maths(解方程二 : 代數方法)

2009-09-12 7:29 am

P.209
1. 解 (12/x + 1) (x - 3) = 9
2. 解 (1 / x - 3) - (3 / x + 2) = 1
p. 210
1. 試解二次方程 x^2 - 10x + 9 = 0
2. 考慮四次方程 x^4 -10x^2 + 9 = 0
(a) 多項式 x^4 -10x^2 + 9 有多少項? 有沒有常數項
(b) 設 y = x^2
(i) 原來的四次方程可變換成什麼方程
(ii) 試求出 y 的值
(iii) 對應的 x值是什麼
3. 解 x^4 - 5x^2 + 4 = 0
p. 211
1. 解 x^5 - 7x^3 + 6x = 0
p.212
1.. 考慮方程開方x-1 - 3 = 0,把等號兩邊同時取平方,可以把根號消去嗎

回答 (1)

用戶9112

2009-09-12 9:08 am

✔ 最佳答案

1. (12/x + 1)(x - 3) = 9
=> (12 + x)(x - 3) = 9x
=> 12x - 36 + x2 - 3x = 9x
=> x2 - 36 = 0
=> (x + 6)(x - 6) = 0
=> x = -6 或x = 6
2. (1 / x - 3) - (3 / x + 2) = 1
=> 1 - 3x - 3 - 2x = x
=> -2 = 6x
=> x = -1/3
1. x2 - 10x + 9 = 0
=> x2 - x - 9x + 9 = 0
=> x(x - 1) - 9(x - 1) = 0
=> (x - 9)(x - 1) = 0
=> x = 1 或x = 9
2. x4 - 10x2 + 9 = 0
(a) 多項式 x4 -10x2 + 9 有3項. 9 為常數項
(b) 設 y = x2
(i) 原來的四次方程可變換成 y2 - 10y + 9 = 0
(ii) y = 1 或 9
(iii) y = 1 => x2 = 1 => x = 1或 -1
y = 9 => x2 = 9 => x = 3或 -3
3. x4 - 5x2 + 4 = 0
=> (x2 - 4)(x2 - 1) = 0
=> (x - 2)(x + 2)(x - 1)(x + 1) = 0
x = -2, -1, 1或2
1. x5 - 7x3 + 6x = 0
x(x4 - 7x2 + 6) = 0
x(x2 - 6)(x2 - 1) = 0
x(x - √6)(x + √6)(x - 1)(x + 1) = 0
x = -√6, -1, 0, 1或 √6

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