設g(x)=6x^4-2cx+1
a)求g(d)和g(-d)
b)若對於任何實數,d,g(d)=g(-d),求b的值
16..中四MATH..請詳列步驟..
2006-10-31 1:54 am
回答 (5)
2006-11-01 2:24 am
✔ 最佳答案
設g(x)=6x^4 - 2cx + 1a)求g(d)和g(-d)
g(d)=6d^4 - 2cd + 1
g(-d)=6(-d)^4- 2c(-d) + 1
=6d^4 + 2cd + 1
b)若對於任何實數,d,g(d)=g(-d),求b的值
g(d)=g(-d)
6d^4 - 2cd + 1= 6d^4 + 2cd + 1
6d^4 - 6d^4 - 2cd + 1= 2cd + 1
-2cd + 1=2cd + 1
2cd + 2cd=0
4cd=0
cd=0
c=0
Put c=0 into g(x)=6x^4 - 2cx + 1
g(x)=6x^4+1
g(-1)=6(-1)^4+1
g(-1)=6+1
g(-1)=7
參考: me
2006-10-31 2:09 am
g(d)=6d^4-2cd+1
g(-d)=6(-d)^4-2c(-d)+1
=6d^4-2cd+1
b)唔明b點出黎...
g(-d)=6(-d)^4-2c(-d)+1
=6d^4-2cd+1
b)唔明b點出黎...
2006-10-31 2:06 am
a) g(d) = 6d^4-2cd+1
g(-d) = 6d^4+2cd+1
g(d) = g(-d)
6d^4-2cd+1 = 6d^4+2cd+1
-2cd=2cd
2cd+2cd=0
4cd=0
但係 b 的值.........unknown
g(-d) = 6d^4+2cd+1
g(d) = g(-d)
6d^4-2cd+1 = 6d^4+2cd+1
-2cd=2cd
2cd+2cd=0
4cd=0
但係 b 的值.........unknown
2006-10-31 2:06 am
A)g(d)=6d^4-2cd+1
g(-d)=6(-d)^4-2c(-d)+1
=6d^4+2cd+1
b) g(d)=g(-d)
6d^4-2cd+1=6d^4+2cd+1
6d^4-6d^4-2cd+1=2cd+1
-2cd+1=2cd+1
- 2cd - 2cd=0
之後自己計
g(-d)=6(-d)^4-2c(-d)+1
=6d^4+2cd+1
b) g(d)=g(-d)
6d^4-2cd+1=6d^4+2cd+1
6d^4-6d^4-2cd+1=2cd+1
-2cd+1=2cd+1
- 2cd - 2cd=0
之後自己計
2006-10-31 2:05 am
a) g(d) = 6d^4-2cd+1
g(-d) = 6(-d)^4-2c(-d)+1
= 6d^4+2cd+1
b) =.=*我穩唔到邊度有b......
希望我係咁而幫到下你啦!!!
g(-d) = 6(-d)^4-2c(-d)+1
= 6d^4+2cd+1
b) =.=*我穩唔到邊度有b......
希望我係咁而幫到下你啦!!!
參考: ps 我都係form4
收錄日期: 2021-04-12 22:54:55
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20061030000051KK04224