1. (x^2-x)^2-8(x^2-x)+12=0
2. x-1/4=x/x+3
F.5 Quadratic equation
2012-09-10 7:56 am
回答 (2)
2012-09-10 9:13 am
✔ 最佳答案
1.Let u = x² - x
(x² - x)² - 8(x² - x) + 12 = 0
u² - 8u + 12 = 0
(u - 2)(u - 6) = 0
(x² - x - 2)(x² - x - 6) = 0
(x + 1)(x - 2)(x + 2)(x - 3) = 0
x = -1 or x = 2 or x = -2 or x = 3
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2.
(x - 1)/4 = x/(x + 3)
(x - 1)(x + 3) = 4x
x² + 2x - 3 = 4x
x² - 2x - 3 = 0
(x + 1)(x - 3) = 0
x = -1 or x = 3
參考: micatkie
2012-09-10 8:13 am
1)
(x^2-x)^2-8(x^2-x)+12 = 0
(x^2-1-6)(x^2-x-2) = 0
x^2-7=0 or (x-2)(x+1)=0
x=±√7 or 2 or -1
2)
x-1/4 = x/(x+3)
4x-1 = 4x/(x+3)
4x^2+12x-4x = x+3
4x^2+7x-3 = 0
x = (-7±√(97))/8
(x^2-x)^2-8(x^2-x)+12 = 0
(x^2-1-6)(x^2-x-2) = 0
x^2-7=0 or (x-2)(x+1)=0
x=±√7 or 2 or -1
2)
x-1/4 = x/(x+3)
4x-1 = 4x/(x+3)
4x^2+12x-4x = x+3
4x^2+7x-3 = 0
x = (-7±√(97))/8
參考: 神風亨
收錄日期: 2021-04-13 18:57:51
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