| x |, x以外的符號 | | 是所謂的絕對值. 這表示不管x是一個正數或負數的值，取絕對值得結果是恆正. 例如 | x | = 1, 則 x = 1 或 x = -1, | x | > 1, 則 x > 1 或 x < -1, 且 | x | < 1, 則 x < 1 或 x > -1. 從幾何的角來看, | x | < 1可解示成以x=0為中心點距離小於1的所有點的集合. 絕對值的題目可以蠻進階的. 例如求 | 3x + 1| > | x – 4 | 中x的範圍? 此問題的解答, 在本文章的最下面. 首先我們來討論如何從邏輯推理的角度來探討這個問題. | x |, the symbol , | |, outside of x is called the absolute value. It means that no matter x is a negative or a positive value,| x | is always positive. For example, if | x | = 1 then x = 1 or x = -1, | x | > 1 then x > 1 or x < -1, and | x | < 1 then x < 1 or x > -1. Geometrically, | x | < 1 means that the set of all of points that the distance from the point x = 0 is less than 1. The problems on the absolute value could get very advanced, for example, find the range of x for | 3x + 1| > | x – 4 |. The answer of this problem is at bottom of this article. Now, let’s find the procedure that we can solve this problem logically.
解題, 首先必先知道如何問自已一連串比此問題較簡單的問題. 就以求 | 3x + 1| > | x – 4 | 中x的範圍為例子. 以下一連串問題必須先瞭解. (1) | x | > 1 (2) | x + 1 | > 1 (3) | 2x + 1 | > 1 (4) | 1/x | > 1 (5) | 1 + 1/x | > 1 (6) | 1 + 1/x | > x (7) | 1 + 1/x | > | x |. 從這一連串由淺入深問題中來分析及瞭解它們邏輯的變化. 進而問自已比就以求 | 3x + 1| > | x – 4 | 中x的範圍更難的問題. 例如| x + 1/x | > | x |和| x + 1/x | +| 2x - 2/x | > | x2+x+1 |. 從以上的訓練, 可以增強邏輯推理的能力.
Before solving the problem to find the range of x for | 3x + 1| > | x – 4 |, one should ask oneself how to solve a series of problems that simpler than the original one is very essential. For example, how to solve and understand the logic between the following problems from easier to complex ones? Moreover, understand the reason why the problems become easier or tougher. (1) | x | > 1 (2) | x + 1 | > 1 (3) | 2x + 1 | > 1 (4) | 1/x | > 1 (5) | 1 + 1/x | > 1 (6) | 1 + 1/x | > x (7) | 1 + 1/x | > | x |. After understanding the reason how come the problem is tougher or easier then one should be able to derive even more complex problems as | x + 1/x | > | x | and | x + 1/x | +| 2x - 2/x | > | x2+x+1 | to ask oneself. By doing the above exercise, one will contribute to one’s logical thinking ability.
