c) h(n+1)
d) h(n+1)-h(n)
e) h(x-1)
請問係唔係要展開?
如果展開後我又不懂代入
> <幫下我,唔該大家
更新1:
SORRY 我打"溜"左...h(x)=2x^2-x
函數運算 希望能幫助解答,萬分感激!~
2013-01-15 2:19 pm
已知(x)=2x^2-x 求
c) h(n+1)
d) h(n+1)-h(n)
e) h(x-1)
請問係唔係要展開?
如果展開後我又不懂代入
> <幫下我,唔該大家
c) h(n+1)
d) h(n+1)-h(n)
e) h(x-1)
請問係唔係要展開?
如果展開後我又不懂代入
> <幫下我,唔該大家
回答 (4)
2013-01-15 4:59 pm
✔ 最佳答案
(c)h(n + 1)
= 2(n + 1)^2 - (n + 1)
= 2(n^2 + 2n + 1) - (n + 1)
= 2n^2 + 4n + 2 - n - 1
= 2n^2 + 3n + 1
(d)
h(n + 1) - h(n)
= (2n^2 + 3n + 1) - (2n^2 - n)
= 2n^2 + 3n + 1 - 2n^2 + n
= 4n + 1
(e)
h(x - 1)
= 2(x - 1)^2 - (x - 1)
= 2(x^2 - 2x + 1) - (x - 1)
= 2x^2 - 4x + 2 - x + 1
= 2x^2 - 5x + 3
參考: knowledge
2013-01-15 7:51 pm
建議先factorize h(x)。
h(x) = 2x²-x = x(2x-1)
c)
h(n+1)
= (n+1)[2(n+1)-1]
=(n+1)(2n+1)
d)
h(n+1) - h(n)
= (n+1)(2n+1) - n(2n-1)
= (2n²+3n+1) - (2n²-n)
= 4n+1
e)
h(x-1)
= (x-1)[2(x-1)-1]
= (x-1)(2x-3)
2013-01-15 14:21:47 補充：
其實part d我就咁做既:
h(n+1) - h(n)
= (n+1)(2n+1) - n(2n-1)
= n(2n+1) +(2n+1) -n(2n-1)
= n[(2n+1)-(2n-1)] + (2n+1)
= 4n+1
h(x) = 2x²-x = x(2x-1)
c)
h(n+1)
= (n+1)[2(n+1)-1]
=(n+1)(2n+1)
d)
h(n+1) - h(n)
= (n+1)(2n+1) - n(2n-1)
= (2n²+3n+1) - (2n²-n)
= 4n+1
e)
h(x-1)
= (x-1)[2(x-1)-1]
= (x-1)(2x-3)
2013-01-15 14:21:47 補充：
其實part d我就咁做既:
h(n+1) - h(n)
= (n+1)(2n+1) - n(2n-1)
= n(2n+1) +(2n+1) -n(2n-1)
= n[(2n+1)-(2n-1)] + (2n+1)
= 4n+1
2013-01-15 4:19 pm
SORRY 我打"溜"左...h(x)=2x^2-x
2013-01-15 2:40 pm
你確定是"(x)"而不是"h(x)？
收錄日期: 2021-04-28 14:21:22
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20130115000051KK00036