若x+y+z=0,求x2+y2+z2-xy-xz-zy的值
由x+y+z=0得(x+y+z)2=0;于是可得x2+y2+z2+2(xy+yz+zx)=0;所以有xy+yz+zx=-(x2+y2+z2)\/2;则有:x2+y2+z2-xy-yz-zx=x2+y2+z2-(xy+yz+zx)=x2+y2+z2-[-(x2+y2+z2)\/2]=x2+y2+z2+(x2+y2+z2)\/2=3(x2+y2+z2)\/2 ...
若x+y+z=0,则x2+y2+z2=
如果x+y+z=0,则x2+y2+z2=0,x2是2x 答案就是0.
已知x+y+z=0,x2+y2+z2=1,求xy+yz+xz,x4+y4+z4的解
x+y+z=0,x2+y2+z2=1,所以(x+y+z)^2=X^2+Y^2+Z^2+2*(xy+yz+xz)即0=1+2*(xy+yz+xz)所以xy+yz+xz=-0.5 因为(xy+yz+xz)^2=X^2Y^2+Y^2Z^2+Z^2X^2+(x+y+z)*xy 所以(-0.5)^2= X^2Y^2+Y^2Z^2+Z^2X^2+0 X^2Y^2+Y^2Z^2+Z^2X^2=0.25...
分解因式x2+y2+z2-xy-yz-xz
x2+y2+z2-xy-yz-xz=1\/2(2x*2+2y*2+2z*2-2xy-2yz-2xz)=1\/2[(x-y)*2+(y-z)*2+(x-z)*2]
已知x+y+z=0,x2+y2+z2=1,则x(y+z)+y(x+z)+z(x+y)的值为___
∵(x+y+z)2=x2+y2+z2+2xy+2zx+2zy=0,x2+y2+z2=1,∴2xy+2zx+2zy=-1,则原式=xy+xz+xy+yz+zx+zy=2(xy+zx+zy)=-2.故答案为:-2.
设x+y+z=0,x2+y2+z2=10
解:x+y+x=0 x²+y²+x²=10 假设x=0 y=√(5)z=-√(5)假设x=√(5)y=0 z=-√(5)
当x+y+z=0,x的立方加y的立方加z的立方等于3xyz怎么证明
x+y+z=0 左右两边同时乘X2+y2+z2-xy-xz-yz (X2+y2+z2-xy-xz-yz)乘(x+y+z)=0 打开化简得 x的立方加y的立方加z的立方等于3xyz
当x+y+z=0,x的立方加y的立方加z的立方等于3xyz怎么证明
解答:由公式:x³+y³+z³-3xyz=﹙x+y+z﹚﹙x²+y²+z²-xy-yz-zx﹚得:x³+y³+z³-3xyz=0 ∴x³+y³+z³=3xyz。
联立方程组x+y+z=0,x2+y2+z2=1。求偏x\/偏z,偏y\/偏z
x+y+z=0 (1)x²+y²+z²=1 (2)(1)两边对z求偏导:∂x\/∂z+∂y\/∂z=-1 (3)(2)两边对z求偏导:x∂x\/∂z+y∂y\/∂z=-z (4)以 ∂x\/∂z、∂y\/∂z 为未知数求解方...
已知:x+y+z=1,x2+y2+z2=2,x3+y3+z3=3,试求:(1)xyz的值;(2)x4+y4+z...
(1)由条件可得 (x+y+z)2=x2+y2+z2+2(xy+yz+xz)=1,即 1=2+2(xy+yz+xz),∴xy+yz+xz=-12.再根据 x3+y3+z3-3xyz=(x+y+z)(x2+y2+z2-xy-yz-zx),即3-3xyz=2+12,∴xyz=16.(2)由题意可得 (x2+y2+z2)2=x4+y4+z4+2x2?y2+2y2?z2+2x2?