Question

几道因式分解，急！！！！！！！！！！1

（1）x^2*y^2+xy-x^2-y^2+x+y+2
(2)x^5-x^4+x^3-x^2+x-1
(3)4*x^4*y^2-%x^2y^2-9*y^2
(4)a^3-3a+2
能做一题是一题，各位好心人帮帮忙！！！

Answer 1

（1）x^2*y^2+xy-x^2-y^2+x+y+2
解：原式=(x²y²-x²-y²+1)+(xy+x+y+1)
=[x²(y²-1)-(y²-1)]+[x(y+1)+(y+1)]
=(x²-1)(y²-1)+(x+1)(y+1)
=(x+1)(x-1)(y+1)(y-1)+(x+1)(y+1)
=(x+1)(y+1)[(x-1)(y-1)+1]
=(x+1)(y+1)(xy-x-y+2)
(2)x^5-x^4+x^3-x^2+x-1
解：原式=x^4(x-1)+x^2(x-1)+(x-1)
=(x-1)(x^4+x^2+1)
(3)4*x^4*y^2-%x^2y^2-9*y^2
解：原式=y^2(4x^4-5x^2-9)
=y^2(4x^2-9)(x^2+1)
=y^2(2x-3)(2x+3)(x^2+1)(4)a^3-3a+2
解：原式=a^3-3a+2
=a^3-1-(3a-3)
=(a-1)(a^2+a+1)-3(a-1)
=(a-1)(a^2+a-2)
=(a-1)^2×(a+2)
不懂，请追问，祝愉快

Answer 2

（1）x^2*y^2+xy-x^2-y^2+x+y+2
=(x²y²-x²-y²+1)+(xy+x+y+1)
=[x²(y²-1)-(y²-1)]+[x(y+1)+(y+1)]
=(x²-1)(y²-1)+(x+1)(y+1)
=(x+1)(x-1)(y+1)(y-1)+(x+1)(y+1)
=(x+1)(y+1)[(x-1)(y-1)+1]
=(x+1)(y+1)(xy-x-y+2)
(2)x^5-x^4+x^3-x^2+x-1
=x^3(x^2-x+1)-(x^2-x+1)
=(x^3-1)(x^2-x+1)
=(x-1)(x^2+x+1)(x^2-x+1)

(3)4*x^4*y^2-%x^2y^2-9*y^2
题目有误， %？？？

(4)a^3-3a+2
=a^3+a^2-a^2-3a+2
=a^2(a-1)+(a-2)(a-1)
=(a^2+a-2)(a-1)
=(a+2)(a-1)(a-1）
=(a+2)(a-1)^2

追问

好地，还有一题要请教一下
求方程xy+2x+y=o的所有整数解，
好像是要通过拆项补项之类的方法将原式因式分解，再写出两个因数的所有可能性。

追答

xy+2x+y=o
xy+2x+y+2=2
y(x+1)+2(x+1)=2
（x+1)(y+2)=2
x，y是整数
x+1，y+2是整数
x+1=1 y+2=2
x=0 y=0
或x+1=2 y+2=1
x=1 y=-1
或x+1=-1 y+2=-2
x=-2 y=-4
或x+1=-2 y+2=-1
x=-3 y=-3