x y%=(x z)*2% z:x=(y-z):2

(x+y+z)²=1,x²+2xy+y²+2(x+y)z+z²=1,x²+y²+z²+2(x+y)z+2xy=1xy+yz+xz=

x-3=y-2x-y=1y-2=z-1y-z=1x-3=z-1z-x=-2x^2+y^2+z^2-xy-yz-xz=x(x-y)+y(y-z)+z(z-x)=x+y-2zx-3=z-1y-2=z-12

记√x=a,√y=b,√z=c,代入原方程得：a^2bc+b^2ac+a^2b^2=39-->ab(ab+ac+bc)=39b^2ac+c^2ab+b^2c^2=52-->bc(ab+ac+bc)=5

x^2+y^2+z^2-xy-yz-xz=0(1/2)*2(x^2+y^2+z^2-xy-yz-xz)=0(1/2)*(x^2+y^2-2xy+z^2+y^2-2zy+x^2+z^2-2xz)=0(x

可是X+Y+Z=2,XY+YZ+XZ=-5,求X^2+Y^2+Z^2（X+Y+Z）^2=X^2+Y^2+Z^2+2(XY+YZ+XZ)=-6

(x+y+z)²=1²x²+y²+z²+2xy+2yz+2xz=1x²+y²+z²+2(xy+yz+xz)=1x&sup

thedragon53的错了,(1)-(2)得2xz-yz=4,而不是2xz+yz=4正确的做法：xy=xz+3.①,yz=xy+xz-7.②（x,y,z均为正整数）由①得到y=z+3/x,由于x,y

2^x=10^z所以(2^x)^y=(10^z)^y2^(xy)=10^yz5^y=10^z(5^y)^x=(10^z)^x5^xy=10^xz所以2^xy*5^xy=10^yz*10^xz(2*5)

证明命题错误满足xy=xz=yz必须要x=y=z带如原式显然不成立

令x/3=y/2=z/5=k则x=3ky=2kz=5k∴(xy+yz+zx)/(x²+y²+z²)=(6+10+15)k²/(9+4+25)l²=31

该题可以进行图形辅助解析由x²+y²+xy=25/4x²+z²+xz=169/4y²+z²+yz=36=144/4  &

左式可化为[(xy)^3+(xz)^3+(yz)^3]/xyz+6xyz;然后[(xy)^3+(xz)^3+(yz)^3]/xyz>=3xyz(这一步是将分子利用(a+b+c)>=3*(abc)^(1

题目是这样吧1=xy/(x+y),2=yz/(y+z),3=xz/(x+z)倒数法,写成每个式子的倒数；1=1/x+1/y,(1)1/2=1/y+1/z,(2)1/3=1/x+1/z(3)三式相加,得

假设x,y,z>0.那么由算数几何不等式推出sqrt[3]{xyz}=3*sqrt[3]{x/y/z*y/z/x*z/x/y}=3*sqrt[3]{1/xyz}.把(1)代入上式,就得到左边>=3*3

（x+y+z）^2=x^2+y^2+z^2+2(xy+yz+xz)=25x^2+y^2+z^2=25-14=11

x^2+y^2+z^2=xy+yz+xz2(x^2+y^2+z^2)=2(xy+yz+xz)x^2+y^2-2xy+x^2+z^2-2xz+y^2+z^2-2yz=0(x-y)^2+(y-z)^2+(

x^2+y^2+z^2=xy+yz+xz2(x^2+y^2+z^2)=2(xy+yz+xz)x^2+y^2-2xy+x^2+z^2-2xz+y^2+z^2-2yz=0(x-y)^2+(y-z)^2+(

答案是：(2*X)/((X-Z)*(X+Z))再问：解题过程给我写下1再答：=（2X+Z-Y)/[(x-y)(x+z)]-(y-z)/[(x-z)(x-y)]=[(2x+z-y)(x-z)-(y-z)

令2/x=3/y=7/z=k∴x=2/ky=3/kz=7/k∴(xy+xz+yz)/(x^2+y^2+z^2)=(2/k*3/k+2/k*7/k+3/k*7/k)/(4/k²+9/k

再答：再问：范围不对o再答：？再问：范围不对哦再问：【-1，2】再答：再答：能看清吗？