plot3(x,y,z)绘图程序

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plot3(x,y,z)绘图程序
(x+y-z)(x-y+z)=

[x+(z-y)][x-(z-y)]=x-(z-y)记得采纳啊

(y-x)/(x+z-2y)(x+y-2z)+(z-y)(x-y)/(x+y-2z)(y+z-2x)+(x-z)(y-z

∑是循环和例如∑a=a+b+c∑a^2=a^2+b^2+c^2∑(z-y)(x-y)/(x+y-2z)(y+z-2x)=∑(z-y)(x-y)(x+z-2y)/(x+y-2z)(y+z-2x)(x+z

①(x+y+z)(-x+y+z)(x-y+z)(x+y-z)

(1)原式=x+y+z)(-x+y+z)(x-y+z)(x+y-z)=[(x+y+z)(x+y-z)]*{[z+(x-y)][z-(x-y)]}=[(x+y)^2-z^2][z^2-(x-y)^2]=

数学 多项式(x+y-z)(x-y+z)-(y+z-x)(z-x-y)公因式

(x+y-z)(x-y+z)-(y+z-x)(z-x-y)=(x+y-z)(x-y+z)+(y+z-x)(x+y-z)所以公因式是(x+y-z)

化简(y-x)(z-x)/(x-2y+z)(x+y-2z)+(z-y)(x-y)/(x-2z+y)(y+z-2x)+(x

∵x-2y+z=(x-y)-(y-z),x+y-2z=(y-z)-(z-x),y+z-2x=(z-x)-(x-y).设x-y=a,y-z=b,z-x=c,则原式=-ac/(a-b)(b-c)+(-ba

x,y,z正整数 x>y>z证明 x^2x +y^2y+z^2z>x^(y+z)*y^(x+z)*z^(x+y)

正整数?取对数即证:2xlnx+2ylny+2zlnz>(y+z)lnx+(x+z)lny+(x+y)lnzx>y>z,lnx>lny>lnz由排序不等式得xlnx+ylny+zlnz>ylnx+zl

matlab绘图三角函数 x=(1+cos(u))cos(u);y=(1+cos(u))sin(u);z=sin(u)怎

试试这样行不行;clear all;clc;u=0:pi/40:3*pi;x=(1+cos(u)).*cos(u);y=(1+cos(u)).*sin(u);z=sin(u);plot3(x

matlab的三维绘图,方程为x^2-y^2-4*(z^2)=4,只知道画出来是个双叶双曲面.

可用ezsurf绘制ezsurf('2*sec(u)','2*tan(u)*cos(v)','tan(u)sin(v)',[-pi/2,3*pi/2,0,2*pi])axisauto即将x^2-y^2

matlab二维绘图x=0:0.1*pi:4*pi;y=sin(x);z=cos(x);v=[0,15,-2,2];ax

x=0:0.1*pi:4*pi;y=sin(x);z=cos(x);v=[0,15,-2,2];axis(v)subplot(2,2,1);plot(x,y,'--r');xlabel

z/(x-y) × y/(x+y)

z/(x-y)×y/(x+y)=zy/(x-y)(x+y)=zy/(x²-y²)再问:还有两道题!麻烦你了!1.已知x-1/x=2,求x²/x四次方-x²+12

方向 X Y Z

X--水平横向方向;Y--水平竖向方向;Z--垂直竖向方向.

(x+y+z)^5-(x+y-z)^5-(x+z-y)^5-(z+y-x)^5,

(x+y+z)^5-(x+y-z)^5-(x+z-y)^5-(z+y-x)^5=80xyz(x^2+y^2+z^2)注:x^5,y^5,z^5之类的是被消掉了.我的结果100%是正确的,你再算算吧.朝

X+Y+Z=?

X+Y+Z

化简(y-x)(z-x)/(x-2y+z)(x+y-2z)+(z-y)(x-y)/(xy-2z)(y+z-2x)+(x-

第二个分母写错了?(y-x)(z-x)/(x-2y+z)/(x+y-2z)+(z-y)(x-y)/(x+y-2z)/(y+z-2x)+(x-z)(y-z)/(y+z-2x)/(x-2y+z)=1

f(x,y,z,w)=x*(x+y)*(x+y+z)*(x+y+z+w)

f=x+1f+u=2x+3f+u+c=3x+8f+u+c+k=4x+15f(f,u,c,k)=(x+1)(2x+3)(3x+8)(4x+15)

如何化简(x+y+z)(x+y-z)(x-y+z)(-x+y+z)

应用平方差公式a^2-b^2=(a+b)(a-b)(x+y+z)(x+y-z)(x-y+z)(-x+y+z)=[(x+y)^2-z^2][z^2-(x-y)^2]=-z^4+[(x+y)^2+(x-y

(x+y+z)(-x+y+z)(x-y+z)(x+y-z)怎么算

(x+y+z)(-x+y+z)(x-y+z)(x+y-z)=-[(x+y+z)(x+y-z)][(x-y+z)(x-y-z)]=-[(x+y)²-z²]*[(x-y)²-