已知xyz=1,x+y+z=2,x^2+y^2+z^2=16,求
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+2y)
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.
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,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
已知xyz=1,x+y+z=2,x^2+y^2+z^2=16,求
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....
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...
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=13.原式=[(x-2)+(y-2)+(z-2)]\/(x-2)(y-...
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.....
【初二数学题】已知xyz=1,x+y+z=2,x⊃2;+y⊃2;+z⊃2;=16,求1\/...
已知xyz=1,x+y+z=2,x²+y²+z²=16，求1\/(xy+2z)+1\/(yz+2x)+1\/(zx+2y)的值.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...
已知xyz=1,x+y+z=2,x^2+y^2+z^2=16,则1\/(xy+2z)+1\/(yz+2x)+1\/(zx+...
xy + xz + yz = ((x+y+z)^2 - (x^2+y^2+z^2))\/2 = -6 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.x2+y2+z2=16.求1\/xy+2z+1\/yz+2x+1\/xz+2y的值
如果是 xyz=1,x+y+z=2,x^2+y^2+z^2=16,求1\/xy+2z+1\/yz+2x+1\/xz+2y 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+...
【初二数学题】已知xyz=1,x+y+z=2,x²+y²+z²=16,求1\/(xy+2z...
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,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)(...
已知xyz=1,x+y+z=2,x2+y2+z2=16,求1\/x+y+1\/y+z+1\/x+z
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