能不能用-配方法-解方程24(R)(1-R)=6
24(R)(1-R)=6
能不能用配方法解方程 24(R)(1-R)=6
2007-01-01 4:06 am
回答 (5)
2007-01-01 4:19 am
✔ 最佳答案
24(R)(1-R)=6 (R)(1-R)=1/4
R^2-R=-1/4
R^2-R+(1/2)^2=-1/4+(1/2)^2
(R+1/2)^2=0
R=-1/2(repeated)
不知道會不會錯~
但以我所知,所有二元一次方程式都可以用配方法解,
所以才會有quadratic formula
2006-12-31 20:30:18 補充：
不好意思~其實R=1/2......我搞錯左負號.....
2007-01-01 4:14 am
24(R)(1-R)=6
24R(1-R) = 6
24R*1 - 24R^2 = 6
-24R^2 + 24R -6= 0
24R^2 -24R+6 = 0 <-- all divided by 6
4R^2-4R+1 = 0
(2R - 1)(2R-1)= 0
--> 2R-1=0
2R =1
R= 1/2 (REPEATED)
24R(1-R) = 6
24R*1 - 24R^2 = 6
-24R^2 + 24R -6= 0
24R^2 -24R+6 = 0 <-- all divided by 6
4R^2-4R+1 = 0
(2R - 1)(2R-1)= 0
--> 2R-1=0
2R =1
R= 1/2 (REPEATED)
2007-01-01 4:13 am
24(R)(1-R)=6
24R^2-24R+6=0
24( R^2 - R + 1/4 )
24( R- 1/2 )^2
24R^2-24R+6=0
24( R^2 - R + 1/4 )
24( R- 1/2 )^2
2007-01-01 4:12 am
yes you can
24R(1-R)=6
4R(1-R)=1
-4R^2+4R=1
-4R^2+4R-1=0
R=(-4±√(4^2-4(-4)(-1)))/2(-4)
R=(-4±√0)/(-8)
R=1/2
2006-12-31 20:13:27 補充：
任何二次方程都可以用配方法解求解公式都係咁黎
24R(1-R)=6
4R(1-R)=1
-4R^2+4R=1
-4R^2+4R-1=0
R=(-4±√(4^2-4(-4)(-1)))/2(-4)
R=(-4±√0)/(-8)
R=1/2
2006-12-31 20:13:27 補充：
任何二次方程都可以用配方法解求解公式都係咁黎
參考: me
2007-01-01 4:11 am
R-R^2=1/4
R^2-R+1/4=0
(2R-1)^2=0
R=1/2
R^2-R+1/4=0
(2R-1)^2=0
R=1/2
收錄日期: 2021-04-12 21:32:52
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20061231000051KK04149