矩阵matlab x^2 y^2 z^2=16

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(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

设矩阵A+=(1 x 0,2 y 0,3 z 1),且矩阵A与矩阵B相似,矩阵B的特征值为1,2,3,则x.y.z各等于

相似矩阵行列式值相等;主对角线元素之和相等[1x0][2y0][3z1]1+y+1=1+2+3;所以y=4;|A|=y-2x=|B|=6;所以x=-1;再计算|E-A|=0;可以算出z

用矩阵解方程组15x + 4y + 3z = 22x + 3y + 4z = 52,求a值2x - y + 3z = 4

5x+4y+3z=254322x+3y+4z=523452,求a值2x-y+3z=42-134x+y+az=311a3x+ay+z=31az3

试证明(x+y-2z)+(y+z-2x)+(z+x-2y)=3(x+y-2z)(y+z-2x)(z+x-2y)

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

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

x2-(y+z)2=(x+y+z)(x-y-z)=8x+y+z=2所以x-y-z=8÷2=4

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

在R^3上定义的线性变换T(x,y,z)=(x,x+2y,y-z),求T在另一组基(x1,x2,x3)下的矩阵

T(1,1,1)=(1,3,0)=x1+2x2-3x3T(0,1,1)=(0,2,0)=2x2-2x3T(0,0,1)=(0,0,-1)=-x3故所求矩阵为100220-3-2-1

因式分解 (x+y+z)^2+yz(y+z)+xyz

=(x+y+z)^2+yz(y+z+x)=(x+y+z)(x+y+z+yz)

方程组用逆矩阵来解?x+2y+z=0;2x+9y+2z=0;3x+y+z=-2;

x=-1,y=0,z=1再问:b是什么意思?A怎么变成的A^-1?再答:b是方程组右边的常数项组成的列向量,称为增广矩阵。求矩阵的逆一般是用初等行变换(A,E)~(E,A^-1),刚开始学的时候也会用

求用矩阵解三元方程x+y-z=3-z+y+2x=42y+z+2x-12=0suppose inverse of the

这个题很简单啊,上面的三元方程组看成Ax=bA是系数矩阵,b=(3,4,12)TAx=b两边左乘A^(-1)x=A^(-1)b=|-110||4/3-11/3|*(3,4,12)T|-2/301/3|

①(2y-z)²[2y(z+2y)+z²]²=

解①原式={(2y-z)×[2y(z+2y)+z²]}²={(2y-z)[(2y)²+2yz+z²]}²=[(2y)³-z³]&#

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

4X^2-8X+1+8X^2+8X=12X^2-48X+48+348X=54X=9/8

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

分式约分:(y+z-x)/{x^2-(y+z)^2}

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

分解因式:f(x,y,z)=x^2(y-z)+y^2(z-x)+z^2(x-y)

=x²(y-z)+y²(z-x)+z²(x-z+z-y)=(y-z)(x²-z²)+(z-x)(y²-z²)=(y-z)(x-z)

设x、y、z为整数,证明:x^4*(y-z)+y^4*(z-x)+z^4*(x-y)/(y+z)^2+(z+x)^2+(

x^4(y-z)+y^4(z-x)+z^4(x-y)=xy(x^3-y^3)+yz(y^3-z^3)+zx(z^3-x^3)=xy(x^3-y^3)+yz(y^3-z^3)-zx[(x^3-y^3)+

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

根据公式(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ac公式展开:得到(x^2+y^2+z^2=2xy-2yz-2xz)-(x^2+y^2+z^2-2xy-2yz+2xz)合并同类项