一元两次方程,用因式分解法
来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/08/24 04:53:35
一元两次方程,用因式分解法
(1)2x²-3X-14=o
(2)3(X-5)²=2(5-x)
(3)22y²-49y-15=0
(1)2x²-3X-14=o
(2)3(X-5)²=2(5-x)
(3)22y²-49y-15=0
(1)2x²-3X-14=o
(2x-7)(x+2)=0
2x-7=0 x=7/2
x+2=0 x=-2
(2)3(X-5)²=2(5-x)
3(x-5)²-2(5-x)=0
3(x-5)²+2(x-5)=0
(x-5)[3(x-5)+2]=0
(x-5)(3x-15+2)=0
(x-5)(3x-13)=0
x-5=0 x=5
3x-13=0 x=13/3
(3)22y²-49y-15=0
(11y+3)(2y-5)=0
11y+3=0 y=-3/11
2y-5=0 y=5/2
(2x-7)(x+2)=0
2x-7=0 x=7/2
x+2=0 x=-2
(2)3(X-5)²=2(5-x)
3(x-5)²-2(5-x)=0
3(x-5)²+2(x-5)=0
(x-5)[3(x-5)+2]=0
(x-5)(3x-15+2)=0
(x-5)(3x-13)=0
x-5=0 x=5
3x-13=0 x=13/3
(3)22y²-49y-15=0
(11y+3)(2y-5)=0
11y+3=0 y=-3/11
2y-5=0 y=5/2