设z=x^3f(xy,y x)-搜答案-智慧作业帮

xy+xz=8-x²yx+yz=12-y²zy+zx=-4-z²x(x+y+z)=8y(x+y+z)=12z(x+y+z)=-4(x+y+z)²=8+12-4=

若x+y=4，xy=3，求yx

∵x+y=4，xy=3，∴原式=x2+y2xy=(x+y)2−2xyxy=16−63=103．

设u=xy,v=lnx+g(xy),则x(∂z/∂x)-y(∂z/∂y)=∂f/∂v.原因如下：dz=(∂f/

对方程两边求全微分得：(e^z-1)dz+y^3dx+3xy^2dy=0（方法和求导类似）移项,有dz=-(y^3dx+3xy^2dy)/(e^z-1)

e^z-z+xy^3=0偏z/偏x：z'e^z-z'+y^3=0y^3=z'(1-e^z)z'=y^3/(1-e^z)偏z/偏y:z'e^z-z'+3xy^2=0z'=3xy^2/(1-e^z)偏z/

x^2+y^2+z^2-3xyz=0两边对x求偏导,2x+2z*dz/dx-3yz-3xydz/dx=0从中解得：dz/dx=(3yz-2x)/(2z-3xy)(1)同理：dz/dy=(3xz-2y)

题好像不太对啊,第二个式子貌似不对.初中生?给的49,36,25,我觉得应该是X^2＋2xy＋Y^2＝49,这样的话（x＋y）^2＝49,x＋y＝7,同理y＋z＝6,x＋z＝5,三个式子相加得2（＋y

x=6-3y &nbs

令u=xy,v=x+yz=f(u,v)az/ax=y(fu)+(fv)a^2z/axay=a(az/ax)/ay=a(y(fu)+(fv))/ay=(fu)+y(a(fu)/ay)+a(fv)/ay=

x+4y+3z=0x=-4y-3z3x-2y-5z=03x=2y+5zx=(2y+5z)/3所以-4y-3z=(2y+5z)/3-12y-9z=2y+5z14y=-14zy=-z所以x=-4*(-z)

xy-3xy+2yx-yx

=xy-3xy+2xy-xy=-xy

设u=xy,v=y/x,则z=f(u,v),所以ðz/ðx=f'1*ðu/ðx+f'2*ðv/ðx=yf'1-yf'2/x^2,注意到f'1

题目有问题,yx/(y+x)=4/3应该是yz/(y+z)=4/3xy/(x+y)=-2(x+y)/(xy)=-1/21/x+1/y=-1/2(1)yz/(y+z)=4/3(y+z)/(yz)=3/4

设u=xy,v=y/x,则z=x³f(u,v),au/ax=y,av/ax=-y/x²故az/ax=3x²f(u,v)+x³f'u(u,v)(au/ax)+x&

1.z'x=3x²y²z'y=2x³y2.z'x=4x³z'y=3y³3.z'x=ye^(xy)+2xyz'y=xe^(xy)+x²4.u'

xy+yx=10x,y是多少

xy+yx=10x+y+10y+x=11x+11y=100+x10x=100-11yx=10-1.1y所以y只能是0

∵x-y=4xy，∴2x+3xy-2yx-2xy-y=2(x-y)+3xyx-y-2xy=8xy+3xy4xy-2xy=112．故答案为：112．

令u=xy,v=e^(x+y)Z'x=Z'u*U'x+Z'v*V'x=f'u*y+f'v*e^(x+y)Z'y=Z'u*U'y+Z'v*V'y=f'u*x+f'v*e^(x+y)

XY=XZ+YX?那么也就是XY=X（Z+Y）咯,Y=Z+Y?无法证明的.题抄错啦~`

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