题目

把下列各式分解因式.(1)a²-2ab+b²-a²(2)x²-y²+6y-9
(3)x²-6xy-1+9y²（4）2xy-x²-y²+1

提问时间：2020-12-19

答案

(1)a²-2ab+b²-a²
=-2ab+b²
=b(b-2a)
(2)x²-y²+6y-9
=x²-(y²-6y+9)
=x²-(y-3)²
=(x+y-3)(x-y+3)
(3)x²-6xy-1+9y²
=x²-6xy+9y²-1
=(x-3y)²-1²
=(x-3y+1)(x-3y-1)
（4）2xy-x²-y²+1
=1-(x²-2xy+y²)
=1-(x-y)²
=(1+x-y)(1-x+y)