已知z=(1 xy)^y求dy

(x+y+z)²=1,x²+2xy+y²+2(x+y)z+z²=1,x²+y²+z²+2(x+y)z+2xy=1xy+yz+xz=

1=xy/(x+y)两边倒数1/x+1/y=1同理1/y+1/z=1/21/z+1/x=1/3联合三个方程得1/x=5/121/y=7/121/z=-1/12即x=12/5y=12/7z=-12x+y

解析2xdx+ydx+xdy+3y²dy=0(2x+y)dx+(x+3y²)dy=0(2x+y)dx=-(x+3y²)dydy/dx=(2x+y)/-(x+3y²

xyz=1所以z=1/xyxz=1/yyz=1/xx/(xy+x+1)+y/(yz+y+1)+z/(xz+z+1)=x/(xy+x+1)+y/(1/x+y+1)+(1/xy)/(1/y+1/xy+1)

dz/dx=arctan(xy)+xy/[1+(xy)^2](dz/dx)|(1,1)=π/4+1/2(dz/dy)|(1,1)=x^2/[1+(xy)^2]=1/2

z=(x+y)^2*cos(x^2*y^2)dz/dx=2*(x+y)*cos(x^2*y^2)-2*(x+y)^2*sin(x^2*y^2)*x*y^2dz/dy=2*(x+y)*cos(x^2*y

∵(x+y+z)(x²+y²+z²)=x³+y³+z³+x²(y+z)+y²(x+z)+z²(x+y)∴1*2

(xy+yz+xz)²=x²y²+x²z²+y²z²+2xyz²+2x²yz+2xy²z=1=x&#

y=-12;一共是三个方程,因为xy/(x+y)=3推出（x+y）/(xy)=1/3-------方程1；同理：（y+z）/(yz)=1/2-------方程2；（x+z）/(xz)=1-------

(x+y+z)²=1²x²+y²+z²+2xy+2yz+2xz=1x²+y²+z²+2(xy+yz+xz)=1x&sup

令u=x/y,则dx/dy=u+ydu/dy原式化为u+ydu/dy=-u/y+2u+1(即变量y因变量u的一次线性非齐次方程)整理得du/dy-(1/y^2-1/y)u=1/y先求齐次方程du/dy

1.z=3y/2把:z=3y/2代入x+y+z=3y得:x+y+3y/2=3y整理后得:x=y/2所以:x/(x+y+z)=(y/2)/(y/2+y+3y/2)=1/62.因为1/x-1/y=3,则1

就是把这dydx转为求导前的式子,然后再求导一遍验证一下对错.再问：就是算到最后有个积分搞不出来。求过程。

由已知得dy/dx=(e^y+z)/(e^x+z),dz/dx=(z^2-e^(x+y))/(e^x+z),dz/dy=(z^2-e^(x+y))/(e^y+z),所以可以得到三式,e^ydx+zdx

dy/dx=dy/du*du/dx+dy/dv*dv/dx=v*e^(x+y)+u*y/x=ln(xy)*e^(x+y)+e^(x+y)*y/x=e^(x+y)[ln(xy)+y/x]所以dy=e^(

xy/(x+y)=1=>xy=x+y=>1/x+1/y=1--式一yz/(y+z)=2=>yz=2y+2z=>1/y+1/z=1/2--式二xz/(x+z)=3=>xz=3x+3z=>1/x+1/z=

同学,xyz=1吧?这样的话,原式=x/(xy+x+xyz)＋y/(yz+y+xyz)＋z/(xz+z+xyz)=1/(y+1+yz)+1/(z+1+xz)+1/(x+1+xy)=xyz/(y+xyz

y+xy'=（1+y'）e^（x+y）则y'=（y-e^(x+y)）/(e^(x+y)-x),dy=（y-e^(x+y)/(e^(x+y)-x)dx

u=x^2+y∂u/∂x=2x∂u/∂y=1du=(∂u/∂x)dx+(∂u/∂y)dy=2xdx+dy

(x+y+z)²=x²+y²+z²+2xy+2yz+2xz所以可得：xy+yz+xz=[(x+y+z)²-(x²+y²+z