分解因式:(1)a的平方-2ab+b的平方-1 (2)1-x的平方+2xy-y的平方

2024-12-28 06:22:54
推荐回答(4个)
回答1:

原式=(a²-2ab+b²)-1
=(a-b)²-1²
=(a-b-1)(a-b+1)
原式=1-(x²-2xy+y²)
=1²-(x-y)²
=(1-x+y)(1+x-y)

回答2:

a^2-2ab+b^2-1
=(a-b)^2-1
=(a-b+1)(a-b-1)
1-x^2+2xy-y^2
=1-(x^2-2xy+y^2)
=1-(x-y)^2
=(1+x-y)(1-x+y)

回答3:

a^2-2ab+b^2-1=(a-b)^2-1=(a-b+1)(a-b-1)
1-x^2+2xy-y^2=1-(x^2-2xy+y^2)=1-(x-y)^2=(1-x+y)(1+x-y)

回答4:

(1)解:原式=(a+b)平方-1
=(a+b+1)(a+b-1)
(2)解:原式=1-(x-y)平方
=(1+x-y)(1-x+y)