已知xyz=1,x+y+z=2,x2+y2+z2=16?求[1÷(xy+2z)+1÷(yz+2x)+1÷(zx+2y)]的值

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已知xyz=1,x+y+z=2,x2+y2+z2=16?求[1÷(xy+2z)+1÷(yz+2x)+1÷(zx+2y)]的值
已知xyz=1,x+y+z=2,x2+y2+z2=16?求[1÷(xy+2z)+1÷(yz+2x)+1÷(zx+2y)]的值

已知xyz=1,x+y+z=2,x2+y2+z2=16?求[1÷(xy+2z)+1÷(yz+2x)+1÷(zx+2y)]的值
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