一條數(香港青少年數學精英選拔賽)

2007-01-31 1:47 am
If a and b are real numbers such that a+b<0, ab<0 and a<b, arrange a, -a, b and -b
in ascending order of magnitude.

若 a 及 b 為實數, 且a+b<0, ab<0 及 a<b, 請將 a, -a, b 及 -b 由小至大排列

請大家幫忙!

回答 (3)

2007-01-31 2:04 am
✔ 最佳答案
a + b < 0………….(1)
ab < 0………..(2)
a < b…………(3)

from (1)
a + b < 0
a < -b……………….(4)
also by (1) x -1
-a > b………………..(5)

From (2)
ab < 0
that means one is +ve and the other is negative
if a> 0 and b < 0
then b < 0 < a, which will contradict with (3)

therefore, b > 0 and a < 0
b > 0
then –b < 0
-b < b…………..(6)

By (3), (4), (5), (6)
a < -b < b < -a
參考: me
2007-01-31 7:15 pm
a<-b<-a

2007-01-31 13:22:19 補充:
a

2007-01-31 13:22:52 補充:
a

2007-01-31 13:24:04 補充:
a<-b<b<-a
2007-01-31 2:10 am
(1) If a+b<0, than a<-b and b<-a.

(2) If ab<0, either (a<0 and b>0) or (a>0 or b<0). That is,
(i) a<0

2007-01-30 18:14:02 補充:
(1) If a b < 0, than a

2007-01-30 18:15:38 補充:
(1) If a b < 0, than a < -b and b < -a.(2) If ab < 0, either (a < 0 and 0 < b) or (0 < a or b < 0). That is,(i) a < 0 < b, or(ii) b < 0 < a.

2007-01-30 18:15:54 補充:
(3) If a < b, then from (2), only (i) is satisfied.From (1), (2) and (3), we have a < -b, b < -a and a < 0 < b.Therefore, a < -b < 0 < b < -a, or a < -b < b < -a.


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