已知y z分之x=a

令a分之x=b分之y=c分之z=kx=ak,y=bk,z=ck代入就可以了

x^2+y^2+z^2-xy-yz-xz=(1/2)[(x-y)^2+(z-y)^2+(z-x)^2]x-y=1/2+根号3z-y=1/2-根号3所以y=x-(1/2+根号3)=z-(1/2-根号3)

4x-3y-3z=0(1)x-3y-z=0(2)(1)-(2)3x-2z=0z=(3/2)x(1)-(2)×34x-3y-3x+9y=0x-6y=0y=(1/6)x所以xz=x*(3/2)x=(3/2

设x/2=y/3=z/4=a则:x=2a;y=3a;z=4a代入得：(xy+yz+zx)/(x^2+y^2+z^2)=(6a^2+12a^2+8a^2)/(4a^2+9a^2+16a^2)=26a^2

设2分之x=3分之y=4分之z=k∴x=2k,y=3k,z=4kx2+y2+z2分之xy+yz+zx=(6k²+12k²+8k²)/(4k²+9k²+

∵1/x+1/y=1/6,1/y+1/z=1/9,1/z+1/x=1/15∴（1/x+1/y）+（1/y+1/z）+（1/z+1/x）=1/6+1/9+1/152（1/x+1/y+1/z）=15/90

xy/(x+y)=51/x+1/y=1/5yz/(y+z)=7/21/y+1/z=2/7zx/(z+x)=41/x+1/z=1/4(xy+yz+zx)分之xyz=1/(1/x+1/y+1/z)=280

令x/3=y/4=z/5=kx=3ky=4kz=5k原式=(3k)^2+(4k)^2+(5k)^2/3k*4k+4k*5k+3k*5k=50k^2/47k^2=50/47

∵x/3=y/4=z/5∴x:y:z=3:4:5设x=3a则y=4az=5a代入式子：x2+y2+z2/xy+yz+zx=(3a)²+(4a)²+(5a)²/(3a)(4

设x/2=y/3=z/4=k则x=2k,y=3k,z=4k（x²+y²+z²）/（xy+yz+zx）=（4k²+9k²+16k²）/（6k&

令3分之x=4分之y=6分之z=kx=3k,y=4kz=6k(xy+yz+xz)/(x^2+y^2+z^2)=k^2(12+18+24)/k^2(36+16+9)=54/61

x+y+z=a(x+y+z)^2=a^2(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)x^2+y^2+z^2=(x+y+z)^2-2(xy+yz+zx)=a^2-2

由已知得(x+y)/(xy)=1(y+z)/(yz)=1/2(z+x)/(zx)=1/3变形：1/x+1/y=1(1)1/y+1/z=1/2(2)1/z+1/x=1/3(3)[(1)+(2)+(3)]

3x-4y＝z,2x+y＝8z,解得：x=3z,y=2zxy+yz分之x二次方+y二次方-z二次方=(x^2+y^2-z^2)/(xy+yz)=(9z^2+4z^2-z^2)/(6z^2+2z^2)=

答案11分之14看图

答案是11分之14. 求采纳

x分之3=y分之1x=3yy分之1=z分之2z=2yxy+yz+zx分之2x²-2y²+5z²=[2(3y)²-2y²+5(2y)²]/(3

令3/x=4/y=6/z=1/k则x=3ky=4kz=6k(x²+y²+z²)/(xy+yz+xz)=(9k²+16k²+36k²)/(12

(X+Y+Z)^2=x^2+y^2+z^2+2(xy+yz+xz)=a^2=x^2+y^2+z^2+2b所以x^2+y^2+z^2=a^2-2

1/x+1/y=3(1)1/y+1/z=2(2)1/x+1/z=1(3)(1)+(2)+(3)2(1/x+1/y+1/z)=61/x+1/y+1/z=3(4)由（1）1/z=0题目有误,请核对,或者更