令x-y=m
y/x=n
y=nx
x-nx=m
x=m/(1-n)
y=mn/(1-n)
f(m,n)=x^2-y^2
=(x+y)(x-y)
=[m/(1-n)+mn/(1-n)]m
=m/(1-n)(1+n)*m
=m^2(1+n)/(1-n)
所以
f(x,y)=x^2(1+y)/(1-y)
令x-y=m
y/x=n
y=nx
x-nx=m
x=m/(1-n)
y=mn/(1-n)
f(m,n)=x^2-y^2
=(x+y)(x-y)
=[m/(1-n)+mn/(1-n)]m
=m/(1-n)(1+n)*m
=m^2(1+n)/(1-n)
所以
f(x,y)=x^2(1+y)/(1-y)对了。。。哈哈
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