已知实数x,y,z,满足3^x=4^y

x+2y-z=6①x-y+2z=3②，①×2+②，得x+y=5，则y=5-x③，①+2×②，得x+z=4，则z=4-x④，把③④代入x2+y2+z2得，x2+（5-x）2+（4-x）2=3x2-18x

用x来表示y和z解方程组y-z=-x-y+2z=-3x两式相加得z=-4x把z=-4x代入y-z=-x中,得y=-5x所以x:y:z＝x:(-5x):(-4x)=1:(-5):(-4)或x:y:z＝-

三式相加：x+y+z+1/x+1/y+1/z=22/3三式相乘：xyz+y+x+1/z+z+1/x+1/y+1/xyz=28/3将1式代入2式得到xyz+22/3+1/xyz=28/3即：xyz+1/

实数x,y,z,满足那么x+y=6,z^2=xy-9,∴xy=z^+9,(x-y)^=(x+y)^-4xy=-4z^>=0,∴z=0,(x+y)^z=6^0=1.

上式等于(x-3y)^2+(y+2)^2++∣Z²-3Z+2∣=0由于上面三个都是非负数,所以这三个得解都是0解得x=3y=-6y=-2z=2或1得（X+Y)的Z次方的值=-8或64

x²-6xy+10y²+4y+|z²-3z+2|+4=0(x²-6xy+9y²)+(y²+4y+4)+|z²-3z+2|=0(x-

用中学方法还是大学方法?再问：中学再答：柯西不等式：(x²+y²+z²)(2²+3²+1²)≥(2x+3y+z)=1因此x²+y&

x-y=5x=5+yz^2=-xy-y-9=-(5+y)y-y-9=-y^2-6y-9=-(y+3)^2所以,z=0,y+3=0z=0,y=-3x=5+y=5-3=2x-2y+3z=2-2*(-3)+

x+2y-z=6所以2x+4y-2z=12因为x-y+2z=3两边相加3x+3y=15x+y=5带回去得到y=5-xz=4-x带回x^2+y^2+z^2=3x^2-18x+41=3(x^2-6x+9)

设x^2+y^2+z^2=t由3x+2y+2z=17得：y+z=(17-3x)/2又y^2+z^2=t-x^2可变得：yz=（17-3x)^2/8+(x^2-t)/2y,z可以看成m^2-[(17-3

x=5-yz2=(5-y)y+y-9=6y-y2-9=-(9-6y+y2)=-(y-3)2由题意,只有当该项为0时等式成立得y=3那么z=0x=2即原式=2+6+0=8

2x-3y-z=0..（1）x-2y+z=0...（2）（1）+（2）：3x-5y=03x=5yy=3/5x将y=3/5x代入（1）z=2x-3y=2x-3*3/5x=x/5x：y：z=x：3/5x：

x+3y=6,2xy-z^2=6考虑到z^2=2xy-6>=0x+3y=6==>x=6-3y2（6-3y）y-6>=0==>y^2-2y+1y=1带回去,所以x=3,z=0

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+

∵正实数x，y，z满足2x（x+1y+1z）=yz，∴x2+x（1y+1z）=12yz，∴（x+1y）（x+1z）=x2+x（（1y+1z）+1yz=12yz+1yz≥212=2．当且仅当yz=2，取

x+2y+3z=1的话,x=1/14;y=1/7;z=3/14三个数平方和最小值则为：1/14

|4x-4y+1|+1/3*√(2y+z)+(z^2-z+1/4)=0|4x-4y+1|+1/3*√(2y+z)+(z-1/2)^2=0则4x-4y+1=02y+z=0z-1/2=0解得z=1/2y=

移项,整理[(x-5)-4√(x-5)+4]+[(y-4)-4√(y-4)+4]+[(z-3)-4√(z-3)+4]=0[√(x-5)-2]²+[√(y-4)-2]²+[√(z-3

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-------------------------------------