题目内容
12、分解因式:x2-y2+3x-y+2=
(x+y+2)(x-y+1)
.分析:将-y拆项,即-2y+y,3x拆项,x+2x,再分组,x2-y2+3x-y+2=(x2-y2)+(x+y)+(2x-2y+2),再运用平方差公式和提公因式分解即可.
解答:解:x2-y2+3x-y+2 (x2-y2)+(x+y)+(2x-2y+2)
=(x+y)(x-y)+(x+y)+2(x-y+1)
=(x+y)(x-y+1)+2(x-y+1)
=(x-y+1)(x+y+2);
故答案为(x+y+2)(x-y+1).
=(x+y)(x-y)+(x+y)+2(x-y+1)
=(x+y)(x-y+1)+2(x-y+1)
=(x-y+1)(x+y+2);
故答案为(x+y+2)(x-y+1).
点评:本题考查了因式分解--分组分解法,拆项是因式分解常用的方法,但难度较大.
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