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...