----------- > 4
x + 1
, 謝謝
更新1:
另外, 可否將解題的原理, 詳細說一次, 因不明白, 掌握不到, 謝謝.
求 x 的區間 (interval)?
回答 (3)
2012-11-02 8:36 pm
✔ 最佳答案
http://maeducation.edu.hk/Yahoo/7012110200058.pdfMethod 1 乘(x+1)^2，必為正數，所以不影響大小號，不過會令次方變大
Method 2 負正得負，正負得負的概念
Method 3 分範圍
個人建議 Method 3，但對於這題，建議用Method 1
參考: maeducation.edu.hk
2012-11-02 4:20 pm
Sol
(2x+3)/(x+1)>4
(2x+3)/(x+1)－4>0
[(2x+3)－4(x+1)]/(x+1)>0
(－2x－1)/(x+1)>0
(2x+1)/(x+1)<0
(2x+1)(x+1)<0，x+1<>0
(x+1/2)(x+1)<0，x+1<>0
(x+1/2)(x+1)<0
－1<－1/2
(2x+3)/(x+1)>4
(2x+3)/(x+1)－4>0
[(2x+3)－4(x+1)]/(x+1)>0
(－2x－1)/(x+1)>0
(2x+1)/(x+1)<0
(2x+1)(x+1)<0，x+1<>0
(x+1/2)(x+1)<0，x+1<>0
(x+1/2)(x+1)<0
－1<－1/2
2012-11-02 3:00 pm
(2x + 3)(x + 1) > 4 (x + 1)^2
2x^2 + 5x + 3 > 4x^2 + 8x + 4
2x^2 + 3x + 1 < 0
(2x + 1)(x + 1) < 0
so - 1 < x < - 1/2
Note : Square of any real number is always positive, so (x + 1)^2 is always positive, multiplying both side by (x + 1)^2 will not change the sign.
2x^2 + 5x + 3 > 4x^2 + 8x + 4
2x^2 + 3x + 1 < 0
(2x + 1)(x + 1) < 0
so - 1 < x < - 1/2
Note : Square of any real number is always positive, so (x + 1)^2 is always positive, multiplying both side by (x + 1)^2 will not change the sign.
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原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20121102000051KK00058