1.x^6-y^62.（x^2+4x+8）^2+3x^3-14x^2+24x3.2x^2+3xy-2y^2-x+8y-64.已知x+y=3,x^2+y^2-xy=4求x^4+y^4+x^3y+xy^35.已知x^2+y^2+z^2-2x+4y-6z+14=0求x+y+z6.方程x^3-y^3+x^2y-xy^2=32的正整数解的个数

Question

1.x^6-y^6
2.（x^2+4x+8）^2+3x^3-14x^2+24x
3.2x^2+3xy-2y^2-x+8y-6
4.已知x+y=3,x^2+y^2-xy=4 求x^4+y^4+x^3y+xy^3
5.已知x^2+y^2+z^2-2x+4y-6z+14=0 求x+y+z
6.方程 x^3-y^3+x^2y-xy^2=32的正整数解的个数

Answer 1

1.x^6-y^6
=(x³+y³)(x³-y³)
=(x+y)(x²-xy+y²)(x-y)(x²+xy+y²)
2.题目是否出错了?
（x^2+4x+8）^2+3x^3+14x^2+24x
=(x²+4x+8)²+2x(x²+4x²+8)+x²+(x³+5x²+8x)
=(x²+4x+8+x)²+x(x²+5x+8)
=(x²+5x+8)²+x(x²+5x+8)
=(x²+6x+8)(x²+5x+8)
=(x+2)(x+4)(x²+5x+8)
3.2x^2+3xy-2y^2-x+8y-6
因为 2x²+3xy-2y²=(2x-y)(x+2y)
所以设 2x^2+3xy-2y^2-x+8y-6=(2x-y+a)(x+2y+b)
=2x²+3xy-2y²+ax+2ay+2bx-by+ab
=2x²+3xy-2y²+(a+2b)x+(2a-b)y+ab
所以 a+2b=-1,2a-b=8,ab=-6
解得 a=3,b=-2
所以 2x^2+3xy-2y^2-x+8y-6==(2x-y+3)(x+2y-2)
4.已知x+y=3,x^2+y^2-xy=4 求x^4+y^4+x^3y+xy^3
x^4+y^4+x^3y+xy^3
=(x^4-2x²y²+y^4)+(x³y+2x²y²+xy³)
=(x²-y²)²+xy(x+y)²
=(x-y)²(x+y)²+xy(x+y)²
=(x+y)²(x²-2xy+y²+xy)
=(x+y)²(x²-xy+y²)
=3²×4
=36
5.已知x^2+y^2+z^2-2x+4y-6z+14=0 求x+y+z
x^2+y^2+z^2-2x+4y-6z+14=0
(x²-2x+1)+(y²+4y+4)+(z²-6z+9)=0
(x-1)²+(y+2)²+(z-3)²=0
所以 x-1=0,y+2=0,z-3=0
所以 x=1,y=-2,z=3
所以 x+y+z=2
6.方程 x^3-y^3+x^2y-xy^2=32的正整数解的个数
(x³-y³)+(x²y-xy²)=32
(x-y)(x²+xy+y²)+xy(x-y)=32
(x-y)(x+y)²=32
因为 x,y为正整数,
所以 (x-y)(x+y)²是32的因数,且 x+y＞x-y
故只有 x+y=4,x-y=2 符合要求
此时 x=3,y=1
所以只有一对正整数解
希望不要晚了