几个分解因式法解方程,
来源:学生作业帮 编辑:百度作业网作业帮 分类:数学作业 时间:2024/08/18 03:14:13
几个分解因式法解方程,
(4x-1)(5x+7)=0 3x(x-1)=2-2x (2x+3)²=4(2x+3) 2(x-3)²=x²-9 5(x²-x)=3(x²+x) (x-2)²=(2x+3)² (x-2)(x-3)=12 2x+6=(x+3)² 2y²+4y=y+2 2(x+3)²=x(x+3) x²-2根号5x+2=0 (x+1)²-3(x+1)+2+0
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(4x-1)(5x+7)=0 3x(x-1)=2-2x (2x+3)²=4(2x+3) 2(x-3)²=x²-9 5(x²-x)=3(x²+x) (x-2)²=(2x+3)² (x-2)(x-3)=12 2x+6=(x+3)² 2y²+4y=y+2 2(x+3)²=x(x+3) x²-2根号5x+2=0 (x+1)²-3(x+1)+2+0
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![几个分解因式法解方程,](/uploads/image/z/3695892-60-2.jpg?t=%E5%87%A0%E4%B8%AA%E5%88%86%E8%A7%A3%E5%9B%A0%E5%BC%8F%E6%B3%95%E8%A7%A3%E6%96%B9%E7%A8%8B%2C)
(4x-1)(5x+7)=0
4x-1=0或5x+7=0
x1=1/4,x2=-7/5
3x(x-1)-(2-2x)=0
3x(x-1)-2(1-x)=0
3x(x-1)+2(x-1)=0
(x-1)(3x-2)=0
x-1=0或3x-2=0
x1=1,x2=2/3
(2x+3)²=4(2x+3)
(2x+3)²-4(2x+3)=0
(2x+3)(2x+3-4)=0
(2x+3)(2x-1)=0
2x+3=0或2x-1=0
x1=-3/2,x2=1/2
2(x-3)²=x²-9
2(x-3)²-(x²-9)=0
2(x-3)²-(x-3)(x+3)=0
(x-3)[2(x-3)-(x+3)]=0
(x-3)(2x-6-x-3)=0
(x-3)(x-9)=0
x-3=0或x-9=0
x1=3,x2=9
5(x²-x)=3(x²+x)
5(x²-x)-3(x²+x)=0
x(5x-5)-x(3x+3)=0
x[(5x-5)-(3x+3)]=0
x(5x-5-3x-3)=0
x(2x-8)=0
x=0或2x-8=0
x1=0,x2=4
(x-2)²=(2x+3)²
(x-2)²-(2x+3)²=0
[(x-2)+(2x+3)][(x-2)-(2x+3)]=0
(x-2+2x+3)(x-2-2x-3)=0
(3x+1)(-x-5)=0
3x+1=0或-x-5=0
x1=-1/3,x2=-5
(x-2)(x-3)=12
x²-3x-2x+6=12
x²-5x+6-12=0
x²-5x-6=0
(x+1)(x-6)=0
x+1=0或x-6=0
x1=-1,x2=6
2x+6=(x+3)²
(2x+6)-(x+3)²=0
2(x+3)-(x+3)²=0
(x+3)[2-(x+3)]=0
(x+3)(2-x-3)=0
(x+3)(-x-1)=0
x+3=0或-x-1=0
x1=-3,x2=-1
2y²+4y=y+2
(2y²+4y)-(y+2)=0
2y(y+2)-(y+2)=0
(y+2)(2y-1)=0
y+2=0或2y-1=0
y1=-2,y2=1/2
2(x+3)²=x(x+3)
2(x+3)²-x(x+3)=0
(x+3)[2(x+3)-x]=0
(x+3)(2x+6-x)=0
(x+3)(x+6)=0
x+3=0或x+6=0
x1=-3,x2=-6
x²-2根号5x+2=0
x²-(2√5)x+2+3=0+3
x²-(2√5)x+5=3
(x-√5)²=(±√3)²
x-√5=±√3
x-√5=√3或x-√5=-√3
x1=√5+√3,x2=√5-√3,如果你的题没有错,这就是正解
(x+1)²-3(x+1)+2=0,把(x+1)看成整体,用十字相乘法分解
[(x+1)-1]][(x+1)-2]=0
(x+1-1)(x+1-2)=0
x(x-1)=0
x=0或x-1=0
x1=0,x2=1
4x-1=0或5x+7=0
x1=1/4,x2=-7/5
3x(x-1)-(2-2x)=0
3x(x-1)-2(1-x)=0
3x(x-1)+2(x-1)=0
(x-1)(3x-2)=0
x-1=0或3x-2=0
x1=1,x2=2/3
(2x+3)²=4(2x+3)
(2x+3)²-4(2x+3)=0
(2x+3)(2x+3-4)=0
(2x+3)(2x-1)=0
2x+3=0或2x-1=0
x1=-3/2,x2=1/2
2(x-3)²=x²-9
2(x-3)²-(x²-9)=0
2(x-3)²-(x-3)(x+3)=0
(x-3)[2(x-3)-(x+3)]=0
(x-3)(2x-6-x-3)=0
(x-3)(x-9)=0
x-3=0或x-9=0
x1=3,x2=9
5(x²-x)=3(x²+x)
5(x²-x)-3(x²+x)=0
x(5x-5)-x(3x+3)=0
x[(5x-5)-(3x+3)]=0
x(5x-5-3x-3)=0
x(2x-8)=0
x=0或2x-8=0
x1=0,x2=4
(x-2)²=(2x+3)²
(x-2)²-(2x+3)²=0
[(x-2)+(2x+3)][(x-2)-(2x+3)]=0
(x-2+2x+3)(x-2-2x-3)=0
(3x+1)(-x-5)=0
3x+1=0或-x-5=0
x1=-1/3,x2=-5
(x-2)(x-3)=12
x²-3x-2x+6=12
x²-5x+6-12=0
x²-5x-6=0
(x+1)(x-6)=0
x+1=0或x-6=0
x1=-1,x2=6
2x+6=(x+3)²
(2x+6)-(x+3)²=0
2(x+3)-(x+3)²=0
(x+3)[2-(x+3)]=0
(x+3)(2-x-3)=0
(x+3)(-x-1)=0
x+3=0或-x-1=0
x1=-3,x2=-1
2y²+4y=y+2
(2y²+4y)-(y+2)=0
2y(y+2)-(y+2)=0
(y+2)(2y-1)=0
y+2=0或2y-1=0
y1=-2,y2=1/2
2(x+3)²=x(x+3)
2(x+3)²-x(x+3)=0
(x+3)[2(x+3)-x]=0
(x+3)(2x+6-x)=0
(x+3)(x+6)=0
x+3=0或x+6=0
x1=-3,x2=-6
x²-2根号5x+2=0
x²-(2√5)x+2+3=0+3
x²-(2√5)x+5=3
(x-√5)²=(±√3)²
x-√5=±√3
x-√5=√3或x-√5=-√3
x1=√5+√3,x2=√5-√3,如果你的题没有错,这就是正解
(x+1)²-3(x+1)+2=0,把(x+1)看成整体,用十字相乘法分解
[(x+1)-1]][(x+1)-2]=0
(x+1-1)(x+1-2)=0
x(x-1)=0
x=0或x-1=0
x1=0,x2=1