已知y z分之x x z分之y x y分之z=1
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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)
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
y=3x/5原式=x/(x+3x/5)+(3x/5)/[x-3x/5]-(9x^3/25)/(x^3-9x^3/25)=8/3-3/2-9/16=29/48
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
设x/2=y/3=z/4=k∴x=2k,y=3k,z=4k∴(xy+yz+zx)/(x²+y²+z²)=(6k²+12k²+8k²)/(4k
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
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)]
【x+y】分之xy=-2,xy分之【x+y】=-1/21/x+1/y=-1/2(1)【y+z】分之yz=3分之4,yz分之【y+z】=3/41/y+1/z=3/4(2)【z+x】分之zx=-3分之4,
z²/x²再问:怎么得的再答:原式=x/y²*yz/x²*yz/x=z/xy*yz/x=z²/x²
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
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