Question

f（x+y,y/x）=x^2-y^2，求f（x,y）=?

Answer 1

令
x+y=t (1)
y/x=s (2)
则y=sx,代入（1）,得
x(1+s)=t,x=t/(1+s)
从而 y=ts/(1+s)
所以 f(t,s)=x² -y²=t²/(1+s)²-t²s²/(1+s)²=t²(1-s²)/(1+s)²=t²(1-s)/(1+s)
从而f(x,y)=x²(1-y)/(1+y)
http://zuoye.baidu.com/question/81d647fe22e014874ab625f8d01ec001.html

Answer 2

y=sx
x=t-y=t-sx
(s+1)x=t
x=t/(s+1)
y=sx=st/(s+1)
f(x+y,y/x)=x²-y²
f(t，s)=[t/(s+1)]²-[st/(s+1)]²
=t²/(s+1)²-s²t²/(s+1)²
=(t²-s²t²)/(s+1)²
=t²(1-s²)/(1+s)²
=t²(1+s)(1-s)/(1+s)²
=t²(1-s)/(1+s)
将t换成x，s换成y，得
f(x，y)=x²(1-y)/(1+y)

Answer 3

令a=x+y,b=y/x,得x=a/(1+b),y=ab/(1+b)
故f(a,b)=x^2-y^2=[a/(1+b)]^2-[ab/(1+b)]^2