2x + 1.......1 + x
─── - ───
2x - 1.......2 - 4x
(.......) 只係想將D數隔開
幫手計條數.................
2007-08-08 6:55 pm
回答 (6)
2007-08-08 7:12 pm
✔ 最佳答案
...2x + 1..... 1 + x=─── - ─────
...2x - 1..... -2(-1+2x)
...2(2x + 1).........1 + x
=──── + ────
...2(2x - 1).........2(2x-1)
...4x + 2 + 1 + x
=──────
....2(2x-1)
...5x +3
=────
...4x-2
2007-08-08 7:34 pm
2x + 1.......1 + x
─── - ───
2x - 1.......2 - 4x
=(2x+1)/(2x-1) - (1+x)/[-2(2x-1)]
=-2(4x+2)-(1+x)/[-2(2x-1)]
=-8x-4-1-x/(2-4x)
=(9x+5)/(4x-2)
2007-08-08 11:36:12 補充：
2x + 1.......1 + x─── - ───2x - 1.......2 - 4x=(2x+1)/(2x-1) - (1+x)/[-2(2x-1)]=-2(2x+1)-(1+x)/[-2(2x-1)]=-4x-2-1-x/(2-4x)=(5x+3)/(4x-2)
─── - ───
2x - 1.......2 - 4x
=(2x+1)/(2x-1) - (1+x)/[-2(2x-1)]
=-2(4x+2)-(1+x)/[-2(2x-1)]
=-8x-4-1-x/(2-4x)
=(9x+5)/(4x-2)
2007-08-08 11:36:12 補充：
2x + 1.......1 + x─── - ───2x - 1.......2 - 4x=(2x+1)/(2x-1) - (1+x)/[-2(2x-1)]=-2(2x+1)-(1+x)/[-2(2x-1)]=-4x-2-1-x/(2-4x)=(5x+3)/(4x-2)
參考: me
2007-08-08 7:22 pm
= (5x+3) / (4x-2)
2007-08-08 7:19 pm
2x + 1.......1 + x
─── - ───
2x - 1.......2 - 4x
= (2x+1)/(2x-1) + (1+x)/[2(2x-1)]
= (4x+2+1+x)/[2(2x-1)]
= (5x+3) / (4x-2)
─── - ───
2x - 1.......2 - 4x
= (2x+1)/(2x-1) + (1+x)/[2(2x-1)]
= (4x+2+1+x)/[2(2x-1)]
= (5x+3) / (4x-2)
2007-08-08 7:12 pm
= (2x+1)(2-4x) - (1+x)(2x-1)
--------------------------------------
(2x-1)(2-4X)
= 4x-8x^2 +2-4x - (2x-1+2x^2 -X)
--------------------------------------------
4x-8x^2-2+4x
= -10x^2 +3 - x
------------------
8x-8x^2-2
Shadow.
--------------------------------------
(2x-1)(2-4X)
= 4x-8x^2 +2-4x - (2x-1+2x^2 -X)
--------------------------------------------
4x-8x^2-2+4x
= -10x^2 +3 - x
------------------
8x-8x^2-2
Shadow.
2007-08-08 7:04 pm
2x + 1.......1 + x
─── - ───
2x - 1.......2 - 4x
=(2x+1)/(2x-1) + (1+x)/[2(2x-1)]
=(4x+2+1+x)/[2(2x-1)]
=(5x+3) / (4x-2)
─── - ───
2x - 1.......2 - 4x
=(2x+1)/(2x-1) + (1+x)/[2(2x-1)]
=(4x+2+1+x)/[2(2x-1)]
=(5x+3) / (4x-2)
參考: me
收錄日期: 2021-04-23 20:00:07
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20070808000051KK01399