Question

设x,y,z是不等于1的数,且[x(y+z-x)]/lgx = [y(z+x-y)]/lgy =[z(x+y-z)]/l

Answer 1

令[x(y+z-x)]/lgx = [y(z+x-y)]/lgy =[z(x+y-z)]/lgz=a

lg函数要求x>0、y>0、z>0；同时y+z-x≠0、z+x-y≠0、x+y-z≠0,否则,若有任何一个为0,则全部为0,即y+z-x=z+x-y=x+y-z=0,容易解出x=y=z=0,矛盾.因此a≠0.

x(y+z-x) = a·lgx (1)

y(z+x-y) = a·lgy (2)

z(x+y-z) = a·lgz (3)

(1)*y+(2)*x,得：

xy(y+z-x)+xy(z+x-y) = ay·lgx+ax·lgy

即：y·lgx+x·lgy = 2xyz/a

x^y·y^x = 10^(2xyz/a)

同理,y^z·z^y = 10^(2xyz/a),x^z·z^x = 10^(2xyz/a)

x^y·y^x = y^z·z^y = x^z·z^x