代數不等式

2014-06-15 9:53 am
解不等式﹕l x+2 l + l 2x-4 l <=10 應該點計好,請各位師兄幫忙

回答 (4)

2014-06-15 10:06 am
✔ 最佳答案
這類題可以分情況考慮。

|x + 2| + |2x - 4| ≤ 10

|x + 2| + 2|x - 2| ≤ 10 ...(★)

情況一: x < -2 ...(1)
此時,|x + 2| = -(x + 2) 及 |x - 2| = -(x - 2)
(★) 即
-(x + 2) - 2(x - 2) ≤ 10
-x -2 -2x + 4 ≤ 10
-3x ≤ 8
x ≥ -8/3
配合 (1) 即 -8/3 ≤ x < -2 ...(A)

情況二: -2 ≤ x < 2 ...(2)
此時,|x + 2| = x + 2 及 |x - 2| = -(x - 2)
(★) 即
-(x + 2) + 2(x - 2) ≤ 10
-x -2 +2x -4 ≤ 10
x ≤ 16
配合 (2) 即 -2 ≤ x < 2 ...(B)

情況三: x ≥ 2 ...(3)
此時,|x + 2| = x + 2 及 |x - 2| = x - 2
(★) 即
(x + 2) + 2(x - 2) ≤ 10
x + 2 +2x - 4 ≤ 10
3x ≤ 12
x ≤ 4
配合 (3) 即 2 ≤ x ≤ 4 ...(C)

綜合答案(A)、(B)、(C)為:
-8/3 ≤ x < -2 或 -2 ≤ x < 2 或 2 ≤ x ≤ 4
即 -8/3 ≤ x ≤ 4

2014-06-16 17:51:27 補充:
Jacky, 那個是普通括號嗎?

我以為是絕對值 absolute value。

2014-06-16 19:41:37 補充:
HK~,願聞其詳。

|x + 2| + |2x - 4| ≤ 10

(x + 2)² + 2|x + 2|×|2x - 4| + (2x - 4)² ≤ 10²

Then?
2014-06-18 7:14 am
這類題可以分情況考慮。

|x + 2| + |2x - 4| ≤ 10

|x + 2| + 2|x - 2| ≤ 10 ...(★)

情況一: x < -2 ...(1)
此時,|x + 2| = -(x + 2) 及 |x - 2| = -(x - 2)
(★) 即
-(x + 2) - 2(x - 2) ≤ 10
-x -2 -2x + 4 ≤ 10
-3x ≤ 8
x ≥ -8/3
配合 (1) 即 -8/3 ≤ x < -2 ...(A)

情況二: -2 ≤ x < 2 ...(2)
此時,|x + 2| = x + 2 及 |x - 2| = -(x - 2)
(★) 即
-(x + 2) + 2(x - 2) ≤ 10
-x -2 +2x -4 ≤ 10
x ≤ 16
配合 (2) 即 -2 ≤ x < 2 ...(B)

情況三: x ≥ 2 ...(3)
此時,|x + 2| = x + 2 及 |x - 2| = x - 2
(★) 即
(x + 2) + 2(x - 2) ≤ 10
x + 2 +2x - 4 ≤ 10
3x ≤ 12
x ≤ 4
配合 (3) 即 2 ≤ x ≤ 4 ...(C)

綜合答案(A)、(B)、(C)為:
-8/3 ≤ x < -2 或 -2 ≤ x < 2 或 2 ≤ x ≤ 4
即 -8/3 ≤ x ≤ 4

2014-06-18 00:18:40 補充:
呀呀呀呀呀呀呀呀呀呀呀呀呀呀呀呀呀呀呀呀呀呀呀呀!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
呀呀呀呀呀呀呀呀呀呀呀呀呀呀呀呀呀呀呀呀呀呀呀呀!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!呀呀呀呀呀呀呀呀呀呀呀呀呀呀呀呀呀呀呀呀呀呀呀呀!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
呀呀呀呀呀呀呀呀呀呀呀呀呀呀呀呀呀呀呀呀呀呀呀呀!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!呀呀呀呀呀呀呀呀呀呀呀呀呀呀呀呀呀呀呀呀呀呀呀呀!!!!!!!!!!!!!!!!
參考: 知足常樂
2014-06-17 2:54 am
其實仲有第2個方法計,

不過呢個方法不見議使用...

因為呢1個方法就係2邊都2次方 2次

好麻煩 0.0

2014-06-22 23:21:19 補充:
sorry..

唔記得左睇tim

I mean
|x + 2| + |2x - 4| ≤ 10
|x + 2| ≤ 10 - |2x - 4|
(x+2)^2 ≤ 100 - 20 |2x - 4| + (2x - 4)^2
(x+2)^2 -100 - (2x - 4)^2 ≤ - 20 |2x - 4|

之後兩面2次方,最後計到個ans就check番個ans適唔適合原本條式入面,
因為呢個方法幾麻煩下,
所以唔見議使用 -.-
2014-06-17 12:17 am
這題很簡單= =
(x+2)+(2x-4)<=10
x+2+2x-4<=10
3x+2-4<=10
3x<=10+4-2
3x<=12
x<=4


收錄日期: 2021-04-16 16:31:32
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