matlab x平方 y平方 z平方=0 画图

原式=x(x²+xy+xz+yz)=x[x(x+y)+z(x+y)]=x(x+y)(x+z(

x+2y-9z=0,——（1）x-2y-5z=0——（2）（1）＋（2）2x-14z=0x=7z将x＝7z代入（1）7z+2y-9z=0y=z(2x^2+3y^2+7z^2)/(x^2-4y^2+4z

X+Y+Z=0（x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)=1+2(xy+yz+zx)=0xy+yz+zx=-1/2(xy+yz+zx)^2=x^2y^2+y^2z^2+z^2x

=[2(x+y+z)]^2-[3(x-y-z)]^2=[2(x+y+z)+3(x-y-z)]*[2(x+y+z)-3(x-y-z)]=(5x-y-z)(5y+5z-x)

(x+y+z)(-x-y+z)=z^2-(x+y)^2=z^2-x^2-2xy-y^2再问：确定吗？、再答：确定

根据均值不等式,只有当x=y=z时,他们的平方和有最小值.所以x=y=z=2,他们的平方和为12

x^2(y-z)+y^2(z-x)+z^2(x-y)=x^2y-x^2z+y^2z-y^2x+z^2x-z^2y=y(x^2-z^2)-xz(x-z)-y^2(x-z)=y(x+z)(x-z)-xz(

(z²-x²-y²)-4x²y²=(z²-x²-y²-2xy)(z²-x²-y²+2xy)

题目是否错了应该改为(x+y+z)^2-4(x-y-z)^2(x+y+z)^2-4(x-y-z)^2=[(x+y+z)-2(x-y-z)][(x+y+z)+2(x-y-z)]=(3Y+3Z-X)(3X

设a=x-y,b=y-z,-a-b=z-x(y-z)平方+(x-y)平方+(z-x)平方=(y+z-2x)平方+(z+x-2y)平方+(x+y-2z)平方b^2+a^2+(-a-b)^2=(-a-b-

(4x-3y)²-16y²=(4x-3y+4y)(4x-3y-4y)=(4x+y)(4x-7y)x²+2x(y-z)+(z-y)²=x²+2x(y-z

[(x+y)^2-z^2]/(x+y+z)=[(x+y+z)(x+y-z)]/(x+y+z)=x+y-z

1先把约束条件图像画出（课本上写得详细）,这个区域实在两条线之间的左上区域,不取等（描述的不科学,2.约束函数其实是一个圆（圆心是原点,半径是根z）的形式.问题就转化为,以条件区域内的点为圆上的点,以

题目有问题吧,我感觉应该是(x²+y²-z²)²-4x²y²(x²+y²-z²)²-4x²

(2x-3y-4z)的平方-(2x+3y-4z)的平方=(2x-3y-4z+2x+3y-4z)(2x-3y-4z-2x-3y+4z)=(4x-8z)(-6y)=48yz-24xy

解,设,x3=a,y3=b,z3=c.原式=（a+b）3+（c-a）3-（b+c)3令a+b=m,c-a=n原式=m3+n3-（m+n）3=-3mn（m+n）=-3（a+b）（c-a）（b+c）=-3

[x²-(y-z)²]/[(x+y)²-z²]=(x+y-z)(x-y+z)/(x+y+z)(x+y-z)=(x-y+z)/(x+y+z)再问：谢谢你！但是最终

感觉是你写错了,应该是下面的这个吧(x+y-z)²-(x-y+z)²=[(x+y-z)-(x-y+z)][(x+y-z)+(x-y+z)]=(x+y-z-x+y-z)(x+y-z+

(x+y+z)²=[(x+y)+z]²=(x+y)²+z²+2(x+y)z=x²+y²+z²+2xy+2yz+2zx

(x+y-z)(x+y+z)=(x+y)^2-z^2=x^2+2xy+y^2-z^2（a-b+c）（a+b+c)=[(a-b)+c][(a+b)+c]=(a-b)(a+b)+c(a-b)+c(a+b)