Question

在线等待!一道关于多元函数的题目

Answer 1

令z=x+y, 则x=z-y

f(z,y)=f(x+y,y)=x^2-y^2=(z-y)^2-y^2

=z^2-2yz+y^2-y^2

=z^2-2yz

因此

f(x,y)=x^2-2xy