(1)设p/(x^2-yz)=q/(y^2-zx)=r/(z^2-xy)=k,
所以p=k(x^2-yz),q=k(y^2-zx),r=k(z^2-xy),
因为p+q+r=9,所以k(x^2+y^2+z^2-xy-yz-zx)=9,
而(px+qy+rz)/(x+y+z)
=[kx(x^2-yz)+ky(y^2-zx)+kz(z^2-xy)]/(x+y+z)
=k(x^3+y^3+z^3-3xyz)/(x+y+z),
注意到(x^3+y^3+z^3-3xyz)/(x+y+z)=x^2+y^2+z^2-xy-yz-zx,
所以(px+qy+rz)/(x+y+z)=k(x^2+y^2+z^2-xy-yz-zx)=9,
所以(px+qy+rz)/(x+y+z)=9；
(2)因为x+4y=1,
所以x=1-4y,
代入第一个式子,得2(1-4y)+my=4,
所以(m-8)y=2,
因为2=1*2=(-1)*(-2),
所以m-8=1,y=2,
或m-8=2,y=1,
或m-8=-1,y=-2,
或m-8=-2,y=-1,
所以m=9或m=10或m=7或m=6,
所以当m取6,7,9,10时,x和y都是整数；
(3)因为1/(1+y+yz+yzt)
=1/(xyzt+y+yz+yzt)
=1/[y(1+z+zt+ztx)]
=1/[y(xyzt+z+zt+ztx)]
=1/[yz(1+t+tx+txy)]
=1/[yz(xyzt+t+tx+txy)]
=1/[yzt(1+x+xy+xyz)],
同理,1/(1+z+zt+ztx)=1/[zt(1+x+xy+xyz)],
1/(1+t+tx+txy)=1/[t(1+x+xy+xyz),
所以1/(1+x+xy+xyz)+1/(1+y+yz+yzt)+1/(1+z+zt+ztx)+1/(1+t+tx+txy)
=yzt/[yzt(1+x+xy+xyz)]+1/[yzt(1+x+xy+xyz)]+y/[yzt(1+x+xy+xyz)+
yz/[yzt(1+x+xy+xyz)]
=(1+y+yz+yzt)/[yzt(1+x+xy+xyz)]
=(1+y+yz+yzt)/(yzt+xyzt+xyzt*y+xyzt*yz)
=(1+y+yz+yzt)/(yzt+1+y+yz)
=(1+y+yz+yzt)/(1+y+yz+yzt)
=1.
所以1/(1+x+xy+xyz)+1/(1+y+yz+yzt)+1/(1+z+zt+ztx)+1/(1+t+tx+txy)=1.
