z=x3-y3-12x 6y 5的极值
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二次型是三元二次型,本来就有三个变量x1,x2,x3,使用的变换自然应该有三个式子,前两个式子是由配方以后的结果决定的,第三个有无穷多种取法,只要保证变换是可逆的即可,这里使用y3=x3,矩阵C是上三
x3+y3=100(x+y)(x^2-xy+y^2)=100因x+y=1所以x^2-xy+y^2=100(x+y)^2-3xy=1001-3xy=100xy=-33x^2+y^2=(x+y)^2-2x
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
x3+y3-x2y-xy2=(x+y)(x2-xy+y2)-xy(x+y)=(x+y)(x2-2xy+y2)=(x+y)(x2+2xy+y2-4xy)=(x+y)[(x+y)2-4xy]=10×(10
∵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)(x²+y²+z²)=x³+y³+z³+x²(y+z)+y²(x+z)+z²(x+y)∴1*2
(x+y+z)^3-x^3-y^3-z^3=3yx^2+3xy^2+3xz^2+3yz^2+3zx^2+3zy^2+6xyz=3xy(x+y)+3z^2(x+y)+3z(x^2+y^2+2xy)=3x
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
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
x3(y-z)+y3(z-x)+z3(x-y)=x3(y-z)+y3(z-x)-z3(y-z)-z3(z-x)=(x3-z3)(y-z)+(y3-z3)(z-x)=(x-z)(y-z)(x2+xz+z
x+y=1(x+y)^2=x^2+2xy+y^2=1(x+y)^3=x^3+y^3+3xy(x+y)=1而x^3+y^3=1/3,代入得:3xy=2/3xy=2/9由于x=1-y;故代入xy=2/9;
A-B=(x3+2y3-xy2)-(﹣y3+x3+2xy2)=x³+2y³-xy²+y³-x³-2xy²=3y³-3xy²
x^3+y^3+z^3+(x+y)^3+(y+z)^3+(z+x)^3=[x^3+(y+z)^3]+[y^3+(z+x)^3]+[z^3+(x+y)^3]=(x+y+z)(x^2-xy+y^2-xz+
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,
x3+3xy-y3=(x-y)(x^2+y^2+xy)+3xy=-x^2-y^2+2xy=-(x-y)^2=-1