解题思路:先对(x2-2x+1-y2)进行因式分解得:(x2-2x+1-y2)=(x-1+y)(x-1-y),再去相除得即可求出结果.
(x2-2x+1-y2)÷(x+y-1),
=(x-1+y)(x-1-y)÷(x+y-1),
=x-y-1.
故应填:x-y-1.
点评:
本题考点: 整式的除法.
考点点评: 主要考查多项式除法运算,先把被除式进行因式分解是解题的关键,也是难点.
解题思路:先对(x2-2x+1-y2)进行因式分解得:(x2-2x+1-y2)=(x-1+y)(x-1-y),再去相除得即可求出结果.
(x2-2x+1-y2)÷(x+y-1),
=(x-1+y)(x-1-y)÷(x+y-1),
=x-y-1.
故应填:x-y-1.
点评:
本题考点: 整式的除法.
考点点评: 主要考查多项式除法运算,先把被除式进行因式分解是解题的关键,也是难点.