Question

设f(x+y,y/x)＝x∧2-y∧2,则f(x,y)＝?

Answer 1

解：
令x+y=m，y/x=n，则x=m/(n+1)，y=mn/(n+1)
f(x+y，y/x)=f(m，n)
=[m/(n+1)]²-[mn/(n+1)]²
=(m²-m²n²)/(n+1)²
=m²(1-n²)/(n+1)²
=m²(1+n)(1-n)/(n+1)²
=m²(1-n)/(1+n)
将m换成x，n换成y，得
f(x，y)=x²(1-y)/(1+y)
分式有意义，1+y≠0，y≠-1
函数解析式为f(x，y)=x²(1-y)/(1+y)，(y≠-1)

Answer 2

f(x+y,y/x)=x2-y2
另①x+y=m,②y/x=n
所以由②得：y=nx,代入①得：x+nx=m
所以x=m/(n+1)
代入②得：y=nx=nm/(n+1)
所以
f(m,n)=[m/(n+1)]2-[nm/(n+1)]2
=(m2)(1-n)/(1+n)
即：所求为：f(x,y)==(x2)(1-y)/(1+y)

Answer 3

Answer 4