已知xy均为整数xy x y=17

XY+(x+y)=17,xy²+x²y=xy(x+y)=66可知xy,x+y是方程a²-17a+66=0的两根（a-11）（a-6)=0a1=11;a2=6即xy=11,

x=2+m/3;y=2m/3-2∵xy小于0∴（6+m）×（2m-6）＜0-6＜m＜3又∵x、y都为整数∴m为3的整数倍,所以m为-3,0

令F（XY）=1/XY+XY,当XY=1的时候,F（XY)=2,最小.（可由函数图形象得出）.XY趋于正无穷大的时候F(XY)趋于正无穷大,XY无限趋于零的时候F(XY)趋于正无穷大.所以XY越接近1

因为4x+y=m3x-y=5所以7x=m+5/7x=m+5/7化为同系数项12x+3y=3m12x-4y=20所以7y=3m-30y=3m-30/7因x>0y03m-20

xy=6/(x+y)xy(x+y)=6,已知x和y是整数,则xy和(x+y)也是整数,而6=2*3=1*6当xy=3时,x+y=2,无整数解当xy=1时,x+y=6,无整数解当xy=6时,x+y=1,

将xyz+xy+yz+xz+x+y+z-2变形xyz+xy+y+xz+z+yz+x-2=xyz+y(x+1)+z(x+1)+yz+x-2=xyz+(y+z)(x+1)+yz+x-2=yz(x+1)+(

x^-2xy+y^-x+y-2=0(x-y)^2-(x-y)-2=0(x-y+1)(x-y-2)=0x-y=-1x-y=2又周长=2（x+y）=16x+y=8联立方程组得x-y=-1x+y=8解x=3

7!再问：为什么再答：再问：thanks再答：it'sdoesn't.mater再答：netter再答：metter

（1）∵1/x+1/y=3∴(y+x)/xy=3∴x+y=3xy(3X-4XY+3y)/(X+2XY+Y)=[3(x+y)-4xy]/[(x+y)+2xy]=(9xy-4xy)/(3xy+2xy)=5

分解因式(x-y)(x+2y)=7因为7是质数,因此只有两种情况：x-y=1;x+2y=7以及x-y=7;x+2y=1解此方程组得x=3y=2或x=5y=-2

由已知：xy+x+y=17,xy(x+y)=66,可知xy和x+y是方程t2-17t+66=0的两个实数根,得：t1=6,t2=11.即xy=6,x+y=11,或xy=11,x+y=6.x2+y2=（

x,y,z都是整数而2010里有个质数67还剩因子2,3,5试一下,是5,6,67使xy+yz+zx最小为767上面是在正整数范围内考虑的,如果要把负整数也算上,那就-1,-1,2010,那个式子最小

x平方+2xy-3y平方=17(x+3y)(x-y)=17=1*17因x、y是整数所以：x+3y=17x-y=1y=4x=5或者x+3y=1x-y=17y=-4x=13

有两种情况：X=21或29时Y为整数,Y=4,所以X+Y=25或33

(x-y)^2-(x-y)=2===>(x-y-1)(x-y)=2将x=8-y带入(8-2y-1)(8-2y)=2==>(8-2y-1)(4-y)=1由于Y为整数,(8-2y-1)和(4-y)也都为整

3xy=2x+3y+5(3y-2)(x-1)=7所以3y-2=7x-1=1得y=3x=2xy=6或3y-2=1x-1=7得y=1x=6xy=6所以xy=6

xy-3x-5y+15=92(x-5)(y-3)=92=1×92=2×46=4×23所有整数对（x,y)的组数为12

x-y-x^2+2xy-y^2+2=0(x-y)-(x-y)^2+2=0(x-y)^2-(x-y)-2=0(x-y-2)(x-y+1)=0x=2+yx=y-12x+2y=204+2y+2y=20y=4

能问具体点吗?是求此函数有最大值或有最小值还是求XY的最大值或者最小值再问：我就是想说是最大值还是最小值啦。今天考完对答案。同学写的是最小值。我写的是最大值。再答：最小值，因为是非负整数再问：我汗。。