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)
(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)
x2-4x=42 2
原方程可化为:x2-2
x-4=02
∵△=(-2
)2-4×1×(-4)=24>02
∴x=
=2
±2 24 2
±2 6
即x1=
+2
,x2=6
-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