问题 解答题
解下列方程:
(1)(y-1)(y+2)=2y(1-y);
(2)(x-2)(x+3)=66;
(3)
2
x2-4x=4
2

(4)x2-b2=a(3x-2a+b).
答案

(1)(y-1)(y+2)+2y(y-1)=0

(y-1)(y+2+2y)=0

y-1=0或3y+2=0

解得:y1=1,y2=-

2
3

(2)原方程可化为:x2+x-72=0

(x-8)(x+9)=0

解得:x1=8,x2=-9

(3)

2
x2-4x=4
2

原方程可化为:x2-2

2
x-4=0

△=(-2

2
)2-4×1×(-4)=24>0

x=

2
2
±
24
2
=
2
±
6

x1=

2
+
6
x2=
2
-
6

(4)原方程化为一般形式为:x2-3ax+2a2-ab-b2=0

即x2-3ax+(2a+b)(a-b)=0

[x-(2a+b)][x-(a-b)]=0

解得:x1=2a+b,x2=a-b

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