x y-2z=3,3x 2y z=-1,x-z=0
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xy/x+y=-2,取倒数得1/x+1/y=-1/2①yz/y+z=3/4取倒数得1/y+1/z=4/3②zx/z+x=-3/4取倒数得1/x+1/z=-4/3③①+②+③得2(1/x+1/y+1/z
2x-3y-z=0(1)x+3y-14z=0(2)(1)+(2)3x-15z=0x=5z(2)*2-(1)6y-28z+3y+z=09y=27zy=3z代入(4x^2-5xy+z^2)/(xy+yz+
算数平方根有意义,xy同号.x²+4y²+z²-3xy=2z√(xy)x²+4y²+z²-2z√(xy)-3xy=0x²-4xy+
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
由2x-3y-z=0,x+3y-14z=0,且x,y,z不全为0解得x=5zy=3z将x=5zy=3z带入4x平方-5XY+Z的平方/xy+yz+zx得4*25z平方-5*5Z*3z+Z的平方/5z*
xy+yz+xz=1/2x(y+z)+1/2y(x+z)+1/2z(x+y)=(1/2x)*(1/2yz)+1/2y*(1/3zx)+1/2z*(xy)=11/12xyz应该知道答案了吧
x=6-3y &nbs
∂Z/∂x=y*cos(xy)-2cos(xy)*sin(xy)*y=y*cos(xy)-y*sin(2xy)∂Z/∂y=x*cos(xy)-2cos(
X=1,Y=2,Z=3其实很简单!
z=x²+4y²-3xy≥4xy-3xy=xy所以xy/z≤1.xy/z取得最大值时xy=z且x=2y,所以z=2y².2/x+1/y-2/z=1/y+1/y-1/y
3x-4y-z=02x+y-8z=08x+4y-32z=03x-4y-z+8x+4y-32z=011x-33z=0x=3zy=2zx²+2xy+z²/xy+yz+zx=(3z)^2
题目有点问题,z/(xy)没有最大值.由条件z=x²+4y²-3xy,故z/(xy)=x/y+4y/x-3.取x=1,当y趋于0时,可知右端趋于正无穷.正确的说法可能是z/(xy)
3[-(x+y)+2xy²-z]-2[(x+y)-xy²+z]-5[-3(x+y)-z]=3(-x-y+2xy²-z)-2(x+y-xy²+z)-5(-3x-3
2x-3y-z=0(1)x+3y-14z=0(2)(1)+(2)3x-15z=0x=5z(2)*2-(1)6y-28z+3y+z=09y=27zy=3z代入(4x^2-5xy+z^2)/(xy+yz+
2x²+2xy+y²-4x+z-2√z-3+2=0对其化简:(x²+2xy+y²)+(x²-4x+4)+(z-2√z-3-2)=0(x+y)²
解方程组:{2x-3y-z=0.(1){x+3y-14z=0.(2)(1)+(2)得:3x-15z=0即:x=5z,代入(1)式得y=3z所以:(4x²-5xy+z²)/(xy+y
解题思路:本题的关键是将三个方程两边取倒数,化简后分别将方程等号左边和右边相加,得到1/x+1/y+1/z的值,最后将要求的分式化简,把1/x+1/y+1/z的值带入即可。解题过程:
|x-3|+|y+z|+|2z+1|=0则|x-3|=0x=3|y+z|=0y=-z=1/2|2z+1|=0z=-1/2xy-yz=3x1/2-1/2x(-1/2)=7/4
3x-y=-2zx+2y=-3z那么:x=-z,y=-z(3x^-xy+2y^)/(2x^+4xy+y^)=(3z^2-z^2+2z^2)/(2z^2+4z^2+z^2)=4z^2/7z^2=4/7
x+2=0x=-23y-1=03y=1y=1/3z-2=0z=2(-3xy)*(-x²z)*6xy=[-3*(-2)*1/3]*[-(-2)²*2]*6*(-2)*1/3=2*(-