证明由方程F（x-az，y-bz）=0确定的函数z=z（x，y）应满足a（z⼀x）+b（z⼀y）=1

解：设u=x-az,v=y-bz
则，原方程写为 F(u,v)=0
方程F(u，v)=0 两端分别对x,y求偏导得
ðF/ðx=ðF/ðu*(ðu/ðx+ðu/ðz*ðz/ðx)+ðF/ðv*(ðv/ðz*ðz/ðx)
=ðF/ðu*(1-a*ðz/ðx)+ðF/ðv*(-b*ðz/ðx)
=ðF/ðu-a*ðF/ðu*ðz/ðx-b*ðF/ðv*ðz/ðx
=ðF/ðu-(a*ðF/ðu+b*ðF/ðv)ðz/ðx
=0
得：ðz/ðx=a*(ðF/ðu)/(a*ðF/ðu+b*ðF/ðv)
ðF/ðy=ðF/ðu*(ðu/ðz*ðz/ðy)+ðF/ðv*(ðv/ðy+ðv/ðz*ðz/ðy)
=ðF/ðu*(-a*ðz/ðy)+ðF/ðv*(1-b*ðz/ðy)
=-a*ðF/ðu*ðz/ðy+ðF/ðv-b*ðF/ðv*ðz/ðy
=ðF/ðv-(a*ðF/ðu+b*ðF/ðv)*ðz/ðy
得：ðz/ðy=a*(ðF/ðv)/(a*ðF/ðu+b*ðF/ðv)
则，a(ðz/ðx)+b(ðz/ðy)
=a*(ðF/ðu)/(a*ðF/ðu+b*ðF/ðv)+b*(ðF/ðv)/(a*ðF/ðu+b*ðF/ðv)
=(a*ðF/ðu+b*ðF/ðv)/(a*ðF/ðu+b*ðF/ðv)
=1

用代入法求

不知道 不好意思