1/x^2(x-2)^2 + 2/x(x-2) - 3 = 0
圖 : http://postimg.org/image/r7ttk60rv/
F. 5 數學求解
2013-12-29 10:48 pm
回答 (2)
2013-12-29 11:23 pm
✔ 最佳答案
設 u = 1/[x(x - 2)]1/[x²(x - 2)²] + 2/[x(x - 2)] - 3 = 0
u² + 2u - 3 = 0
(u + 3)(u - 1) = 0
u + 3 = 0 或 u - 1 = 0
u = -3 或 u = 1
1/[x(x - 2)] = -3 或 1/[x(x -2)] = 1
1/[x² - 2x] = -3 或 1/[x² - 2x] = 1
-3x² + 6x = 1 或 x² - 2x = 1
3x² - 6x + 1 = 0 或x² - 2x - 1 = 0
x = (3 ± √6)/3 或 x = 1 ± √2
參考: fooks
2013-12-29 11:36 pm
收錄日期: 2021-04-13 19:52:40
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20131229000051KK00081