Question

分解因式两道

(x+y-2xy)(x+y-2)+(xy-1)²
(x+1)(x+2)(x+3)(x+6)+x²

Answer 1

1。 (x+y-2xy)(x+y-2)+(xy-1)²=(x+y-2-2xy+2)(x+y-2)+(xy-1)² 令x+y-2=a xy-1=b
原式=（a-2b）a+b²=a²-2ab+b²
=（a-b）²然后将a，b代回即可
2. 你可以令x+3=a 则x²=（a-3）² 带入题中，与1 是相同的方法，不过还有其他的方法可以解，请参考

Answer 2

解：
1.令x+y=m,xy=n,则：原式为：
(m-2n)(m-2)+(n-1)^2
=m^2+n^2-2mn-2(m-n)+1
因为m^2+n^2-2mn = (m-n)^2
所以令：m^2+n^2-2mn-2(m-n)+1 = (m-n+a)(m-n+b)
可得：a=b=-1
所以：原式=(x+y-xy-1)^2 = (x-1)^2(y-1)^2
2.观察(x+2)(x+3)和(x+1)(x+6)有相同项，所以：
原式=(x^2+6+5x)(x^2+6+7x)+x^2
=(x^2+6)^2 + 12x(x^2+6)+36x^2
=[(x^2+6)+6x]^2
=(x^2+6x+6)^2

Answer 3

解：(x+1)(x+2)(x+3)(x+6)+x²
=(x+1)(x+6)(x+2)(x+3)+x²
=(x²+6+7x）（x²+6+5x） + x²
=（x²+6）²+12x（x²+6）+35x²+x²
=（x²+6+6x）²
=（x+3）^4

解：(x+y-2xy)(x+y-2)+(xy-1)²
=（x+y）²+2xy（x+y）-2（x+y）+4xy+（xy）²-2xy+1
=（x+y）²+2xy（x+y）+（xy）²-2[（x+y）-xy]+1
=（x+y+xy）²-2[（x+y）-xy]+1
=（x+y-xy-1）²
=（x-1）²（y-1）²

Answer 4

解：(x+y-2xy)(x+y-2)+(xy-1)²
=(x+y)²-2(x+y)-2xy(x+y)+4xy+x²y²-2xy+1
=(x+y)²-2(xy+1)(x+y)+x²y²+2xy+1
=(x+y)²-2(xy+1)(x+y)+(xy+1)²
=[x+y-(xy+1)]²
=(x+y-xy-1)²
=(x-1)²(y-1)²

(x+1)(x+2)(x+3)(x+6)+x²
=(x+1)(x+6)(x+2)(x+3)+x²
=(x²+7x+6)(x²+5x+6)+x²
=[(x²+6x+6)+x][(x²+6x+6)-x]+x²
=(x²+6x+6)²-x²+x²
=(x²+6x+6)²

Answer 5

x+y-2XY)(x+y-2)+(xy-1)^2
= [(x+y)-2xy][(x+y)-2]+(xy-1)^2
= (x+y)^2-2(xy+1)(x+y)+4xy+x^2y^2-2xy+1
= (x+y)^2-2(xy+1)(x+y)+x^2y^2+2xy+1
= (x+y)^2-2(xy+1)(x+y)+(xy+1)^2
= [(x+y)-(xy+1)]^2
= (x+y-xy-1)^2
= [x(1-y)-(1-y)]^2
= (x-1)^2 (1-y)^2
= (x-1)^2 (y-1)^2

(x+1)(x+2)(x+3)(x+6)+x^2
= [(x+1)(x+6)][(x+2)(x+3)]+x^2
= (x^2+7x+6)(x^2+5x+6)+x^2
= {(x^2+6)+7x}{(x^2+6)+5x}+x^2
= (x^2+6)^2 + 12x(x^2+6) + 35x^2+x^2
= (x^2+6)^2 + 12x(x^2+6) + 36x^2
= (x^2+6+6x)^2
= (x^2+6x+6)^2