因式分解,第五题
2个回答

5、(1)a²b²-2ab+1=(ab-1)²

(2)9-12a+4a²=(3-2a)²

(3)4x²+4x+1=(2x+1)²

(4)x²-6xy+9y²=(x-3y)²

6、（1）(a+b)²+6(a+b)+9=[(a+b)+3]²=(a+b+3)²

(2)16-24(x-y)+9(x-y)²=[4-3(x-y)]²=(4-3x+3y)²

(3)a²-2a(b+c)+(b+c)²=[a-(b+c)]²=(a-b-c)²

(4)4x²-4x(y-1)+(y-1)²=[2x-(y-1)]²=(2x-y+1)²

7、(1)-a+2a²-a³=-a(1-2a+a²)=-a(1-a)²

(2)3ax²-3ay²=3a(x²-y²)=3a(x+y)(x-y)

(3)16a²b-16a³-4ab²=-4a(4a²-4ab+b²)=-4a(2a-b)²

(4)x²(y²-1)+(1-y²)=x²(y²-1)-(y²-1)=(y²-1)(x²-1)=(y+1)(y-1)(x+1)(x-1)

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