z=x^2-xy y^2-2x

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

1／x＝p1／y＝q1／z＝rpq+qr+pr=1(y+x)/z+(y+z)/x+(z+x)/y≥2(1/x+1/y+1/z)^2为(pq+qr+pr)[r/p+r/q+q/r+q/p+p/r+p/q

原式=[x-y(x-y)2-y(x+y)(x+y)(x-y)]•xyy-1=（1x-y-yx-y）•xyy-1=1-yx-y•xyy-1=-xyx-y．故答案是：-xyx-y．

∑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

我去,这什么题啊

x+y-z=6y+z-x=2z+x-y=0三式相加得x+y+z=8-得2z=2z=1-得2x=6x=3-得2y=8y=4x=3y=4z=1

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-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

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

题目是√x^3+X^2y+1/4xy+√(1/4x^3)-X^2y+xy^2如果是：√x^3+X^2y+1/4xy+√(1/4x^3)-X^2y+xy^2=(3/2)√x^3+xy/4+xy^2=(3

原式=[(x+y)2(x-y)(x+y)+-4xy(x-y)(x+y)]×(x+3y)(x-3y)(x+3y)(x-y)=x-3yx+y，由已知得（3x-2y）（x+y）=0，因为x+y≠0，所以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-

x^2/C1+y^2/C2=1两边对x求导：2x/c1+2yy'/c2=0x/c1=-yy'/c2(yy')/x=-c2/c1两边对x求导：[(y'^2+yy'')x-yy']/x^2=0xyy''+

产生4种配子xyxyyyx有一种y有两种xy1y2xy1xy2y1y2产生的配子如上y1、y2是一个意思所以分为一组xyxyyy=1221

将(x+y+z)²展开有(x+y+z)²=x²+y²+z²+2xy+2xz+2yz=x²+y²+z²所以2xy+2xz+

x=1,y=0,z=9首先x、y、z都是个位数xyy可以写成100x+10y+y同理,zz可以写成10z+zyx写成10y+x等式重新代入以上化解后的式子,就是：100x+11y-11z=10y+x合

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

3元一次方程,好像是初一的问题哦.根据前面两个等式可以得出x=3zy=z(平方)/32x+3y+4z=2*(3z)+3*(z方/3)+4z现在变成了一元二次方程,你应该会解吧.

设x/2=y/3=z/5=ax=2ay=3az=5a是不是求的是：(x+3y-z)/(x-3y+z)?若是,如下：(x+3y-z)/(x-3y+z)=(2a+9a-5a)/(2a-9a+5a)=-3

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