我想知道點樣移項呀，有冇人知道呀?
2x－1 3－x
-------- － --------- =2
3 ║ ５
我完全唔明點樣移項，請大家詳細告訴我。
方程式(移項題目)
2006-11-16 7:15 am
回答 (4)
2006-11-16 7:29 am
✔ 最佳答案
1) 通分母[5(2x-1) - 3(3-x)]/15 = 2
2) 15移項變 乘15
[5(2x-1) - 3(3-x)] = 2(15)
3) 展開
10x - 5 - 9 + 3x = 30
13x - 14 = 30
4) -14移項變 + 14
13x = 30 + 14 = 44
5) 13移項變 除13
x = 44/13
2006-11-16 7:30 am
2x-1 3-x
------ -- ------- = 2
3 5
2x-1 3-x
------ x 15 -- ------- x 15 = 2 x 15
3 5
5 (2x-1) -- 3(3-x) = 30
10x -- 5 -- 9 + 3x = 30
13x = 30 + 5 + 9
13x = 49
x = 49
----
13
10
x = 3 ----
13
請check 一次題目有冇抄錯
------ -- ------- = 2
3 5
2x-1 3-x
------ x 15 -- ------- x 15 = 2 x 15
3 5
5 (2x-1) -- 3(3-x) = 30
10x -- 5 -- 9 + 3x = 30
13x = 30 + 5 + 9
13x = 49
x = 49
----
13
10
x = 3 ----
13
請check 一次題目有冇抄錯
2006-11-16 7:22 am
(2x-1)/3 - (3-x)/5=2
[(2x-1)/3 - (3-x)/5]*3*5=2*3*5 (將左邊分數既分母約簡)
(2x-1)/3*3*5-(3-x)/5*3*5=30 (拆括號)
(2x-1)*5-(3-x)*3=30
(10x-5)-(9-3x)-30
10x-5-9+3x=30(拆括號,小心負負得正)
10x+3x-5-9=30(將同項 放左一起)
13x-14=30
13x=44
x=44/13 or 3又5/13
[(2x-1)/3 - (3-x)/5]*3*5=2*3*5 (將左邊分數既分母約簡)
(2x-1)/3*3*5-(3-x)/5*3*5=30 (拆括號)
(2x-1)*5-(3-x)*3=30
(10x-5)-(9-3x)-30
10x-5-9+3x=30(拆括號,小心負負得正)
10x+3x-5-9=30(將同項 放左一起)
13x-14=30
13x=44
x=44/13 or 3又5/13
2006-11-16 7:20 am
Multiple both sides by 15
then we get
5(2x-1) - 3(3-x) = 30
5x - 5 - 9 + 9x = 30
14x = 44
x = 44/14
then we get
5(2x-1) - 3(3-x) = 30
5x - 5 - 9 + 9x = 30
14x = 44
x = 44/14
收錄日期: 2021-04-12 22:31:39
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20061115000051KK04948