实数x，y，z满足x+y+z=0，且xyz＞0，设M＝1x+1y+1z，则（）A．M＞0B．M＜0C．M=0D．M可正可负

Question

实数x，y，z满足x+y+z=0，且xyz＞0，设M＝(1)/(x)+(1)/(y)+(1)/(z)，则（　　）

A．M＞0
B．M＜0
C．M=0
D．M可正可负

Answer 1

∵xyz＞0，∴x，y，z均不为零．
M＝(1)/(x)+(1)/(y)+(1)/(z)=(yz+xz+xy)/(xyz)=(1)/(2)((x+y+z)2−(x2+y2+z2))/(xyz)
=−(1)/(2)(x2+y2+z2)/(xyz)．
由已知可得x²+y²+z²＞0，又xyz＞0，∴−(1)/(2)(x2+y2+z2)/(xyz)＜0，即M＜0．
故选：B．