至於如何求 | 3x + 1| > | x – 4 | 中x的範圍, 其有很多方法. 在此提供兩種如下:
(1) 由| 3x + 1| 得 當x ³ -1/3 時 | 3x + 1| = 3x + 1; 當x < -1/3 時| 3x + 1| = -(3x + 1). 由| x – 4 |得 當x ³ 4 時 | x – 4 | = x – 4; 當x < 4 時| x – 4 | = -( x – 4). 故此題可分成x³ 4, x<4 且 x ³ -1/3,和x < -1/3三個區間來討論. 當x ³ 4 時, | x – 4 | = x – 4且| 3x + 1| = 3x + 1 Þ | 3x + 1| > | x – 4 | = 3x + 1 > x – 4 Þ 2x > - 5 Þ x > -5/2. 故x ³ 4….(A) 為甚麼? 當x<4 且 x ³ -1/3時| x – 4 | = -(x – 4) ;| 3x + 1| = 3x + 1 Þ | 3x + 1| > | x – 4 | = 3x + 1 > -( x – 4 ) Þ 3x + 1 > -x + 4 Þ 4x > 3 Þ x > ¾. 故4 > x > ¾…(B) 為甚麼? 當x<-1/3時| x – 4 | = -(x – 4) ;| 3x + 1| = -(3x + 1) Þ | 3x + 1| > | x – 4 | = -(3x + 1) > -(x – 4) Þ -3x –1 > -x + 4 Þ -5 > 2x Þ x < -5/2. 故 x < -5/2…(C) 為甚麼? 由(A)(B)(C) 得x > ¾ 或 x < -5/2. 為甚麼? From | 3x + 1| we get that when x ³ -1/3 then | 3x + 1| = 3x + 1 and when x < -1/3 then | 3x + 1| = -(3x + 1). From| x – 4 | we get that when x ³ 4 then | x – 4 | = x – 4 and x < 4 then| x – 4 | = -( x – 4). From these two results we can discuss this problems in three intervals, namely x³ 4, x<4 and x ³ -1/3, and x < -1/3. When x ³ 4, | x – 4 | = x – 4 and | 3x + 1| = 3x + 1 Þ | 3x + 1| > | x – 4 | = 3x + 1 > x – 4 Þ 2x > - 5 Þ x > -5/2. Therefore, x ³ 4….(A) why? When x<4 and x ³ -1/3 then | x – 4 | = -(x – 4) ;| 3x + 1| = 3x + 1 Þ | 3x + 1| > | x – 4 | = 3x + 1 > -( x – 4 ) Þ 3x + 1 > -x + 4 Þ 4x > 3 Þ x > ¾. Therefore, 4 > x > ¾…(B) why?When x<-1/3 then | x – 4 | = -(x – 4) ;| 3x + 1| = -(3x + 1) Þ | 3x + 1| > | x – 4 | = -(3x + 1) > -(x – 4) Þ -3x –1 > -x + 4 Þ -5 > 2x Þ x < -5/2. Therefore, x < -5/2…(C) why? From (A)(B)(C) we get x > ¾ or x < -5/2. why?
(2) | 3x + 1| > | x – 4 | Þ Ö (3x + 1)2 > Ö ( x – 4 ) 2 為甚麼? Þ (Ö (3x + 1)2 )2 > (Ö ( x – 4 ) 2)2 為甚麼? Þ (3x + 1)2 > ( x – 4 )2| Þ 9x2+6x+1 > x2-8x+16 Þ 8x2+14x-15 > 0 Þ (4x-3)(2x+5) > 0 Þ 4x-3 > 0 且2x+5 > 0 Þ x > ¾且 x > -5/2 Þ x > ¾. 或4x-3 < 0 且2x+5 < 0 Þ x < ¾且 x < -5/2 Þ x < -5/2. 故x的範圍為x > ¾ 或 x < -5/2. 為甚麼? 最後必須驗算答案才有完整性. 知道如何驗算嗎? **以上的 『Ö』是根號. | 3x + 1| > | x – 4 | Þ Ö (3x + 1)2 > Ö ( x – 4 ) 2 why? Þ (Ö (3x + 1)2 )2 > (Ö ( x – 4 ) 2)2 why? Þ (3x + 1)2 > ( x – 4 )2| Þ 9x2+6x+1 > x2-8x+16 Þ 8x2+14x-15 > 0 Þ (4x-3)(2x+5) > 0 Þ 4x-3 > 0 and 2x+5 > 0 Þ x > ¾ and x > -5/2 Þ x > ¾. Or 4x-3 < 0 and 2x+5 < 0 Þ x < ¾ and x < -5/2 Þ x < -5/2. Therefore, the range of x is x > ¾ or x < -5/2. why? The next task you need to do in order to complete solving the problem is that you need to be able to verify or prove your answer is correct. Do you know how to verify your own answer? ** The symbol 『Ö』means square root.
如果您能了解並作每一個問題, 我相信您已經了解絕對值真正的涵義. If you can go through the above all of the exercises and understand them then you really understand what the absolute value means.
留言列表
