两元一次方程

4x^2-4x-x^2+1=0
3x^2-4x+1=0
b^2-4ac=16-12=4
x=(4+2)/(2*3)=1
x=(4-2)/(2*3)=1/3

解：去括号得
4x^2 - 4x = x^2 -1

移项、合并同类项，得
3x^2 - 4x + 1 = 0

因式分解，得
(3x-1)*(x-1) = 0
于是得
(3x-1)= 0 或)(x-1) = 0
x1 = 1/3 ，x2 = 1 。

4x^2 - 4x = x^2 -1
3x^2 - 4x + 1 = 0
(3x-1)*(x-1) = 0
x1 = 1/3
x2 = 1

解：4X^2-4X=X^2-1
3X^2-4X+1=0
a=3 b=-4 c=1
判别式＝b^2-4ac=(-4)^2-4*3*1=4
X=(-b加减根号判别式)/2a
=(4+(-)2)/6
X1=1 X2=2/2

此题呒解...!用公式法的话就...此题呒解...!

a=1 b=-4 c=3 b的平方-4ac=16-12=4 x=(4±2)÷1 x1=6 x2=2