求e^z-6*z 2*x*y=5的切平面
来源:学生作业帮助网 编辑:作业帮 时间:2024/05/25 02:57:52
把x+y+z=2两边平方得:(x+y+z)2=x2+y2+z2+2xy+2yz+2zx=4,把xy+yz+xz=-5代入得:x2+y2+z2=14.
利用柯西不等式∵(x^2+y^2+z^2)(2^2+3^2+4^2)≥(2x+3y+4z)^2∴x^2+y^2+z^2≥(2x+3y+4z)^2/(2^2+3^2+4^2)=100/29当x/2=y/
根据题意,设x=2k,则y=3k,z=5k,代入x+y+z=20,∴2k+3k+5k=20,得k=2,∴x=4,y=6,z=10;∴2x2+3y2+5z2=2×16+3×36+5×100=640;故选
方法一:特殊值法,假设x=0,y=1,z=-1x2+y2-z2分之一加x2+z2-y2分之一加y2+z2-x2分之一=0方法二:x2+y2-z2分之一=(x2+y2-(x+y))^2分之一=-1/(2
x²+y²+z²-xy-yz-xz=1/2(2x²+2y²+2z²-2xy-2yz-2xz)=1/2(x²-2xy+y²
1.x^2+y^2+z^2=(x+y+z)^2-2*(xy+yz+xz)=82.配方(x-1)^2+(y+2)^2+(z-3)^2=0x=1y=-2z=3x+y+z=2再问:兄弟我敬佩你,如果你能帮我
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
x+y+z=a(x+y+z)^2=a^2(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)x^2+y^2+z^2=(x+y+z)^2-2(xy+yz+zx)=a^2-2
/>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+
设y=biz2=bi+(2-bi)i=b+(2+b)iz1=z2(2x+1)+i=b+(2+b)i所以2x+1=b1=2+bb=-1x=-1z1=-1+iz2=z1=-1+i-------------
∵x-z=(x-y)+(y-z)=2+2=4∴x2-z2=(x+z)(x-z)=14×4=56.
因为x-y=5,所以(x-y)^2=x^2+y^2-2xy=25因为y-z=3所以(y-z)^2=y^2+z^2-2yz=9因为x-z=(x-y)+(y-z)=5+3=8所以(x-z)^2=x^2+y
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+
答案是11分之14. 求采纳
令x/3=y/4=z/6=K则x=3K,y=4K,c=6K(xy+yz+xz)/(X2+y2+z2)=(3K*4K+4K*6K+3K*6K)/(3K*3K+4K*4K+6K*6K)=54K2/61K2
|x+2|>=0,|y-3|>=0所以要使等式成立,必须有x+2=0,y-3=0求得x=-2,y=3(x-2y+z)/2=y/2+x+z(-2-2×3+z)/2=3/2-2+z-5-8+z=-11+2
x+y+z平方得x2+y2+z2+2xy+2xz+2yz吧=9所以x+y+z=3或者-3(ps:x=y=z=1或者x=y=z=-1)
∵x2+y2+z2≤2x+4y-6z-14,∴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-
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,