2007x-2008y=2009 2006x-2007y=2008

方程移项配方得:(x-1)^2+(y+1)^2=0,两非负数之和是零,则这两数均是零,得x=1,y=-1,所以x^2007+y^2008=1^2007+(-1)^2008=2

再问：那么，它到底是不是完全平方数再答：分类就是了，并不是所情况都是的

（x-y）[(x+y)(x+y)-xy]+(x+y)[(x-y)(x-y)+xy]=(x-y)(x+y)^2-（x-y)xy+(x+y)(x-y)^2+(x+y)xy=(x-y)(x+y)(x+y+x

1

①-②得：2x+2y=2即：x+y=1③①-③×2007得x=-1代入③可得：-1+y=1解得,y=2所以,方程组的解为：x=-1y=2二十年教学经验,专业值得信赖!如果你认可我的回答,敬请及时采纳,

[2x(x^2y-xy^2)+xy(xy-x^2)]÷(x^2y)=[2x*xy（x-y）+xy*x（y-x）]÷(x^2y)=[2x^2*y(x-y)-x^2*y(x-y)]÷(x^2y)=x^2*

x=2,y=1两式相加得x-y=1两式相减得X+Y=3再问：过程再答：相加后X,Y前面的系数相等，2008+2007=2007+2008=2009+2006，所以得X-Y=1A两式相减得X+Y=3B再

（x^4-y^4）/（x²-2xy-y²）÷（x²+y²）/（y-x）=（x²-y²）（x²+y²）/（x-y）^2÷（

拿2式减1式,得x-y=3,y=x-3将y=x-3带回1式,得y=-1,x=y+3=2

2006X+2007Y=2005.①2007X+2006Y=2008.②②-①,得:x-y=3.③③×2007,得:2007x-2007y=6021.④①+④,得:4013x=8026x=2代入③,得

两式相加得x+y=1两式相减得x-y=3

解1:2007X+2008Y=20062:2008X+2007Y=2009把1式-2式=Y-X=3把2式+1式=4015X+4015Y=4015=X+Y=1所以X=-1,Y=2所以(X-Y)^4=81

2006（X-Y)+2007(Y-Z)-2008(X-Z)=0……①2006*2006(X-Y)+2007*2007(Y-Z)-2008*2008(X-Z)=2007……②由①式可得-2x+y+z=0

2006（x-y)+2007(y-z)+2008(z-x)=01)2006^2(x-y)+2007^2(y-z)+2008^2(z-x)=20092)化简1）得-2x+y+z=03)2）-1）*200

(x^4-y^4)/(x^2-2xy+y^2)÷(x^2+y^2)/(y-x)=(x^2-y^2)(x^2+y^2)/(x^2-2xy+y^2)*(y-x)/(x^2+y^2)=(x^2-y^2)/(

1.2*(2007x-2008y)=4014x-4016y=4018(4014x-4016y)+3(x+y)+4018=4017x-4013y+4018=4018+3+4018=80392.3*(x-

方程y^2-2xy+2008=0y=(2x±sqrt(4x^2-4*2008))/2=x±sqrt(x^2-2008),解出的是y方程x^2-2xy+2008=0x=(2y±sqrt(4y^2-4*2

(x+y-z)/z=(y+z-x)/x=(z+x-y)/y[x+y]/z-1=[y+z]/x-1=[z+x]/y-1[x+y]/z=[y+z]/x=[z+x]/y设[x+y]/z=[y+z]/x=[z

(x²+y²)(x²+y²-2008)-2009=0(x²+y²)²-2008(x²+y²)-2009=0(x

原式=（x+y）²/x(5x-4y)x(5x-4y)/(x+y)+(x²-y）/x=(x+y)/x+(x²-y）/x=(x+x²）/x=1+x把x=2007,y