直线l X 2 -2=Y 3 -7=Z 3与平面4x-2y-2z=0的位置关系是
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∵y3-z3=(y-z)(y2+yz+z2)(立方差公式)又∵y3-z3-y2-yz-z2=0∴(y-z-1)(y2+yz+z2)=0(提取公因式)∵y、z是正实数∴y-z-1=0即y-z=1∵x-y
x+y+z=0,z=-(x+y)代入x^3+y^3+z^3=0x^3+y^3-(x+y)^3=x^3+y^3-x^3-3X^2y-3xy^2-y^3=-3xy(x+y)=3xyz=0所以xyz=03a
∵x+y+z=0,∴z=(-x-y)x^3+y^3+z^3=x^3+y^3-(x+y)^3=x^3+y^3-x^3-y^3-3x^2y-3xy^2=-3xy(x+y)=3xyz
因为:X3-Y3-Z3=3XYZ所以:X3+(-Y)3+(-Z)3-3X(-Y)(-Z)=0(X-Y-Z)(X2+Y2+Z2+XY+XZ-YZ)=0所以:1.X-Y-Z=02.X2+Y2+Z2+XY+
由(x+y+z)2-(x2+y2+z2)可得xy+xz+yz=-5x3+y3+z3-3xyz=(x+y+z)(x2+y2+z2-xy-yz-zx)可得xyz=14再问:谢谢,我看一下其他的答案在采纳再
(x+y+z)²-(x²+y²+z²)=2(xy+yz+zx)=-1,xy+yz+zx=-1/2x3+y3+z3=3xyz+(x+y+z)(x²+y&
∵(x+y+z)(x²+y²+z²)=x³+y³+z³+x²(y+z)+y²(x+z)+z²(x+y)∴1*2
你题目里Z1^2+Z2^2+Z3^2=Z1*Z2+Z2*Z3+Z3*Z1的Z1,Z2,Z3是什么啊?我按自己理解来就当是边长了;(1)正推,因为是正三角形,所以Z1=Z2=Z3,所以等式左边=3*Z1
|AB|=√[(X2-X1)^2+(Y2-Y1)^2+(Z2-Z1)^2],|AC|=√[(X3-X1)^2+(Y3-Y1)^2+(Z3-Z1)^2],|BC|=√[(X3-X2)^2+(Y3-Y2)
如果你的X2是x的平方,X3是x的三次方那么答案是:-(x-y+z)*(x-y-z)*(x+y-z)
请想想直线方程通式y=kx+b三个点都在直线上,分别代入方程5=3k+b-------b=5-3k7=kx2+b-------kx2=7-5+3k=2+3k-----k=2----x2=4y3=-1k
7-5=2(x2-3),x2=4y3-5=2(-1-3),y3=-3
A*B=x1y1z1x2+2x3y2+2y3z2+2z3x2-2x3y2-2y3z2-2z3B*A=x1y1+z12y1-2z1x2y2+z22y2-2z2x3y3+z32y3-2z3
x3+y3-z3+3xyz,=[(x+y)3-3x2y-3xy2]-z3+3xyz,=[(x+y)3-z3]-(3x2y+3xy2-3xyz),=(x+y-z)[(x+y)2+(x+y)z+z2]-3
错题,我做不出来.做不出来的题都是错题,
x^3+y^3+z^3-3xyz=[(x+y)^3-3x^2y-3xy^2]+z^3-3xyz=[(x+y)^3+z^3]-(3x^2y+3xy^2+3xyz)=(x+y+z)[(x+y)^2-(x+
根据√x/y+√y/z+√z/xx,y,z应全>0或全0将题中两式相减得:x^3-x^2+y^3-y^2+z^3-z^2=1(x-1)x^2+(y-1)y^2+(z-1)z^2=1因为x,y,z>0,