Please solve 3x^2-4x+1=0?
回答 (5)
2008-07-11 3:53 pm
✔ 最佳答案
3x^2 - 4x + 1 = 0(3x - 1)(x - 1) = 0
x = 1/3 OR x = 1
2008-07-11 10:48 pm
3x^2 - 4x + 1 = 0
3x^2 - x - 3x + 1 = 0
(3x^2 - x) - (3x - 1) = 0
x(3x - 1) - 1(3x - 1) = 0
(3x - 1)(x - 1) = 0
3x - 1 = 0
3x = 1
x = 1/3
x - 1 = 0
x = 1
â´ x = 1/3 , 1
3x^2 - x - 3x + 1 = 0
(3x^2 - x) - (3x - 1) = 0
x(3x - 1) - 1(3x - 1) = 0
(3x - 1)(x - 1) = 0
3x - 1 = 0
3x = 1
x = 1/3
x - 1 = 0
x = 1
â´ x = 1/3 , 1
2008-07-11 10:34 pm
3x^2 - 4x + 1=0
=> 3x^2 - 3x +x +1=0
=> 3x(x-1) - 1(x-1)=0
=> (x-1) (3x-1)=0
=> x-1=0 or 3x-1=0
=> x=1 or 1/3.......
=> 3x^2 - 3x +x +1=0
=> 3x(x-1) - 1(x-1)=0
=> (x-1) (3x-1)=0
=> x-1=0 or 3x-1=0
=> x=1 or 1/3.......
參考: me
2008-07-11 10:21 pm
3x^2 - 4x + 1 = 0
(3x - 1)(x - 1) = 0
x -1 = 0 --------->change sign on one & move to other side
x = 1
3x - 1 = 0 --------->change sign on one & move to other side
3x = 1--------> divided both sdie by 3
x = 1/3
(3x - 1)(x - 1) = 0
x -1 = 0 --------->change sign on one & move to other side
x = 1
3x - 1 = 0 --------->change sign on one & move to other side
3x = 1--------> divided both sdie by 3
x = 1/3
2008-07-11 10:21 pm
(3x-1)(x-1)
x = 1 and/or 1/3
x = 1 and/or 1/3
收錄日期: 2021-05-01 10:48:44
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