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若x(1-x)=x^2，則ｘ=?
急~~ 若x(1-x)=x^2，則ｘ=?
2007-04-12 1:48 am
回答 (4)
2007-04-12 3:22 am
x=0.5/x=1/2
2007-04-12 2:15 am
x(1-x)=x^2
x - x^2 = x^2
x=2(x^2)
2x^2 - x = 0
x(2x - 1) = 0
只要 x=0 或 (2x-1)=0 便可以使整條方程成立，所以︰
x=0 / 2x-1=0
x=0 / x=0.5
x - x^2 = x^2
x=2(x^2)
2x^2 - x = 0
x(2x - 1) = 0
只要 x=0 或 (2x-1)=0 便可以使整條方程成立，所以︰
x=0 / 2x-1=0
x=0 / x=0.5
參考: 我的大腦
2007-04-12 1:50 am
x(1-x)=x^2
x - x^2 = x^2
2x^2 - x = 0
x(2x - 1) = 0
Now , because when x = 0 OR 2x - 1 = 0
LHS = 0 = RHS ,
so we divide into two cases :
(i)
x = 0
(ii)
2x - 1 = 0
x = 1 / 2
SO the answer is x = 0 OR x = 1 / 2
x - x^2 = x^2
2x^2 - x = 0
x(2x - 1) = 0
Now , because when x = 0 OR 2x - 1 = 0
LHS = 0 = RHS ,
so we divide into two cases :
(i)
x = 0
(ii)
2x - 1 = 0
x = 1 / 2
SO the answer is x = 0 OR x = 1 / 2
收錄日期: 2021-04-12 19:00:11
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20070411000051KK04258