◆〃两道数学题目(整式指数幂)急@!!!!!!~~~~~~

1·由x^2-3y^2=2xy知
（x+y）(x-3y)=0
x=-y或x=3y
x-y分之x+2y=-1/2或5/2
2·由x^2+y^2+z^2-4x-6y-12z<-48，x,y,z为整数
(x-2)^2+(y-3)^2+(z-6)^2<1
x=2,y=3,z=6
(x^-1+y^-1+z^-1)^xyz=1

1.x^2-3y^2=2xy(x>0,y>0)
两边除于x^2得
3（y/x)^2+2(y/x)-1=0
[3(y/x)-1)[(y/x)+1]=0
因为x>0,y>0
3(y/x)-1=0
y/x=1/3
(x+2y)/(x-y)
因为x>0,分子，分母同除x得
[1+2(y/x)]/[1-(y/x)]
=(1+(1/3)*2)/(1-(1/3))
=5/2
2.(1)x^2+y^2+z^2-4x-6y-12z<-48
x^2-4x+4+y^2-6y+9+z^2-12z+36<1
(x-2)^2+(y-3)^2+(z-6)^2<1
因为(x-2)^2+(y-3)^2+(z-6)^2>=0
且x,y,z为整数
所以x=2 y=3 z=6
(2)x^-1+y^-1+z^-1)^xyz
=yz+xz+xy
=3*6+2*6+2*3
=36

1. x^2-3y^2=2xy
x/y-3y/x=2 (两边同除xy ) ，设t=x/y则
t-3/t=2,可解得t

（x-y)/(x+2y)=(t-1/(t+2)==

2.
x^2+y^2+z^2-4x-6y-12z=(x-2)^2+(y-3)^2+(z-6)^2-49<-48
即(x-2)^2+(y-3)^2+(z-6