Question

请教一道初中数学题

请写出解题过程，谢谢

Answer 1

通分，然后合并同类项，最后化简为
2（a-b）/(b-c)
原式=(a+b)²/(a-b)(b-c)-6ab/(a-b)(b-c)+(a²+b²)/(a-b)(b-c)
=[(a+b)²-6ab+(a²+b²)]/(a-b)(b-c)
=(a²+2ab+b²-6ab+a²+b²)/(a-b)(b-c)
=2(a²-2ab+b²)/(a-b)(b-c)
=2(a-b)²/(a-b)(b-c)
=(2a-2b)/(b-c)

简介点就是：
原式=(a+b)^2/(a-b)(b-c)-6ab/(a-b)(b-c)+(a^2+b^2)/(a-b)(b-c)
=[(a+b)^2-6ab+a^2+b^2]/(a-b)(b-c)
=2(a-b)^2/(a-b)(b-c)
=2(a-b)/(b-c)

Answer 2

原式=(a+b)²/(a-b)(b-c)-6ab/(a-b)(b-c)+(a²+b²)/(a-b)(b-c)
=[(a+b)²-6ab+(a²+b²)]/(a-b)(b-c)
=(a²+2ab+b²-6ab+a²+b²)/(a-b)(b-c)
=2(a²-2ab+b²)/(a-b)(b-c)
=2(a-b)²/(a-b)(b-c)
=(2a-2b)/(b-c)

Answer 3

通分，然后合并同类项，最后化简为
2（a-b）/(b-c)