含有絕對值的不等式之小問題

首先, 你說 "他是假設X為負數", 在邏輯理解上, 這句是錯的!

因為正確邏輯是 "他是假設 (X+2) 為負數", 故此, 如不從(5)入手, 便要從 (X+2) 整個數入手, 計算如下

- (X+2) > 5
(-X-2) > 5

明白那分別嗎!

2010-10-21 10:35:19 補充：
在 5>丨X丨這情況下

因為答案 X的值 是介乎 -5 與5之間, 是 "同時" 被兩個 界 (boundaries) 所限制, 故此要用 "and". 因為如果只滿足其中一項限制, 是沒發完成 5>丨X丨的! 舉例,

只考慮 X<5 , 那麼X可以是 - 6, 但丨- 6丨= 6, 6 當然大過5, 那麼 5>丨X丨便不成立!

明白嗎!

2010-10-21 14:35:29 補充：
哇, 你愈問愈多, 好貪喎!

簡單講: 得兩個情況

1) 丨X丨 > A

即 X >A OR X < - A

2) 丨X丨 < A

即 - A < X < A 亦即係 - A < X AND X < A

明嗎?

丨X+2丨>5

Absolute a number means change a number sign be positive
However, the actual value or x+2 can be either positive or negative,
Method 1:(Think it by concept)
|x+2| > 5, (x+2) can be positive or negative.
Hence, x+2>5 or -(x+2)>5
Then, x+2>5 or x+2<-5

Method 2:(Square the both side and use square)
|x+2| > 5
|x+2|^2 > 5^2
x^2 + 4x + 4 > 25
x^2 + 4x - 21 > 0
(x - 3)(x + 7) > 0
Hence, x - 3 > 0 or x + 7 < 0
x > 3 or x < -7
x+2 > 3+2 or x+2 < -7+2
x+2 > 5 or x+2 < -5


2010-10-21 07:14:57 補充：
It shou d be "and"
Since the range should be -5<5
because there is overlapped area, "and"is needed to use, if we use "or" then x can be all real number, which seems to be meaningless.
for x<-5 or x>5, there should be "or", if we use "and", then x will not exist.

2010-10-21 16:17:48 補充：
那麼，分成被正負兩路計算的 (X) ，在實際上，他們的本質其實仍然是同一個數值還是不？
They are the same, because they both represent the domain(the range of x),
Different cases:
if a<=x<=b or a
=x,b<=x or a>x,b

2010-10-21 16:18:40 補充：
if a
x,b

2010-10-21 16:19:29 補充：
if a
x,x

2010-10-21 16:20:25 補充：
if x between a and b, "and " is used as area is overlapped.
if x smaller than a, or x greater than b, "or" is used since no area is overlapped

2010-10-21 16:21:04 補充：
如是，那即是，分開計算的 (X) 到最後是會被正負雙方所得之解所同時限制著，亦即所謂的「AND」?
you need to consider if the final sign is >0 or <0

2010-10-21 16:22:16 補充：
而同時，假設正負雙方所得之解是 <-1 跟 >1 這類的組合吧，那麼，在最後的答案上得出的結果就會是「OR」?
Yes,
since x < -1 , which means x can only be -2,-3,...,-n
x > 1, which means x can only be 2,3,...,n
there is no overlapped number, then "or"is used