因式分解:(1)(x+y)²-4(x+y)+4(2)2a(b-c)²-3b+3c(3)x³-2x²y+xy²-x
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因式分解:(1)(x+y)²-4(x+y)+4(2)2a(b-c)²-3b+3c(3)x³-2x²y+xy²-x因式分解:(1)(x+y)²
因式分解:(1)(x+y)²-4(x+y)+4(2)2a(b-c)²-3b+3c(3)x³-2x²y+xy²-x
因式分解:(1)(x+y)²-4(x+y)+4(2)2a(b-c)²-3b+3c(3)x³-2x²y+xy²-x
因式分解:(1)(x+y)²-4(x+y)+4(2)2a(b-c)²-3b+3c(3)x³-2x²y+xy²-x
(1)(x+y)²-4(x+y)+4
原式=(x+y)²-4(x+y)+2²
=(x+y-2)²
(2)2a(b-c)²-3b+3c
原式=2a(b-c)²-3(b-c)
=(b-c)(2ab-2ac-3)
(3)x³-2x²y+xy²-x
原式=x(x²-1) -xy(2x-y)
=x[x²-1-y(2x-y) ]
=x(x²-2xy+y²-1)
(x+y)²-4(x+y)+4
=(x+y)^2-2*(x+y)*2+2^2
=(x+y-2)^2
2a(b-c)²-3b+3c
=2a(b-c)^2-3(b-c)
=(2ab-2ac-3)(b-c)
x³-2x²y+xy²-x
=x(x^2-2xy+y^2-1)
=x((x-y)^2-1)
=x(x-y+1)(x-y-1)
(1)(x+y)²-4(x+y)+4
=(x+y-2)²
(2)2a(b-c)²-3b+3c
=2a(b-c)²-3(b-c)
=(b-c)(2ab-2ac-3)
(3)x³-2x²y+xy²-x
=x(x²-2xy+y²-1)
=x[(x-y)²-1]
=x(x-y+1)(x-y-1)
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