因式分解法解几道二元一次方程 速度
kuaidi.ping-jia.net 作者：佚名 更新日期：2024-07-04
因式分解法解二元一次方程
(3x-5)(2x-1)=-12x+7
6x^2-13x+5=-12x+7
6x^2-13x+5+12x-7=0
6x^2-x-2=0
(3x-2)(2x+1)=0
3x-2=0,2x+1=0
x1=2/3,x2=-1/2
2.
(6y+5)(6y-5)-24=0
36y^2-25-24=0
36y^2-49=0
(6y+7)(6y-7)=0
6y+7=0,6y-7=0
y1=-7/6,y2=7/6
3.
t^2+2t-35=0
(t+7)(t-5)=0
t+7=0,t-5=0
t1=-7,t2=5
4.
4(x-3)^2-x(x-3)=0
4x^2-24x+36-x^2+3x=0
3x^2-21x+36=0
(3x+9)(x+4)=0
3x+9=0,x+4=0
x1=-3,x2=-4
1 6x^2-13x+5=-12x+7
6x^2-x-2=0
(3x-2)(2x+1)=0
2 36y^2-25-24=0
36y^2-49=0
(6y+7)(6y-7)=0
3 (t+7)(t-5)=0
4 (x-3)(4x-12-x)=0
(x-3)(3x-12)=0
http://zhidao.baidu.com/question/12119388.html上有十字相乘法的题怎么做?
2x^2+1=2√3x
2x^2-2√3x+1=0
x^2-√3x=-1/2
(x^2-√3x+3/4)=3/4-1/2
(x-√3/2)^2=1/4
∴x-√3/2=1/2或x-√3/2=-1/2
∴x=1/2+√3/2
或x=-1/2+√3/2
(3x-5)(2x-1)=-12x+7
6x^2-13x+5=-12x+7
6x^2-13x+5+12x-7=0
6x^2-x-2=0
(3x-2)(2x+1)=0
3x-2=0,2x+1=0
x1=2/3,x2=-1/2
2.
(6y+5)(6y-5)-24=0
36y^2-25-24=0
36y^2-49=0
(6y+7)(6y-7)=0
6y+7=0,6y-7=0
y1=-7/6,y2=7/6
3.
t^2+2t-35=0
(t+7)(t-5)=0
t+7=0,t-5=0
t1=-7,t2=5
4.
4(x-3)^2-x(x-3)=0
4x^2-24x+36-x^2+3x=0
3x^2-21x+36=0
(3x+9)(x+4)=0
3x+9=0,x+4=0
x1=-3,x2=-4
1 6x^2-13x+5=-12x+7
6x^2-x-2=0
(3x-2)(2x+1)=0
2 36y^2-25-24=0
36y^2-49=0
(6y+7)(6y-7)=0
3 (t+7)(t-5)=0
4 (x-3)(4x-12-x)=0
(x-3)(3x-12)=0
http://zhidao.baidu.com/question/12119388.html上有十字相乘法的题怎么做?
