一道初3的题---1元2次方程

ax(a-x)+ab^2=b(bx-x^2)+a^2b
(a-b)[(x-a)(x-b)]=0
x1=a;x2=b

ax(a-x)+ab^2=b(bx-x^2)+a^2b
a^2x-ax^2+ab^2=b^2x-bx^2+a^2b
a^2x-ax^2-b^2x+bx^2=a^2b-ab^2
(a-b)x^2-(a+b)(a-b)x=ab(a-b)
a,b是常数,且a不等于b
两边除以(a-b)
x^2-(a+b)x=ab
x^2-(a+b)x-ab=0
(x-a)(x+b)=0或(x-a)(x+b)=0
x=a,x=-b或x=-a,x=b

ax(a-x)+ab^2=b(bx-x^2)+a^2b
-ax^2+bx^2+a^2x-b^2x+ab^2-a^2b=0
解得x1=a;x2=b