No2. 为通分准备 4=(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)=16+2(xy+yz+zx), 则有 xy+yz+zx=-6 (x-2)(y-2)(z-2)=xyz-2(xy+yz+zx)+4(x+y+z)-8=13
No3. 通分，则有原式=[(x-2)+(y-2)+(z-2)] / [(x-2)(y-2)(z-2)] =-4/13
yz+2x=(y-2)(z-2),zx+2y=(z-2)(x-2).
4=(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)=16+2(xy+yz+zx),
xy+yz+zx=-6.
(x-2)(y-2)(z-2)=xyz-2(xy+yz+zx)+4(x+y+z)-8=13.
原式=[(x-2)+(y-2)+(z-2)]/(x-2)(y-2)(z-2)
=-4/13.本回答被网友采纳
已知XYZ=1,X+Y+Z=2,X的平方+Y的平方+Z的平方=16.求(XY+2Z)分之1+(Y...
No1. 将表达式分母化简，xy+2z=xy+4-2x-2y=(x-2)(y-2), 同理yz+2x=(y-2)(z-2) , zx+2y=(z-2)(x-2).No2. 为通分准备 4=(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)=16+2(xy+yz+zx), 则有 xy+yz+zx=-6 (x-2)(y-2)(z-2)=xyz-2(xy+y...
【初二数学题】已知xyz=1,x+y+z=2,x⊃2;+y⊃2;+z⊃2;=16,求1\/...
解：x≠0,y≠0,z≠0 xy=1\/z, yz=1\/x, zx=1\/y 1\/(xy+2z)+1\/(yz+2x)+1\/(zx+2y)=1\/(1\/z+2z)+1\/(1\/x+2x)+1\/(1\/y+2y)=1\/(（1+2z2）\/z)+1\/(（1+2x2）\/x)+1\/(（1+2y2）\/y)=z\/（1+2z2）+ x\/（1+2x2）+y\/（1+2y2）(z（1+2x2）（1+2y2...
...已知xyz=1,x+y+z=2,x²+y²+z²=16,求1\/(xy+2z)+1\/(yz+2x...
z=2-x-y，所以1\/(xy+2z)=1\/(xy+4-2x-2y)=1\/(x-2)(y-2)设r=x-2,s=y-2,t=z-2 题目变为求1\/rs+1\/st+1\/rt=(r+s+t)\/rst 而由已知可得(r+2)(s+2)(t+2)=1，r+s+t=-4，(r+2)^2+(s+2)^2+(t+2)^2=16 解得(r+s+t)\/rst=-4\/13 ...
已知xyz=1,x+y+z=2,x^2+y^2+z^2=16,则1\/(xy+2z)+1\/(yz+2x)+1\/(zx+...
1\/（xy+2z)=1\/[xy+2(1-1\/xy)]=x^2y^2+2\/xy 通例1\/(yz+2x)=y^2z^2+2\/yz 1\/(zx+2y)=x^2z^2+2\/xz 通分即可，答案是-2
已知xyz=1,x+y+z=2,x^2+y^2+z^2=16,则1\/(xy+2z)+1\/(yz+2x)+1\/(zx+...
x^2+y^2+z^2=16 2xy+2xz+2yz=4-16=-12 1\/(xy+2x)=1\/(xy+4-2x-2y)=1\/(x-2)\/(y-2)所以原式=1\/(x-2)\/(y-2)+1\/(y-2)\/(z-2)+1\/(z-2)\/(x-2)=(x+y+z+6)\/(xyz-2xy-2yz-xz+4x+4y+4z-8)=-4\/13 新人,请多多关照,不知对不对呢.....
已知xyz=1,x+y+z=2,x^2+y^2+z^2=16。求代数式1\\(xy+2z)+1\\(yz+2z...
楼主是不是打错一个字母，第二项数2z应该是2x吧？xy+2z=xy+4-2x-2y=(x-2)(y-2).yz+2x=(y-2)(z-2),zx+2y=(z-2)(x-2).4=(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)=16+2(xy+yz+zx),xy+yz+zx=-6.(x-2)(y-2)(z-2)=xyz-2(xy+yz+zx)+4(x+y+z)-8=...
已知xyz=1,x+y+z=2,x^2+y^2+z^2=16,求代数式1\/xy+2z+1\/yz+2x+1\/zx+...
因为xyz=1所以 X、Y、Z均不为0 所以 1\/xy=Z 1\/yz=x 1\/zx=y 所以原式=z+2z+x+2x+y+2y=3(x+y+z)=3x1=3应该是这样吧
数学题:已知xyz=1,x+y+z=2,x方+y方+z方=16,求xy+1\/2z+yz+1\/2x+zx+1...
x+y+z=2 那么(x+y+z)方=4=x方+y方+z方+2xy+2xz+2yz,xy+1\/2z+yz+1\/2x+xz+1\/2y=xy+yz+zx+1\/2(x+y+z),4=16+2(xy+xz+yz) 综上所述答案为-5
已知xyz=1.x2+y2+z2=16.求1\/xy+2z+1\/yz+2x+1\/xz+2y的值
应该是 原式 = （ 1\/xy + 2z ) + （ 1\/yz + 2x ) + （1\/xz + 2y )通分 = （z+2xyzz)\/xyz + (x+2xxyz)\/xyz + (y+2xyyz)\/xyz 化简 = ( x+y+z+ 2xyz(x+y+z) )\/xyz =6 如果是xyz=1,x+y+z=2,x^2+y^2+z^2=...
已知xyz=1,x+y+z=2,x2+y2+z2=16.则1xy+2z+1yz+2x+1zx+2y=-413-413
∵x+y+z=2，两边平方得，x2+y2+z2+2xy+2yz+2xz=4，∵x2+y2+z2=16，∴xy+yz+xz=-6，又∵z=2-x-y，∴1xy+2z=1xy+4?2x?2y=1(x?2)(y?2)，同理得，1yz+2x=1(y?2)(z?2)，1zx+2y=1(z?2)(x?2)，∴1xy+2z+1yz+2x+1zx+2y，=1(x?2)(y?2)+1(y?2)...
