z=x^2 2Y^2 4X-8Y 2求极值
来源:学生作业帮助网 编辑:作业帮 时间:2024/10/04 12:14:21
解题思路:由已知可得1/x+1/y+1/z=0,如当x=1,y=-2时,z=-2,此时所求代数式的值为:-3/4;而而当x=1,y=2时,z=-(2/3)时,此时所求代数式的值为:-7/4.故所求代数
[x+(z-y)][x-(z-y)]=x-(z-y)记得采纳啊
x2-4x+y2+6y+z+1+13=(x-2)2+(y+3)2+z+1=0,∴x-2=0,y+3=0,z+1=0,即x=2,y=-3,z=-1,则(xy)z=(-6)-1=-16.
∵x2-4x+y2+6y+z−3+13=0,∴(x-2)2+(y+3)2+z−3=0,∴x-2=0,y+3=0,z-3=0,解得x=2,y=-3,z=3,∴(xy)z=[2×(-3)]3=-216.
把z看成已知数,解3x-4y-z=02x+y-8z=0,得x=3z,y=2z原式=(9z^2+4z^2+z^2)/(6z^2+2z^2+3z^2)=14/11
以单式来考虑,执行各式计算后:1.x=10;2.x=5;3.y=21,x=4;4.z=23,y=11,x=4
∵x2+y2+z2-2x+4y-6z+14=0,∴x2-2x+1+y2+4y+4+z2-6z+9=0,∴(x-1)2+(y+2)2+(z-3)2=0,∴x-1=0,y+2=0,z-3=0,∴x=1,y
x=-5y=-1z=15需要过程的话再H我再问:帮我再解一道题,谢谢x+2y=3y+2z=4z+2x=5需要过程
∵(x+y+z)(x²+y²+z²)=x³+y³+z³+x²(y+z)+y²(x+z)+z²(x+y)∴1*2
1.(x-1)^2+(y+2)^2+(z-3)^2=0则x=1,y=-2,z=3x+y+z=22.(3a-2b)(a+b)=0则a=-b或a=2/3×b则a/b-b/a-(a^2+b^2)/ab=(a
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)
请在此输入您的回答,每一次专业解答都将打造您的权威形象
/>x^2+4y^2+z^2-2x+4y-6z+11=0(x²-2x+1)+(4y²+4y+1)+(z²-6z+9)=0(x-1)²+(2y+1)²+
x/(y+z)+y/(z+x)+z/(x+y)=1所以x/(y+z)=1-[y/(z+x)+z/(x+y)]y/(z+x)=1-[x/(y+z)+z/(x+y)]z/(x+y)=1-[x/(y+z)+
等于0.x/(y+z)=1-[y/(z+x)+z/(x+y)]y/(z+x)=1-[x/(y+z)+z/(x+y)]z/(x+y)=1-[x/(y+z)+y/(z+x)]x2/(y+z)+y2/(z+
x+y+z=1xy+yz+zx=21*2=(x+y+z)(xy+yz+zx)=x(xy+yz+zx)+y(xy+yz+zx)+z(xy+yz+zx)=x²y+xyz+zx²+xy&
X+Y+Z
∵(x-2)2+(y+3)2+z+2=0,∴x-2=0,y+3=0,z+2=0,解得x=2,y=-3,z=-2,∴(xy)z=(-6)-2=136.
无数的解把原式化简后为(x+1)^2+(y+1)^2+(z+1)^2=11这个方程是以(-1,-1,-1)为球心,半径为根号11的球面方程.如果是圆的方程,x+y都会有无数的解.对于球的方程更是如此,
3x-4y-z=0,2x+y-8z=0令z=13x-4y=1(1)2x+y=8(2)(2)*4+(1)11x=33x=3,y=2x2+y2+z2/xy+yz+2zx=(9+4+1)/(6+2+6)=1