初二因式分解

Question

1.x4+y4+(x+y)4
2.x4+1999x2+1998x+1999
3.(a+b-2ab)(a+b-2)+(1-ab)2

Answer 1

1）x^4+(x+y)^4+y^4
=(x²+y²)²-2x²y²+(x+y)^4
=[(x+y)²-2xy]²-2x²y²+(x+y)^4
=[(x+y)²]²-4xy(x+y)²+4x²y²-2x²y²+(x+y)^4
=2(x+y)^4-4xy(x+y)²+2x²y²
=2[(x+y)^4-2xy(x+y)²+(xy)²]
=2[(x+y)²-xy]²
=2(x²+xy+y²)²

2）x4+1999x2+1998x+1999
从一般式入手，设n=1998,则n+1=1999
x^4+(n+1)x²+nx+(n+1)
=x^4+nx²+x²+nx+n+1
=(x^4+2x²+1-x²)+(nx²+nx+n)
=(x²+1)²-x²+n(x²+x+1)
=(x²+x+1)(x²-x+1)+n(x²+x+1)
=(x²+x+1)(x²-x+n+1)
再代回
x4+1999x2+1998x+1999
=(x²+x+1)(x²-x+1999)

3)(a+b-2ab)(a+b-2)+(1-ab)2
=(a+b)²-2ab(a+b)-2(a+b)+4ab+(ab)²-2ab+1
=[(a+b)²-2ab(a+b)+(ab)²]+2ab-2(a+b)+1
=(a+b-ab)²+2(ab-a-b)+1
=(ab-a-b)²+2(ab-a-b)+1
=(ab-a-b+1)²
=[a(b-1)-(b-1)]²
=(a-1)²(b-1)²

Answer 2

8x+8y
2002x+5997
（a-1）的平方+（b-1）的平方 -2a²b-2ab²+4ab