设f(x+y,x-y)=x^2+xy,求f(x,y)
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设f(x+y,x-y)=x^2+xy,求f(x,y)设f(x+y,x-y)=x^2+xy,求f(x,y)设f(x+y,x-y)=x^2+xy,求f(x,y)设f(x+y,x-y)=x^2+xy设u=x
设f(x+y,x-y)=x^2+xy,求f(x,y)
设f(x+y,x-y)=x^2+xy,求f(x,y)
设f(x+y,x-y)=x^2+xy,求f(x,y)
设f(x+y,x-y)=x^2+xy
设u=x+y,v=x-y
则x=（u+v）/2,y=(u-v)/2
所以f(u,v)=(u+v)^2/4+(u+v)(u-v)/4
所以f(x,y)=(x+y)^2/4+(x+y)(x-y)/4=(x^2+xy)/2
令a=x+y,b=x-y
两式相加a+b=2x,x=(a+b)/2,y=a-x=a-(a+b)/2=(a-b)/2
代入
f(a,b)=[(a+b)/2]^2+[(a+b)/2][(a-b)/2]
=[(a+b)^2]/4+[(a+b)(a-b)]/4
=2a(a+b)/4=a(a+b)/2
∴f(x,y)=x(x+y)/2
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