已知[1/x+1y+z=12],[1/y+1z+x=13],[1/z+1x+y=14],求[2/x+3y+4z]的值.

1个回答

  • 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).