z 3=x² 4 y² 9 画

∵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

a^3+b^3=(a+b)(a^2-ab+b^2)a^3-b^3=(a-b)(a^2+ab+b^2)(x+y+z)^3-x^3-y^3-z^3=(y+z)[(x+y+z)^2+(x+y+z)x+x^2

以单式来考虑,执行各式计算后:1.x=10;2.x=5;3.y=21,x=4;4.z=23,y=11,x=4

系统给的缺省坐标系与基准面,TOP垂直Z,FRONT垂直.为了需要可建新坐标系,根据基准面.定向..也可根据坐标系建基准面.

∵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

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（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

(x+y+z)^3-x^3-y^3-z^3=3yx^2+3xy^2+3xz^2+3yz^2+3zx^2+3zy^2+6xyz=3xy(x+y)+3z^2(x+y)+3z(x^2+y^2+2xy)=3x

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)

设X反应了KMOL则Y反应了2KMOL生成了Z2MOL又因为n(X)+n(Y)=n(z)所以a-k+b-2k=2kk=(a+b)/5所以Y反应了2(a+b)/5所以转化率为选B

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

应该证z3∈R.z3=(3+4i)z1+(3-4i)z2=(3+4i)(x+yi)+(3-4i)(x-yi)=3x-4y+(4x+3y)i+3x-4y-(4x+3y)i=6x-8y∈Zx^2+y^2=

x3（y-z）+y3（z-x）+z3（x-y）=x3（y-z）+y3（z-x）-z3（y-z）-z3（z-x）=（x3-z3）（y-z）+（y3-z3）（z-x）=（x-z）（y-z）（x2+xz+z

x^3+y^3+z^3+(x+y)^3+(y+z)^3+(z+x)^3=[x^3+(y+z)^3]+[y^3+(z+x)^3]+[z^3+(x+y)^3]=(x+y+z)(x^2-xy+y^2-xz+

因为x^3+y^3+z^3-3xyz=(x+y)^3-3x^y-3xy^2+z^3-3xyz（把x^3+y^3写成(x+y)^3-3x^2y-3xy^2）=[(x+y)^3+z^3]-(3x^2y+3

1:已知a^m=2,a^n=6求a^2m+3na^2m+3n=a^2m*a^3n=(a^m)^2*(a^n)^3=2^2*6^3=4*216=8642:已知a^x+y=6,a^x+y+z=12求a^z

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,