題
2007-12-07 5:05 am
回答 (4)
2007-12-07 5:12 am
✔ 最佳答案
仲以為有三十題 =v=""將a+1同a-1代入f(x)到
f(x)=x^2+x-1
f(a+1)=(a+1)^2+(a+1)-1=a^2+2a+1+a+1-1=a^2+3a+1
f(a-1)=(a-1)^2+(a-1)-1=a^2-2a+1+a-1-1=a^2-a-1
f(a+1)-f(a-1)=a^2+3a+1-(a^2-a-1)
=a^2-a^2+3a+a+1+1
=4a+2
=ans b
2007-12-07 9:39 pm
f(a+1)=(a+1)^2+(a+1)-1
=a^2+2a+1+a+1-1
=a^2+3a+1
f(a-1)=(a-1)^2+(a-1)-1
=a^2-2a+1+a-1-1
=a^2-a-1
So f(a+1)-f(a-1)=4a+2
So B
=a^2+2a+1+a+1-1
=a^2+3a+1
f(a-1)=(a-1)^2+(a-1)-1
=a^2-2a+1+a-1-1
=a^2-a-1
So f(a+1)-f(a-1)=4a+2
So B
2007-12-07 5:12 am
D.....(I am not very sure)
收錄日期: 2021-04-13 14:39:32
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20071206000051KK03544