please solve the following inequalities
(2x+1)^2 is greater than or equal to (3x - 4)^2
(becoz i dunno how to type the symbol ,,, so i use word form)
inequalities
2006-11-02 5:23 pm
回答 (3)
2006-11-02 5:43 pm
✔ 最佳答案
(2x+1)^2 ≥ (3x - 4)^24x^2 + 4x + 1 ≥ 9x^2 - 24x +16
0 ≥ 5x^2 - 28x +15
0 ≥ (5x-3)(x-5)
Therefore,
Case 1:
(5x-3)≤0 and (x-5)≥0
x≤3/5 and x≥5 (Impossible)
Case 2:
(5x-3)≥0 and (x-5)≤0
x≥3/5 and x≤5
2006-11-02 5:49 pm
(2x + 1) ^ 2 } (3x - 4) ^2
2x + 1 } 3x - 4 & 2x + 1 } -(3x - 4)
1 + 4 } 3x - 2x & 2x + 3x } 4 - 1
x { 5 & 5x } 3
3/5 { x { 5
2006-11-02 09:50:14 補充：
} greater than or equal{ smaller than or equal
2x + 1 } 3x - 4 & 2x + 1 } -(3x - 4)
1 + 4 } 3x - 2x & 2x + 3x } 4 - 1
x { 5 & 5x } 3
3/5 { x { 5
2006-11-02 09:50:14 補充：
} greater than or equal{ smaller than or equal
2006-11-02 5:35 pm
(2x+1)^2 is greater than or equal to (3x - 4)^2
4x^2+4x+1 is greater than or equal to 9x^2-24x+16
5x^2-28x+15 is smaller than or equal to 0
(x-5)(5x-3) is smaller than or equal to 0
3/5 is smaller than or equal to x is smaller than or equal to 5
Hopes can help u
4x^2+4x+1 is greater than or equal to 9x^2-24x+16
5x^2-28x+15 is smaller than or equal to 0
(x-5)(5x-3) is smaller than or equal to 0
3/5 is smaller than or equal to x is smaller than or equal to 5
Hopes can help u
參考: myself
收錄日期: 2021-04-12 22:30:51
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20061102000051KK00640