設f(x)=3x³+mx-nx-7。當f(x)除以(x+1)(x-3)時,所得的餘式是2x-4。
(a)求f(-1)和f(3)的值。
(b)設立兩個以m和n表示方程。
(c)求m和n的值。
maths~~~~~~
2007-10-12 5:40 am
回答 (1)
2007-10-12 5:57 am
✔ 最佳答案
a)設: f ( x ) = ( x + 1 )( x - 3 )Q ( x )+ 2x - 4 f ( - 1 ) = ( - 1 + 1 )( - 1 - 3 )Q ( - 1 ) + 2 ( - 1 ) - 4
= -6
f ( 3 ) = ( 3 + 1 )( 3 - 3 ) Q ( 3 ) + 2 ( 3 ) - 4
= 2
b) f ( - 1 ) = -6
3 ( - 1 )^3 + m ( - 1 )^2 - n ( - 1 ) - 7 = -6
m + n = 4
n = 4 - m--- ( 1 )
f ( 3 ) = 2
3 ( 3 )^3 + m ( 3)^2 - n ( 3 ) - 7 = 2
9m - 3n = -72
3m - n = -24 ( 2 )
c) 3m - 4 + m = -24 [式1]
4m = -20
m = -5
n = 4 - ( - 5 ) = 9
參考: My Maths Knowledge
收錄日期: 2021-04-23 20:35:25
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