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yzDx xzDy

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yzdx+xydz+xzdy

解:∵xyz+√(x^2+y^2+z^2)=√2 ∴两边微分,得 d(xyz)+d(√(x^2+y^2+z^2))=d(√2) ==>yzdx+xzdy+xydz+(xdx+ydy+zdz)/√(x...

yzdx+xzdy+xydz+(xdx+ydy+zdz)/√(x²+y²+z²)=0将(x,y,z)=(1,0,-1)代入:-dy+(dx-dz)/√2=0∴dz=dx-√2dy 追思无止境 | 2012-...

设长方体三度为x,y,z.在条件①:2(xy+xz+yz)=a²之下,求V=xyz的最大值。设F(x,y,z,μ)=xyz+μ(2(xy+xz+yz)-a²)②:F′x=yz+2μ(y+...

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