已知2lg[(x-y)/2]=lgx+lgy,则x/y=?已知函数f(x)=a-1/(2^x+b)为奇函数,则a=?

已知2lg[(x-y)/2]=lgx+lgy,则x/y=?已知函数f(x)=a-1/(2^x+b)为奇函数,则a=?
已知2lg[(x-y)/2]=lgx+lgy,则x/y=?
已知函数f(x)=a-1/(2^x+b)为奇函数,则a=?

已知2lg[(x-y)/2]=lgx+lgy,则x/y=?已知函数f(x)=a-1/(2^x+b)为奇函数,则a=?
lg((x-y)/2)^2=lg(xy),化简后得x2-2xy+y2=4xy,同除以4x2得,1/4+y/2x+y2/4x2=y/x令y/x=a,解得a=土2倍根号2+3,1/a=他的倒数,f(0)=0,a=1/(1+b),-f(1)=f(-1)=a-1/(2+b)=1/(0.5+b)-a,a=1,b=0

2lg[(x-y)]/2=lg[(x-y)/2]^2
lgx+lgy=lgxy
2lg[(x-y)/2]=lgx+lgy
所以
lg[(x-y)/2]^2=lgxy
[(x-y)/2]^2=xy
x^2-2xy+y^2=4xy
(x+y)^2=0
x=-y
x/y=-1

2log[(x-y)/2]-2logy=logx-logy
2{log[(x-y)/2]-logy}=log(x/y)
2[log[(x-y)/2y]=log(x/y)
log{[(x-y)/2y]^2}=log(x/y)
x/y=(x-y)^2/4y^2

2lg[(x-y)/2]=lgx+lgy,
lg((x-y)/2)^2=lgxy
(x-y)^2/4=xy
题目是不是有误呀，要(x+y)）才能做下去