一道数学题,谢谢关于分式的已知xyz=1,x+y+z=2,x²+y²+z²=16,求1/(xy+2z)+1/(yz+2x)+1/(zx+2y)的值.

Question

一道数学题,谢谢关于分式的
已知xyz=1,x+y+z=2,x²+y²+z²=16,求1/(xy+2z)+1/(yz+2x)+1/(zx+2y)的值.

Answer 1

因为y=2-x-z,x=2-z-y,z=2-x-y
所以原式=1/(xy+4-2x-2y)+1/(yz+4-2y-2z)+1/(zx+4-2z-2x)=1/(x-2)(y-2)+1/(y-2)(z-2)+1/(z-2)(x-2),（这里是套用公式）,然后通分,得
原式=(z-2)+(x-2)+(y-2)/(z-2)(x-2)(y-2)=(x+y+z-2-2-2)/xyz+4(x+y+z)-2(xz+xy+yz)-8
因为x+y+z=2,xyz=1
所以原式=-4/1-2(xy+xz+yz)
同时(x+y+z)^2=4=(x^2+y^2+z^2)+2(xy+xz+yz)
因为x^2+y^2+z^2=16
所以2(xy+yz+xz)=-12
所以原式=-4/13