已知x,y和z,怎样拟合出z(x,y)

(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

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

方程组两边除以z得4x/z-3y/z=3x/z-3y/z=-1解方程组得x/z=4/3y/z=7/9

x=z(lnz-lny)=zlnz-zlny令F(x,y,z)=zlnz-zlny-xaF/ax=-1aF/ay=-z/yaF/az=lnz+1-lny所以az/ax=-Fx/Fz=1/(lnz+1-

首先,描点；其次,观察点的走向,看看用符合什么曲线,并假设出曲线方程；最后,拟合出曲线方程中的参数.\(^o^)/~再问：具体函数？再答：数据？

2x-3y-4z=01式x+y+z=02式1式+2式×4得到：2x-3y-4z+4x+4y+4z=06x+y=06x=-yx:y=(-1):61式-2式×2得到：2x-3y-4z-2x-2y-2z=0

亲,题目看似很麻烦,仔细想还是有思路的.解:由4x-3y-6z=0,(1式)x+2y-7z=0(2式)(2式)*4得:4x+8y-28z=0(3式)(3式)-(1式)得:4x+8y-28z-(4x-3

(x-4)的平方+x+y+z+的绝对值=0,(x-4)的平方>=0,x+y+z+的绝对值>=0,x-4=0x+y+z=0x=4y+z=-42x+3y+3z=8+3(y+z)=-4

因为x/y+z+y/z+x+z/x+y=1所以x/y+z=1-y/z+x-z/x+y,两边同乘以x得x^2/y+z=x-xy/z+x-xz/x+y同理y^2/x+z=y-xy/z+y-yz/x+y,z

两式相减(后面的减前面的)x+3y=105所以2x+6y=210第一个式子减上式x+y+z=105

设x+y-z/z=x-y+z/y=y+z-x/x=k有x+y-z=kzx-y+z=kyy+z-x=kx三式相加得x+y+z=k（x+y+z）k=1得x+y=（k+1）zx+z=（k+1）yy+z=（k

根据题意得，4x-4y+1=0，2y+z=0，z-12=0，解得x=-12，y=-14，z=12，∴x+z-y=-12+12-（-14）=14，∴x+z−y=14=12．故答案为：12．

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

等于0.x/(y+z)=1-[y/(z+x)+z/(x+y)]y/(z+x)=1-[x/(y+z)+z/(x+y)]z/(x+y)=1-[x/(y+z)+y/(z+x)]x2/(y+z)+y2/(z+

应该求（x+y+z）/y(x+y+z)/(y-x)已知x/3=y/5=z/4x/y=5/3z/y=5/4x/z=4/3代进去就好了比如第一个X/Y+1+Z/Y=5/3+1+4/3=4再问：x/y=5/

X-Z=10X+Z=70X=40Z=30

x+y+z+3x+y-z=80.解得y=40-2x,x+y+z-(3x+y-z)=-20,解得z=x-10,因为x,y,z均为非负数,则y=40-2x＞0,z=x-10＞0,x＞0.解得10＜x＜20

两式相加,得6X-5Z=0即X=5Z/6,即X/Z=5/6.再将X=5Z/6代入式1,得5Y+11Z/2=0得Y/Z=-11/10

6x-2y-6z=0x-6y+6z=0解得x=24y/21,z=17y/21代入x+y+z/x-y+z=62/20=31/10-------------------------------------

已知x乘以y等于z（z不等于0）那么：当z一定时,x和y成：反比例.