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
来源:学生作业帮助网 编辑:作业帮 时间:2024/09/01 07:34:46
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
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
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
(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
(3x-17)(x-5)=0
(2x-5)(11x-6)=0
(1)2x²-3X-14=o
(2x-7)(x+2)=0
∴x1=7/2,x2=-2
(2)3(X-5)²=2(5-x)
3(x-5)²-2(x-5)=0
(x-5)(3x-15-2)=0
∴x1=5,x2=17/3
(3)
22y²-49y-15=0
(11x+3)(2x-5)=0
∴x1=-3/11,x2=5/2.
(1)2x²-3X-14=o
(2x-7)(x+2)=0
x=7/2 x=-2
(2)3(X-5)²=2(5-x)
3(X-5)²+2(x-5)=0
(x-5)(3(x-5)+2)=0
(x-5)(3x-13)=0
x=5 x=13/3
(3)22y²-49y-15=0
(11y+3)(2y-5)=0
y=-3/11 y=5/2
(1)(2x-7)(2x-1)=0 x=7/2或1/2
(2)1.当x=5时符合
2.当x不等于5时 3(x-5)=-2 所以x=13/3
2x²-3X-14=0 (x+2)(x-7)=0 x=-2, 7
3(X-5)²=2(5-x)(5-x)[2-3(5-x)]=0 x=5, 13/3
22y²-49y-15=0 (11x+3)(2x-5)=0 x=-3/11, 5/2
1.x-7)(x+2)=0
x=-2或X=7/2
2.(X-5)[3(X-5)+2]=0
(X-5)(3X-13)=0
x=5或x=13/3
3.(2x-5)(11x+3)=0
x=5/2或x=-3/11