用适当的方法解方程:(1)x2-2x-3=0;(2)2x2-3x-1=0;(公式法解)(3)x(2x+3)=4x+6;(4)(2x+3

2025-02-26 06:43:51
推荐回答(1个)
回答1:

(1)分解因式得:(x-3)(x+1)=0,
x-3=0,x+1=0,
x1=3,x2=-1;

(2)2x2-3x-1=0,
b2-4ac=(-3)2-4×2×(-1)=17,
x=
17
2×2

x1=
3+
17
4
,x2=
3-
17
4


(3)x(2x+3)=4x+6,
x(2x+3)-2(2x+3)=0,
(2x+3)(x-2)=0,
2x+3=0,x-2=0,
x1=-1.5,x2=2;

(4)(2x+3)2=x2-6x+9,
(2x+3)2=(x-3)2
2x+3=±(x-3),
x1=-6,x2=0;

(5)3x2-6x-2=0,
3x2-6x=2,
x2-2x=
2
3

x2-2x+1=
2
3
+1,
(x-1)2=
5
3

x-1=±
5
3

x1=
3+
15
3
,x2=
3-
15
3


(6)x(x+4)=8x+12,
x2-4x-12=0,
(x-6)(x+2)=0,
x-6=0,x+2=0,
x1=6,x2=-2.