x^2-2x+1=0 4x^2+13x+3=0 一般化探索规律,并一般化,写出式子.x^2-2x+1=0(x-1)(x-1)=0x=1x^2-3x+2=0(x-1)(x-2)=0x=1 x=23x^2+x-2=03(x-2/3)(x+1)=0x=2/3,x=-12x^2+5x+2=02(x+1/2)(x+2)=0x=-1/2,x=-24x^2+13x+3=04(x+1/4)(x+3)=0x=-1/4,x=-3以上.结论一般化,写出式子...
问题描述：
x^2-2x+1=0 4x^2+13x+3=0 一般化
探索规律,并一般化,写出式子.
x^2-2x+1=0
(x-1)(x-1)=0
x=1
x^2-3x+2=0
(x-1)(x-2)=0
x=1 x=2
3x^2+x-2=0
3(x-2/3)(x+1)=0
x=2/3,x=-1
2x^2+5x+2=0
2(x+1/2)(x+2)=0
x=-1/2,x=-2
4x^2+13x+3=0
4(x+1/4)(x+3)=0
x=-1/4,x=-3
以上.结论一般化,写出式子...
答
ax^2+(ab+ac)x+abc=0
→a(x+b)(x+c)=0