已知 ,求x y-z的值.

(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=

1=xy/(x+y)两边倒数1/x+1/y=1同理1/y+1/z=1/21/z+1/x=1/3联合三个方程得1/x=5/121/y=7/121/z=-1/12即x=12/5y=12/7z=-12x+y

解|x-1|≥0(y-2)的平方≥0√z+3≥0∴x-1=0,y-2=0,z+3=0∴x=1,y=2,z=-3∴xy+√z的平方=1×2+√9=2+3=5再问：太给力了，你的回答完美解决了我的问题！

答：x+y+z=3y=2zy≠0,则z≠0所以：y=2z/3x+2z/3+z=2zx=z/3令z=3k,y=2k,x=k(xy+yz+zx)/(x²+y²+z²)=(2k

把x=6-y带入z^2-4z+4=xy-9中,得(y-3)^2+(z-2)^2=0,故y-3=0,z-2=0,所以y=3,z=2,x=3.

xyz=1所以z=1/xyxz=1/yyz=1/xx/(xy+x+1)+y/(yz+y+1)+z/(xz+z+1)=x/(xy+x+1)+y/(1/x+y+1)+(1/xy)/(1/y+1/xy+1)

由2x-3y-z=0,x+3y-14z=0,且x,y,z不全为0解得x=5zy=3z将x=5zy=3z带入4x平方-5XY+Z的平方/xy+yz+zx得4*25z平方-5*5Z*3z+Z的平方/5z*

处理这类比例问题,有一个通用方法如果：x:y:z=a:b:c可以设x=aky=bkz=ck带入计算,就行了自己来试试吧~

z²-4z+4=xy-9又x=6-y,代入得z²-4z+4=（6-y）y-9(z-2)²=-(y-3)²(z-2)²+(y-3)²=0所以(

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

（X+Y+Z)²=X²+Y²+Z²+2(XY+YZ+XZ)X²+Y²+Z²=10²-2×8=84

(x+y+z)²=10²x²+y²+z²+2xy+2yz+2xz=100所以x²+y²+z²=100-2(xy+xz+y

(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

1.z=3y/2把:z=3y/2代入x+y+z=3y得:x+y+3y/2=3y整理后得:x=y/2所以:x/(x+y+z)=(y/2)/(y/2+y+3y/2)=1/62.因为1/x-1/y=3,则1

令X=3k,由于x:y:z=3:4:6则:y=4k,z=6k将x=3k,y=4k,z=6k代入（xy+yz+xz）/（x^+y^+z^）则有：（xy+yz+xz）/（x^+y^+z^）=（12k^+2

(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

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

2x+z=6,z-2y=8相减2x+z-z+2y=6-82x+2y=-2x+y=-1x^2+y^2+2xy=(x+y)^2=(-1)^2=1