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thx~~~~~~~
點條數點做?
2007-05-08 4:17 am
回答 (5)
2007-05-08 11:57 pm
✔ 最佳答案
1/(n+3) - (1/3-n)= 1/(n+3) + (1/n-3)
= (n-3+n-3) / [(n+3)(n-3)]
= 2n / n^2 -9
如果double check計得o岩唔o岩,
建議由最後的step check返轉頭,
咁你第一時間就會發現mistake在後一步個9冇乘 -1 啦
2007-05-09 12:28 am
若是依等號後沒東西(即0)時，便可出如下的方程式﹕
(n+3)分之1=(3-n)分之1
由此可見，3+n=3-n
即n=0
(n+3)分之1=(3-n)分之1
由此可見，3+n=3-n
即n=0
2007-05-08 4:57 am
1/(n+3) - 1/(3-n)
= 1x(3-n) / (n+3) x (3-n) - 1x(n+3) / (3-n) x (n+3)
= (3-n) - (n+3) / (n+3) x (3-n)
= (3-n-n-3) / (3n+9-n^2-3n)
= - 2n / (9-n^2)
= 1x(3-n) / (n+3) x (3-n) - 1x(n+3) / (3-n) x (n+3)
= (3-n) - (n+3) / (n+3) x (3-n)
= (3-n-n-3) / (3n+9-n^2-3n)
= - 2n / (9-n^2)
參考: me
2007-05-08 4:30 am
１ １ ３－ｎ ｎ＋３
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ｎ＋３ ３－ｎ ３ｎ－ｎ＊ｎ＋９－３ｎ ３ｎ－ｎ＊ｎ＋９－３ｎ
３－ｎ ｎ＋３
＝ ＿＿＿＿＿ － ＿＿＿＿＿
９－ｎ＊ｎ ９－ｎ＊ｎ
＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝
＿＿＿ － ＿＿＿ ＝ ＿＿＿＿＿＿＿＿＿＿＿ － ＿＿＿＿＿＿＿＿＿＿＿
ｎ＋３ ３－ｎ ３ｎ－ｎ＊ｎ＋９－３ｎ ３ｎ－ｎ＊ｎ＋９－３ｎ
３－ｎ ｎ＋３
＝ ＿＿＿＿＿ － ＿＿＿＿＿
９－ｎ＊ｎ ９－ｎ＊ｎ
＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝
2007-05-08 4:30 am
參考: me
收錄日期: 2021-04-13 14:59:43
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20070507000051KK04153