画出z=√x² y²,z=1,z=2所围成的立体的图形
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![画出z=√x² y²,z=1,z=2所围成的立体的图形](/uploads/image/f/6371421-69-1.jpg?t=%E7%94%BB%E5%87%BAz%3D%E2%88%9Ax%C2%B2+y%C2%B2%2Cz%3D1%2Cz%3D2%E6%89%80%E5%9B%B4%E6%88%90%E7%9A%84%E7%AB%8B%E4%BD%93%E7%9A%84%E5%9B%BE%E5%BD%A2)
(x+y-z)/z=(y+z-x)/x=(z+x-y)/y[x+y]/z-1=[y+z]/x-1=[z+x]/y-1[x+y]/z=[y+z]/x=[z+x]/y设[x+y]/z=[y+z]/x=[z
1/x=p1/y=q1/z=rpq+qr+pr=1(y+x)/z+(y+z)/x+(z+x)/y≥2(1/x+1/y+1/z)^2为(pq+qr+pr)[r/p+r/q+q/r+q/p+p/r+p/q
[x+(z-y)][x-(z-y)]=x-(z-y)记得采纳啊
就是这么一个三角形平面再问:如果只是平面,那x=0,y=0,z=0干什么用的再答:x+y+z=1,形成的是过(0,0,1),(1,0,0),和(0,1,0)3个点的平面x=0,y=0,z=0相当于是3
f=@(x,y,z)(x.^2+9/4*y.^2+z.^2-1).^3-x.^2.*z.^3-9/80*y.^2.*z.^3 implicitmesh(f,[-5 5],150)-
有这样的公式:a^3+b^3+c^2-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)左边减右边,证明:(x+y-2z)^3+(y+z-2x)^3+(z+x-2y)^3-3(x+y
设(x+y-z)/z=(x-y+z)/y=(-x+y+z)/x=k则(1)x+y-z=kz(2)x-y+z=ky(3)-x+y+z=kx(1)+(2)+(3)得x+y+z=k(x+y+z)∴k=1时,
∵y+z÷x=Z+X÷y=X+Y÷z容易发现x,y,z位置互换也成立∴式子与x,y,z值无关∴x=y=z∴(X+Y-Z)÷(X+Y+z)=x/3x=1/3明教为您解答,请点击[满意答案];如若您有不满
设:(x+y-z)/z=(y+z-x)/x=(z+x-y)/y=k{x+y-z=kz(1){y+z-x=kx(2){z+x-y=ky(3)(1)+(2)+(3)得:(x+y+z)=k(x+y+z)(x
令(y+z)/x=(z+x)/y=(x+y)/z=ky+z=kxx+z=kyx+y=kz2(x+y+z)=k(x+y+z)2(x+y+z)=k(x+y+z)(2-k)(x+y+z)=0(x+y+z≠0
x-2y=02y+z=01-2z=0解出来x=-0.5y=-0.25z=0.5快采纳吧~~~再问:给你点个赞再答:O(∩_∩)O哈哈~
因为x/y+z+y/z+x+z/x+y=1所以x/y+z=1-y/z+x-z/x+y,两边同乘以x得x^2/y+z=x-xy/z+x-xz/x+y同理y^2/x+z=y-xy/z+y-yz/x+y,z
1.z²-z+1/4=(z-1/2)².绝对值、根号、平方数都是非负的,而相加为0.所以都为0.即x=y,2y=z,z=1/2.所以x=y=1/4,z=1/2.2.2002x200
1、隐函数对x求导得1+az/ax+yz+xy*az/ax=0,故az/ax=-(1+yz)/(1+xy);F对x求导得aF/ax=e^x*y*z^2+e^x*y*2z*az/ax;当x=0,y=1时
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大于等于2倍根号下xy同理x+z大于等于2倍根号下xzz+y大于等于2倍根号下zy所以(x+y)(y+z)(z+x)大于等于8xyz当取到8xyz时分数值最大为1/8此时x=1/3y=1/3z=
根号x-3+|y-2|+z^2=2z-1根号x-3+|y-2|+(z^2-2z+1)=0根号x-3+|y-2|+(z-1)^2=0由于数值开根号,绝对值和平方数均为大于等于0的数则上式要成立只有X-3
这个题目没有问题么,我是说最后一个式子确定是z+5y+8z=-2?如果没有问题的话:x+y+z=1;①x+3y+7z=-1;②z+5y+8z=-2③①-②2Y+6Z=-2Y=(-2-6Z)/2=-1-
X+Y+Z