试题

题目:
分解因式:(1)(x-1)(x-2)-2(2-x)2
(2)x2-y2-(x+y)2
答案
解:(x)(x-x)(x-2)-2(2-x)2
=(x-x)(x-2)-2(x-2)2
=(x-2)(x-x-2)
=(x-2)(x-3);
(2)x2-y2-(x+y)2
=(x+y)(x-y)-(x+y)2
=(x+y)(x-y-x-y)
=-2y(x+y).
解:(x)(x-x)(x-2)-2(2-x)2
=(x-x)(x-2)-2(x-2)2
=(x-2)(x-x-2)
=(x-2)(x-3);
(2)x2-y2-(x+y)2
=(x+y)(x-y)-(x+y)2
=(x+y)(x-y-x-y)
=-2y(x+y).
考点梳理
因式分解-提公因式法;因式分解-分组分解法.
(1)首先把式子变形为:(x-1)(x-2)-2(x-2)2,再提取公因式(x-2)即可.
(2)首先利用平方差公式把x2-y2进行分解,在提取公因式(x+y)即可.
此题主要考查了提公因式法分解因式与分组分解法分解因式,关键是注意观察找准分解方法.
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