已知xy均为整数,且x

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,

xy-(x+y)=1x+y=xy-1≤[(x+y)/2]^2-1x+y≤(x+y)^2/4-1解得x+y≥2+2sqrt(2)x=y=1+sqrt(2)时,等号成立所以x+y的最小值为2+2sqrt(

x+y=8,x²-2xy+y²-4x+4y+4=0,即(x-y)²-4(x-y)+4=0,(x-y-2)²=0x-y-2=0x=5,y=3长方形的面积=5*3=

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

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

周长16所以2(X+Y)=16X+Y=8(1)即(X-Y)²-4(X-Y)+4=0(X-Y-2)=0X-Y-2=0(2)所以X=5,Y=3.所以面积=XY=15

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

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

有两种情况：X=21或29时Y为整数,Y=4,所以X+Y=25或33

∵2x+2x2−1=2x−1，∴根据题意，得x-1=±1或±2，则x=2或0或3或-1．又x≠±1，则x=0或2或3．

(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

Bxyy那么x为正数,因为负数a为任意有理数a^2等于0所以选B

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

先把它配成（X-Y)^2-2(X-Y)-8=0(X-Y+2)(X-Y-4)=0得到X-Y+2=0或者X-Y-4=0然后联立X+Y=8得到两个方程组,解出两组解,最后检验一下是否都符合题意满意的话望采纳

x+y+xy=52两边同时加1得：xy+x+y+1=53,即(x+1)(y+1)=53由于53是一个质数,则x+1=53,y+1=1,此时x=52,y=0,x-y=52或x+1=1,y+1=53,此时