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f(x+y,y/x)=x^2+y^2
令x+y=t
y/x=w
则 x=t/(1+w)
y=wt/(1+w)
所以
f(t,w)=[t/(1+w)]²+[wt/(1+w)]²
=(t²+w²t²)/(1+w)²
=t²(1+w²)/(1+w)²
f(x-y,xy)
=(x+y)²(1+x²y²)/(1+xy)²
=
令x+y=t
y/x=w
则 x=t/(1+w)
y=wt/(1+w)
所以
f(t,w)=[t/(1+w)]²+[wt/(1+w)]²
=(t²+w²t²)/(1+w)²
=t²(1+w²)/(1+w)²
f(x-y,xy)
=(x+y)²(1+x²y²)/(1+xy)²
=
追问
不好意思了,x^2 与y^2之间是减号,睡觉之前头晕抄错了,麻烦高人把倒数第二个等号的思路详说一下
追答
哦 一样解得
f(x+y,y/x)=x^2-y^2
令x+y=t
y/x=w
则 x=t/(1+w)
y=wt/(1+w)
所以
f(t,w)=[t/(1+w)]²-[wt/(1+w)]²
=(t²-w²t²)/(1+w)²
=t²(1+w)(1-w)/(1+w)²
=t²(1-w)/(1+w)
f(x-y,xy)
=(x-y)²(1-xy)/(1+xy)
=(x+y)²(1+x²y²)/(1+xy)²
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