X-1 2=2-Y 3=Z-3 4
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∵y3-z3=(y-z)(y2+yz+z2)(立方差公式)又∵y3-z3-y2-yz-z2=0∴(y-z-1)(y2+yz+z2)=0(提取公因式)∵y、z是正实数∴y-z-1=0即y-z=1∵x-y
一、先z对x、y分别求偏导数,并令他们分别等零.联立方程求出驻点(x,y).驻点求得:(1,1)、(1,-1)、(-1,-1)、(-1,1)二、再在对z求x、y的二阶偏导和他们的混合偏导.令z对x的二
∵x+y+z=0,∴z=(-x-y)x^3+y^3+z^3=x^3+y^3-(x+y)^3=x^3+y^3-x^3-y^3-3x^2y-3xy^2=-3xy(x+y)=3xyz
因为:X3-Y3-Z3=3XYZ所以:X3+(-Y)3+(-Z)3-3X(-Y)(-Z)=0(X-Y-Z)(X2+Y2+Z2+XY+XZ-YZ)=0所以:1.X-Y-Z=02.X2+Y2+Z2+XY+
由(x+y+z)2-(x2+y2+z2)可得xy+xz+yz=-5x3+y3+z3-3xyz=(x+y+z)(x2+y2+z2-xy-yz-zx)可得xyz=14再问:谢谢,我看一下其他的答案在采纳再
(x+y+z)²-(x²+y²+z²)=2(xy+yz+zx)=-1,xy+yz+zx=-1/2x3+y3+z3=3xyz+(x+y+z)(x²+y&
∵(x+y+z)(x²+y²+z²)=x³+y³+z³+x²(y+z)+y²(x+z)+z²(x+y)∴1*2
偏z/偏y就是求y的偏导把x当作不变量(当作常数)就是f'(u)*3x^2y^2u=x^2y^3e(用这个表示偏导)ez/ey=dz/du*eu/eydz/du=f'(u)eu/ey=3x^2y^2所
证明:因为x2+y2≥2xy≥0(2分)所以x3+y3=(x+y)(x2-xy+y2)≥xy(x+y)(4分)同理y3+z3≥yz(y+z),z3+x3≥zx(z+x)(8分)三式相加即可得2(x3+
设x2=y3=z4=k,则x=2k,y=3k,z=4k,∵2x-3y+4z=22,∴4k-9k+16k=22,∴k=2,∴x+y-z=2k+3k-4k=k=2.
设x−12=2−y3=z−34=k,则x=2k+1,y=-3k+2,z=4k+3,∵x,y,z均为非负实数,∴2k+1≥0−3k+2≥04k+3≥0,解得-12≤k≤23,于是W=3x+4y+5z=3
如果你的X2是x的平方,X3是x的三次方那么答案是:-(x-y+z)*(x-y-z)*(x+y-z)
设x2+y2+z2=t,则∵(x+y+z)2=x2+y2+z2+2(xy+yz+xz),即9=t+2(xy+yz+xz),∴xy+yz+xz=9−t2,∵x3+y3+z3-3xyz=(x+y+z)(x
(x+y)³=x³+y³+3x²y+3xy².记忆方法:各立方,然后3x方y,3xy方(x+y)³=x³-y³-3x
设x+3y3=2y+3z4=2z+2x5=k,则有:x+3y=3k2y+3z=4k2z+2x=5k,解得x=32ky=12kz=k;因此x:y:z=3:1:2.
x3+y3-z3+3xyz,=[(x+y)3-3x2y-3xy2]-z3+3xyz,=[(x+y)3-z3]-(3x2y+3xy2-3xyz),=(x+y-z)[(x+y)2+(x+y)z+z2]-3
3元一次方程,好像是初一的问题哦.根据前面两个等式可以得出x=3zy=z(平方)/32x+3y+4z=2*(3z)+3*(z方/3)+4z现在变成了一元二次方程,你应该会解吧.
由柯西-黎曼条件v'(x)=-u'(y),v'(y)=u'(x)得u'(y)=-6xy,u'(x)=3y²-3x²因而选择B
根据√x/y+√y/z+√z/xx,y,z应全>0或全0将题中两式相减得:x^3-x^2+y^3-y^2+z^3-z^2=1(x-1)x^2+(y-1)y^2+(z-1)z^2=1因为x,y,z>0,