已知X+Y+Z＝2 XY+YZ+XZ＝-5

=[(X+Z)+(X-Y)]/[X(X-Y)+Z(X-Y)]-[(X+Y)+(X+Z)]/[X(X+Y)+Z(X+Y)]=[(X+Z)+(X-Y)]/[(X+Z)(X-Y)]-[(X+Y)+(X+Z)

(x+y+z)²=1,x²+2xy+y²+2(x+y)z+z²=1,x²+y²+z²+2(x+y)z+2xy=1xy+yz+xz=

X^2+Y^2+Z^2=XY+YZ+XZ则有2X^2+2Y^2+2Z^2-2XY-2YZ-2XZ=0==>(X-Y)^2+(Y-Z)^2+(Z-X)^2=0必然X-Y=0,Y-Z=0,Z-X=0==>

x^2+y^2+z^2-xy-yz-xz=0(1/2)*2(x^2+y^2+z^2-xy-yz-xz)=0(1/2)*(x^2+y^2-2xy+z^2+y^2-2zy+x^2+z^2-2xz)=0(x

x-y=5,z-y=10相减z-x=5x²+y²+z²-xy-yz-xz=(2x²+2y²+2z²-2xy-2yz-2xz)/2=[(x&s

y=-12;一共是三个方程,因为xy/(x+y)=3推出（x+y）/(xy)=1/3-------方程1；同理：（y+z）/(yz)=1/2-------方程2；（x+z）/(xz)=1-------

(x+y+z)²=1²x²+y²+z²+2xy+2yz+2xz=1x²+y²+z²+2(xy+yz+xz)=1x&sup

如果可以用排序不等式证明的话x^2+y^2+z^2>=x^1.5y^0.5+y^1.5z^0.5+z^1.5x^0.5=2xxy/2(xy)^0.5+2yyz/2(yz)^0.5+2zzx/2(zx)

(x+y+z)^2=4x^2+y^2+z^2+2xy+2xz+2yz=4x^2+y^2+z^2+2(-5)=4x^2+y^2+z^2=14

x＋y＋z＝5,xy＋yz＋zx＝9所以(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+xz)=25所以x^2+y^2+z^2=25-2×9=25-18=7

由已知可以得出xy=x+y(1)yz=2(y+z)(2)zx=3(z+x)(3)由(3)得z=3x/(x-3)(4)由(1)得y=x/(x-1)(5)把(4)(5)代入(2)解得x=12/5

（x+y+z）^2=x^2+y^2+z^2+2(xy+yz+xz)=25x^2+y^2+z^2=25-14=11

x^2+y^2+z^2=xy+yz+xz2(x^2+y^2+z^2)=2(xy+yz+xz)x^2+y^2-2xy+x^2+z^2-2xz+y^2+z^2-2yz=0(x-y)^2+(y-z)^2+(

x^2+y^2+z^2=xy+yz+xz2(x^2+y^2+z^2)=2(xy+yz+xz)x^2+y^2-2xy+x^2+z^2-2xz+y^2+z^2-2yz=0(x-y)^2+(y-z)^2+(

答案是：(2*X)/((X-Z)*(X+Z))再问：解题过程给我写下1再答：=（2X+Z-Y)/[(x-y)(x+z)]-(y-z)/[(x-z)(x-y)]=[(2x+z-y)(x-z)-(y-z)

令2/x=3/y=7/z=k∴x=2/ky=3/kz=7/k∴(xy+xz+yz)/(x^2+y^2+z^2)=(2/k*3/k+2/k*7/k+3/k*7/k)/(4/k²+9/k

解题思路：本题的关键是将三个方程两边取倒数，化简后分别将方程等号左边和右边相加，得到1/x+1/y+1/z的值，最后将要求的分式化简，把1/x+1/y+1/z的值带入即可。解题过程：

(x+y+z)²=x²+y²+z²+2xy+2yz+2xz所以可得：xy+yz+xz=[(x+y+z)²-(x²+y²+z

xy:yz:zx=3:2:1xy:yz=3:2则x:z=3:2同理y:z=3:1=6:2故(x+y):z=(3+6):2=9:2

①x:y:z因为xy：yz：zx=3：2：1所以xy：yz=3：2所以x：z=3：2同理yz：zx=2：1所以y：x=2：1＝6：3所以x：y：z=3：6：2②x/yz:y/zx＝x^2：y^2＝(x