解下列方程:(1)3x2=12x;(2)x2-x-4=0;(3)(x+1)(x+2)=2x+4;(4)(x-1)2-4(x-1)+4=0

2024-10-31 20:47:40
推荐回答(1个)
回答1:

(1)移项得,3x2-12x=0,
提公因式得,3x(x-4)=0,
解得,x1=0,x2=4.
(2)a=1,b=-1,c=-4,
△=1-4×1×(-4)=17,
∴x1=

1+
17
2×1
=
1+
17
2
;x2=
1?
17
2×1
=
1?
17
2

(3)移项得,(x+1)(x+2)-(2x+4)=0,
提公因式得,(x+1)(x+2)-2(x+2)=0,
移项得,(x+2)(x-1)=0,
解得,x1=-2,x2=1.
(4)原方程可化为,(x-1-2)2=0,
解得,x1=x2=3.