數學 絕對直不等式詳解
│2X+1│<│X-3│
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數學 絕對直不等式詳解
2011-05-28 8:12 am
回答 (3)
2011-05-28 4:10 pm
✔ 最佳答案
因為 0小於 l2x+1l < l x-3l所以, 可以將不等式的兩邊平方
l2x+1l ^2< l x-3l^2
4x^2+4x+1< x^2-6x+9
3x^2+10x - 8<0
(x+4)(3x-2) <0
得到解為 -4 < x < (2/3)
Blog: http://tw.myblog.yahoo.com/math-life
2011-05-28 12:53 pm
The answer is as follows:
| 2x + 1 | < | x - 3 |
| 2x + 1 |^2 < | x - 3 |^2
4x^2 + 4x + 1 < x^2 - 6x + 9
3x^2 + 10x - 8 = 0
(x + 4)(3x - 2) = 0
Ans: -4 < x < 2/3 ...... (ans)
2011-05-29 04:39:58 補充：
In the last second line and last third line, the "=" should be "<', i.e.
3x^2 + 10x - 8 < 0
(x + 4)(3x - 2) < 0
| 2x + 1 | < | x - 3 |
| 2x + 1 |^2 < | x - 3 |^2
4x^2 + 4x + 1 < x^2 - 6x + 9
3x^2 + 10x - 8 = 0
(x + 4)(3x - 2) = 0
Ans: -4 < x < 2/3 ...... (ans)
2011-05-29 04:39:58 補充：
In the last second line and last third line, the "=" should be "<', i.e.
3x^2 + 10x - 8 < 0
(x + 4)(3x - 2) < 0
參考: miraco, miraco
2011-05-28 9:14 am
怪了, 一直沒辦法回答, 你自己去參考以下網址吧!
http://tw.myblog.yahoo.com/jw!JCVbjQyaBRbXTWOakincl1.Wpxbobg--/article?mid=9209
http://tw.myblog.yahoo.com/jw!JCVbjQyaBRbXTWOakincl1.Wpxbobg--/article?mid=9209
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