解题思路:由1x+1y+z=12,1y+1z+x=13,1z+1x+y=14,易得1x=y+z2(x+y+z),1y=z+x3(x+y+z),1z=x+y4(x+y+z),然后代入即可求得答案.
∵[1/x+
1
y+z=
1
2],
∴[x+y+z
x(y+z)=
1/2],
∴x(y+z)=2(x+y+z),
∴x=
2(x+y+z)
y+z,
即:[1/x]=[y+z
2(x+y+z),
同理:
1/y]=[z+x
3(x+y+z),
1/z=
x+y
4(x+y+z)],
∴[2/x+
3
y+
4
z]=
2(y+z)
2(x+y+z)+
3(z+x)
3(x+y+z)+
4(x+y)
4(x+y+z)=[y+z/x+y+z]+[x+z/x+y+z]+[x+y/x+y+z]=
2(x+y+z)
x+y+z=2.
点评:
本题考点: 对称式和轮换对称式.
考点点评: 此题考查了对称式与轮换对称式的知识.此题难度适中,解题的关键是得到:1x=y+z2(x+y+z),1y=z+x3(x+y+z),1z=x+y4(x+y+z).