x*y-z 汇编

(x+y-z)/z=(y+z-x)/x=(z+x-y)/y[x+y]/z-1=[y+z]/x-1=[z+x]/y-1[x+y]/z=[y+z]/x=[z+x]/y设[x+y]/z=[y+z]/x=[z

[x+(z-y)][x-(z-y)]=x-(z-y)记得采纳啊

∑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

x/(y+z)=y/(x+z)=z/(x+y)当x+y+z=0时,x+y=-z(x+y)/z=-z/z=-1当x+y+z≠0时,由x/(y+z)=y/(x+z)=z/(x+y)根据等比性质可得(x+y

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

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

(x+y-z)(x-y+z)-(y+z-x)(z-x-y)=(x+y-z)(x-y+z)+(y+z-x)(x+y-z)所以公因式是(x+y-z)

∵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

z/(x-y)×y/(x+y)=zy/(x-y)(x+y)=zy/(x²-y²)再问：还有两道题！麻烦你了！1.已知x-1/x=2，求x²/x四次方-x²+12

X--水平横向方向；Y--水平竖向方向；Z--垂直竖向方向.

(y+z-2z)这步有错吧

(x+y+z)^5-(x+y-z)^5-(x+z-y)^5-(z+y-x)^5=80xyz(x^2+y^2+z^2)注:x^5,y^5,z^5之类的是被消掉了.我的结果100%是正确的,你再算算吧.朝

DATASSEGMENTxdw11;此处变换X的取值ydw?DATASENDSCODESSEGMENTASSUMECS:CODES,DS:DATASSTART:MOVAX,DATASMOVDS,AXm

嘿嘿俺来帮帮你吧1先要定义一个数据段定义3个一维数组长度分别为5字节2用循环结构实现2个数组的加法运算循环5次完成数组中5个元素的相加并将结果存入另一个数组保存先自己动手做一做吧有疑问可以找我再问：有

X+Y+Z

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

f=x+1f+u=2x+3f+u+c=3x+8f+u+c+k=4x+15f(f,u,c,k)=(x+1)(2x+3)(3x+8)(4x+15)

应用平方差公式a^2-b^2=(a+b)(a-b)(x+y+z)(x+y-z)(x-y+z)(-x+y+z)=[(x+y)^2-z^2][z^2-(x-y)^2]=-z^4+[(x+y)^2+(x-y

（x+y+z)(-x+y+z)(x-y+z)(x+y-z)=-[(x+y+z)(x+y-z)][(x-y+z)(x-y-z)]=-[(x+y)²-z²]*[(x-y)²-