Задача 63771 Доказать равенства Вариант 2...

6204aaca0939d027669c90d0

24.03.2022 07:24:53

Доказать равенства
Вариант 2

5f3ea7e3faf909182968ddd9

24.03.2022 08:56:21

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[m]u=(x-y)(y-z)(z-x)[/m]

[m]\frac{ ∂u }{ ∂x }=u`_{x}=((x-y)(y-z)(z-x))`_{x}=(x-y)(y-z)`_{x}\cdot(z-x)+(x-y)(y-z)\cdot(z-x)`_{x}=[/m]

[m]=(x-y)`_{x}\cdot (y-z)\cdot(z-x)+(x-y)\cdot (y-z)`_{x}\cdot(z-x)+(x-y)(y-z)\cdot(z-x)`_{x}=[/m]

[m]=1\cdot (y-z)\cdot(z-x)+(x-y)\cdot(z-x)\cdot 0+(x-y)(y-z)\cdot(-1)= (y-z)\cdot(z-x)-(x-y)(y-z)[/m]


[m]\frac{ ∂u }{ ∂y }=u`_{x}=((x-y)(y-z)(z-x))`_{y}=(x-y)(y-z)`_{y}\cdot(z-x)+(x-y)(y-z)\cdot(z-x)`_{y}=[/m]

[m]=(x-y)`_{y}\cdot (y-z)\cdot(z-x)+(x-y)\cdot (y-z)`_{y}\cdot(z-x)+(x-y)(y-z)\cdot(z-x)`_{y}=[/m]

[m]=(-1)\cdot (y-z)\cdot(z-x)+(x-y)\cdot(z-x)\cdot 1+(x-y)(y-z)\cdot 0=- (y-z)\cdot(z-x)+(x-y)(y-z)[/m]


[m]\frac{ ∂u }{ ∂z }=u`_{z}=((x-y)(y-z)(z-x))`_{z}=(x-y)(y-z)`_{z}\cdot(z-x)+(x-y)(y-z)\cdot(z-x)`_{z}=[/m]

[m]=(x-y)`_{z}\cdot (y-z)\cdot(z-x)+(x-y)\cdot (y-z)`_{z}\cdot(z-x)+(x-y)(y-z)\cdot(z-x)`_{z}=[/m]

[m]=0 \cdot (y-z)\cdot(z-x)+(x-y)\cdot(z-x)\cdot (-1)+(x-y)(y-z)\cdot 1=-(x-y)(y-z)+(x-y)(y-z)[/m]



Тогда

[m]\frac{ ∂u }{ ∂x }+\frac{ ∂u }{ ∂y }+\frac{ ∂u }{ ∂z }=(y-z)\cdot(z-x)-(x-y)(y-z)- (y-z)\cdot(z-x)+(x-y)(y-z)-(x-y)(y-z)+(x-y)(y-z)=0[/m]