解方程:(1)x2-4x-3=0; ...

2024-11-25 12:22:22
推荐回答(2个)
回答1:

(1)x2-4x-3=0,
b2-4ac=(-4)2-4×1×(-3)=28,
x=

28
2

x1=2+
7
,x2=2-
7
;     
        
(2)(x-3)2+2x(x-3)=0,
(x-3)(x-3+2x)=0,
x-3=0,x-3+2x=0,
x1=3,x2=1;

(3)2x2-10x=3,
2x2-10x-3=0,
b2-4ac=(-10)2-4×2×(-3)=124,
x=
10±
124
2×2

x1=
5+
31
2
,x2=
5?
31
2
;            
        
(4)(x-5)(x+2)=8,
x2-3x-18=0,
(x-6)(x+3)=0,
x-6=0,x+3=0,
x1=6,x2=-3;

(5)3x2+5(2x+1)=0,
3x2+10x+5=0,
b2-4ac=102-4×3×5=40,
x=
?10±
40
2×3

x1=
?5+
10
3
,x2=
?5?
10
3
;          
      
(6)2x2-7x-4=0,
(2x+1)(x-4)=0,
2x+1=0,x-4=0,
x1=-
1
2
,x2=4.

回答2:

解:(1)去分母得:1=x-1-3x+6,
解得:x=2,
经检验x=2是增根,分式方程无解;
(2)这里a=1,b=4,c=-1,
∵△=16+4=20,
∴x=-4±242=-2±5.