Question

若f（x+y，x-2y）=x²-y²，求f(x-y，x/y)。

Answer 1

令x+y=a，x-2y=b，求得x=（2a+b）/3，y=（a-b）/3
带入原方程得f（a，b）=[（2a+b）/3]²-[（a-b）/3]²=(a²+2ab)/3
于是f（x，y）=(x²+2xy)/3
于是f（x-y，x/y）=[（x-y）²+2（x-y）*x/y]/3=⅓[(x²+y²-2xy+(2x²/y)-2x]

Answer 2

f（x+y，x-2y）=x²-y²
x^2-y^2=(x-y)(x+y)
(x-2y)=m(x+y)+n(x-y)=(m+n)x+(m-n)y
m+n=1, (1)
m-n=-2 (2)
(1)+(2), 2m=-1, m=-1/2, n=3/2
x^2-y^2=(x-y)(x+y)=[-1/2 (x+y)+3/2 (x-y)](x+y)
令x+y=X，x-y=Y
f（X，Y）=X（-1/2 X+3/2 Y）
则 f(x-y, xy)=...

Answer 3

解