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急!F5 inequalities unknown 2q2
2012-09-30 8:47 am
回答 (2)
2012-09-30 6:41 pm
✔ 最佳答案
(a)(x - 2)(2 - x) > 0
- (x - 2)(x - 2) > 0
- (x - 2)^2 > 0
Since (x - 2)^2 is always positive, there is no solution for this inequality.
(b)
2x^2 + 5x - 3 < 6x^2 + 5x + 1
0 < 4x^2 + 4
0 < 4(x^2 + 1)
Since x^2 is always positive, the inequality is greater than 0 for all values of x.
2012-10-01 2:49 am
(a)
(x－2)(2－x)>=0
－(x－2)^2>=0
(x－2)^2<=0
(x－2)^2=0
x=2
(b)
(x+3)(2x－1)<6x^2+5x+1
2x^2+5x－3<6x^2+5x+1
4x^2+4>0
The solution is all real number
(x－2)(2－x)>=0
－(x－2)^2>=0
(x－2)^2<=0
(x－2)^2=0
x=2
(b)
(x+3)(2x－1)<6x^2+5x+1
2x^2+5x－3<6x^2+5x+1
4x^2+4>0
The solution is all real number
收錄日期: 2021-04-13 19:00:45
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20120930000051KK00017