maths 挑戰題 - Ans.fyi 參考答案

maths 挑戰題

2006-11-16 1:26 am

好多題都唔識ar!希望你地盡下人事la!!
最好有步驟，
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Q1. 當2x^2 - x - 16 除以ax + b 時，所得的商式是 2x - 7，而餘數是 5。求a和b的值。
Q2. 若 f(y) = 2y^50 + ky^49 + 1 可被 y - 1 整除，則 k = ?
Q3. 已知多項式 f(x) 被 ( x - 3 )( x + 2 ) 所除時的餘式是 2x - 5，則 f(x) 被 x - 3所除時的餘式是?
Q4. F(x) 是一個多項式。當F(x) 除以 x - 1時，所得的餘數是 R。若 F(x) 除以2 - 2x，則餘數是?
Q5.設f(x) = 2x^3 + ax^2 + bx - 6。已知 x - 2 和 2x + 3 都是f(x) 的因式，
則f(x)/ [( x - 2)(2x + 3)]=?
Q6.若x^3 + 3x^2 + ax + b 可被x - 2 整除，則a + b/2 + 9的值是?

回答 (1)

用戶91497

2006-11-16 2:13 am

✔ 最佳答案

1. 2x^2 - x - 16 = (ax + b)(2x - 7) + 5
2x^2 - x - 21 = (x + 3)(2x - 7)
By comparing coeff, a = 1, b = 3.

2. f(y) = 2y^50 + ky^49 + 1
f(1) = 2 + k + 1 = 3 + k
Since f(y) is divisible by y - 1,
f(1) = 0
k = -3.

3. f(x) = g(x) (x - 3)(x + 2) + (2x - 5)
Remainder = f(3) = 2(3) - 5 = 1

4. F(x) = G(x) (x - 1) + R
= (-1/2) G(x) (2 - 2x) + R
The remainder is still R.

5. Let f(x) = 2x^3 + ax^2 + bx - 6 = (x - 2)(2x+3)(m+n)
Look at coeff of:
x^3: 1*2*m = 2
m = 1
constant: -2*3*n = -6
n = 1

f(x)/[(x - 2)(2x + 3)] = x + 1

6. Let f(x) = x^3 + 3x^2 + ax + b
f(2) = 8 + 12 + 2a + b = 0
2a + b + 20 = 0
2a + b + 18 = -2
a + b/2 + 9 = -1.

參考: my mathematical knowledge

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