Question

(x+y-2xy)(x+y-2)+(1-xy)² 因式分解怎么做？悬赏20

还有x3+2x²-5x-6
2x²-5xy+2y²+7x-5y+3

当k为何值时，x²+xy+ky²-2x+11y-15能分解成两个一次因式的乘积?
若多项式2x²-3x3+ax²+7x+b能被多项式x²+x-2整除，求a.b的值?
整数a.b满足6ab=9a-10b+303，则a+b=?
方程6xy+4x-9y-7=0的整数解为______
求证：8x²-2xy-3y²可以化成两个整系数多项式的平方差？

Answer 1

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

x^3+2x^2-5x-6
=x^3+3x^2-x^2-5x-6
=x^2(x+3)-(x+2)(x+3)
=(x+3)(x^2-x-2)
=(x+3)(x-2)(x+1)

2x^2-5xy+2y^2+7x-5y+3
=(x-2y+3)(2x-y+1)

设：
x²+xy+ky²-2x+11y-15
=(x+ay+b)(x+cy+d)
=x^2+cxy+dx+axy+acy^2+ady+bx+bcy+bd
=x^2+(c+a)xy+acy^2+(d+b)x+(ad+bc)y+bd
与上对比，得：
c+a=1
ac=k
d+b=-2
ad+bc=11
bd=-15
解得：
b=-5 d=3或b=3,d=-5
a=(11-b)/(d-b)
a=2,a=-1
a=2时，c=-1 k=-2
a=-1时，c=2 k=-2
因此，k=-2

(6x-9)y=7-4x
y=(7-4x)/(6x-9)
=(1+6-4x)/(6x-9)
=1/(6x-9)-2/3
1/(6x-9)=某整数n+2/3
1/(6x-9)=n+2/3=(3n+2)/3
1/(2x-3)=3n+2

x=(9n+6+1)/(6n+4)
=3/2+1/(6n+4)
1/(6n+4)=某整数m+1/2
n=-(4m+3)/(6m+3)=-2/3-1/(6m+3)
要是整数，只能m=0 ,此时，n=-1
1/(6x-9)=2/3-1=-1/3 x=1
y=-1
方程的整数解为x=1,y=-1

Answer 2

（换元法）
设x+y=a xy=b
(x+y-2xy)(x+y-2)+(1-xy)²
=(a-2b)(a-2)+(1-b)²
=a²+4b-2ab-2a+1-2b+b²
=a²+b²-2ab+2b-2a+1
=(a-b)²-2(a-b)+1
=(a-b-1)²
=(x+y-xy-1)²

x³+2x²+2x²-5x-6
=x³+3x²-x²-5x-6
=x³(x+3)-(x+2)(x+3)
=(x+3)(x²-x-2)
=(x+3)(x-2)(x+1)

2x²-5xy+2y²+7x-5y+3
=(x-2y+3)(2x-y+1)

设：
x²+xy+ky²-2x+11y-15
=(x+ay+b)(x+cy+d)
=x^2+cxy+dx+axy+acy^2+ady+bx+bcy+bd
=x^2+(c+a)xy+acy^2+(d+b)x+(ad+bc)y+bd
与上对比，得：
c+a=1
ac=k
d+b=-2
ad+bc=11
bd=-15
解得：
b=-5 d=3或b=3,d=-5
a=(11-b)/(d-b)
a=2,a=-1
a=2时，c=-1 k=-2
a=-1时，c=2 k=-2
因此，k=-2

(6x-9)y=7-4x
y=(7-4x)/(6x-9)
=(1+6-4x)/(6x-9)
=1/(6x-9)-2/3
1/(6x-9)=某整数n+2/3
1/(6x-9)=n+2/3=(3n+2)/3
1/(2x-3)=3n+2

x=(9n+6+1)/(6n+4)
=3/2+1/(6n+4)
1/(6n+4)=某整数m+1/2
n=-(4m+3)/(6m+3)=-2/3-1/(6m+3)
要是整数，只能m=0 ,此时，n=-1
1/(6x-9)=2/3-1=-1/3 x=1
y=-1
方程的整数解为x=1,y=-1 -5x-6

Answer 3

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

x^3+2x^2-5x-6
=x^3+3x^2-x^2-5x-6
=x^2(x+3)-(x+2)(x+3)
=(x+3)(x^2-x-2)
=(x+3)(x-2)(x+1)

2x^2-5xy+2y^2+7x-5y+3
=(x-2y+3)(2x-y+1)

设：
x²+xy+ky²-2x+11y-15
=(x+ay+b)(x+cy+d)
=x^2+cxy+dx+axy+acy^2+ady+bx+bcy+bd
=x^2+(c+a)xy+acy^2+(d+b)x+(ad+bc)y+bd
与上对比，得：
c+a=1
ac=k
d+b=-2
ad+bc=11
bd=-15
解得：
b=-5 d=3或b=3,d=-5
a=(11-b)/(d-b)
a=2,a=-1
a=2时，c=-1 k=-2
a=-1时，c=2 k=-2
因此，k=-2

(6x-9)y=7-4x
y=(7-4x)/(6x-9)
=(1+6-4x)/(6x-9)
=1/(6x-9)-2/3
1/(6x-9)=某整数n+2/3
1/(6x-9)=n+2/3=(3n+2)/3
1/(2x-3)=3n+2

x=(9n+6+1)/(6n+4)
=3/2+1/(6n+4)
1/(6n+4)=某整数m+1/2
n=-(4m+3)/(6m+3)=-2/3-1/(6m+3)
要是整数，只能m=0 ,此时，n=-1
1/(6x-9)=2/3-1=-1/3 x=1
y=-1
方程的整数解为x=1,y=-1

Answer 4

（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