拜託幫我解一下please~~~
若二元一次方程式 x-12/12 - y+3/13 =0 ，則x/12 - y/13 =?
拜託~~~~求解~~~~
數學二元一次方程式
2015-08-06 8:48 am
回答 (9)
2015-08-07 7:29 am
[(x‒12)/12] ‒ [(y + 3)/13] = 0
[(x/12) ‒ 1] ‒ [(y/13) + (3/13)] = 0
(x/12) ‒ 1 ‒ (y/13) ‒ (3/13) = 0
(x/12)‒ (y/13) = 16/13
[(x/12) ‒ 1] ‒ [(y/13) + (3/13)] = 0
(x/12) ‒ 1 ‒ (y/13) ‒ (3/13) = 0
(x/12)‒ (y/13) = 16/13
2015-08-06 10:46 am
嗯，除非題目打錯。。。
2015-08-06 18:16:58 補充：
土扁 大哥，是否需要補充一下？
2015-08-06 18:16:58 補充：
土扁 大哥，是否需要補充一下？
2015-11-25 2:42 pm
(x/12‒ 12/12) ‒ (y/13 + 3/13) = 0
(x/12) ‒ 1 ‒ (y/13) + 3/13 = 0
(x/12) ‒ 1 ‒ (y/13) ‒ 3/13 = 0
(x/12) ‒ (y/13) = 16/13 (等號左邊 - 1 跟 - 3 / 13 移到等號右邊後)
(x/12) ‒ 1 ‒ (y/13) + 3/13 = 0
(x/12) ‒ 1 ‒ (y/13) ‒ 3/13 = 0
(x/12) ‒ (y/13) = 16/13 (等號左邊 - 1 跟 - 3 / 13 移到等號右邊後)
2015-08-18 7:48 pm
無法解答
2015-08-18 9:20 am
16/13
2015-08-13 8:53 am
題目怪怪的
2015-08-07 9:03 am
無最佳解答
2015-08-07 7:26 am
是太糊塗，看錯了數字。
2015-08-06 9:42 am
[(x‒ 12)/12] ‒ [(y + 3)/13] = 0
[(x/12) ‒ 1] ‒ [(y/13) + 3/13] = 0
(x/12) ‒ 1 ‒ (y/13) ‒ 3/13 = 0
(x/12) ‒ (y/13) = 16/13
應該是這樣0.0?
[(x/12) ‒ 1] ‒ [(y/13) + 3/13] = 0
(x/12) ‒ 1 ‒ (y/13) ‒ 3/13 = 0
(x/12) ‒ (y/13) = 16/13
應該是這樣0.0?
收錄日期: 2021-04-16 17:09:01
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20150806000015KK00097