中四數學問題~快入 - Ans.fyi 參考答案

中四數學問題~快入

2011-05-09 12:35 am

(要步驟)
1. 求下列各多項式的 H.C.F 和 L.C.M
8x²+2x-3 和 32x²-18

2. 設f(x)=x^6+2x^3+1 和 g(x)=(x^2-x)^2-1
a) 因式分解 f(x) 和 g(x)
b) 求 f(x) 和 g(x) 的 H.C.F 和 L.C.M

3. 設f(x)=x^4-x^3-x+1 和 g(x)=x^3-x
a) (i) 證明 x-1 是 f(x) 的因式
(ii)由此，因式分解 f(x)
b) 求f(x) 和g(x)的 H.C.F 和L.C.M

4. 化簡下列各數式
x^2+7x+12
x^2-9

5. x^2-2x x^2+2x
------------------ X ----------
3x^3+6x^2 x^2-4

請各位數學高手幫忙 ! 急~

更新1:

第5題係 x^2-2x x^2+2x ------------------ X ---------- 3x^3+6x^2 x^2-4

更新2:

第5題係 x^2-2x ............ x^2+2x ---------------- X ----------- 3x^3+6x^2.......x^2-4

回答 (1)

pingshek

2011-05-09 1:10 am

✔ 最佳答案

1.
8x² + 2x - 3 = (2x - 1)(4x + 3)
32x² - 18 = 2(4x + 3)(4x - 3)

H.C.F. = (4x + 3)
L.C.M. = 2(2x - 1)(4x + 3)(4x - 3)

= = = = =
2.
a)
f(x) = x^6 + 2x^3 + 1
= (x^3)^2 + 2(x^3) + 1
= (x^3 + 1)^2
= (x + 1)^2 (x^2 - x + 1)^2

g(x) = (x^2-x)^2 - 1
= [(x^2 - x) + 1] [(x^2 - x) - 1]
= (x^2 - x + 1)(x^2 - x - 1)

b)
H.C.F. = (x^2 - x + 1)
L.C.M. = (x + 1)^2 (x^2 - x + 1)^2(x^2 - x - 1)

= = = = =
3.
a) (i)
f(1) = (1)^4 - (1)^3 - (1) + 1 = 0
根據因式定理， x - 1 是 f(x) 的因式。

a) (ii)
用長除法：
(x^4 - x^3 - x + 1) ÷ (x - 1) = x^3 - 1

所以 f(x) = (x - 1)(x^3 - 1)
f(x) = (x - 1)(x - 1)(x^2 + x + 1)
f(x) = (x - 1)^2 (x^2 + x + 1)

(b)
g(x) = x(x^2 - 1)
g(x) = x(x + 1)(x - 1)

H.C.F. = (x - 1)
L.C.M. = x(x + 1)(x - 1)^2 (x^2 + x +1)

= = = = =
4.
(x² + 7x + 12) / (x² - 9)
= (x + 4)(x + 3) / (x + 3)(x - 3)
= (x + 4) / (x - 3)

= = = = =
5.
[(x² - 2x) / (3x³ + 6x²)] * [(x² +2x) / (x² - 4)]
= [x(x - 2) / 3x²(x + 2)] * [x(x + 2) / (x + 2)(x - 2)]
= [(x - 2) / 3x(x + 2)] * [x / (x - 2)]
= x(x - 2) / 3x(x + 2)(x - 2)
= 1 / 3(x + 2)

參考: miraco

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