一元两次方程,用因式分解法
来源:学生作业帮 编辑:搜搜考试网作业帮 分类:数学作业 时间:2024/07/15 19:12:44
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
![一元两次方程,用因式分解法](/uploads/image/z/18719199-63-9.jpg?t=%E4%B8%80%E5%85%83%E4%B8%A4%E6%AC%A1%E6%96%B9%E7%A8%8B%2C%E7%94%A8%E5%9B%A0%E5%BC%8F%E5%88%86%E8%A7%A3%E6%B3%95)
(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