x^2 y^2 z^2=3xyz

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

x:y=2:3=:6;9x:y:z=6:9:10则x＝6/（6+9+10）*50＝12y＝9/（6+9+10）*50＝18z＝10/（6+9+10）*50＝20xyz＝12*18*20＝4320

由|3x-2y+z|≥0,|2x+y+2z|≥0,且|3x-2y+z|+|2x+y+2z|=0,得|3x-2y+z|=|2x+y+2z|=0∴3x-2y+z=2x+y+2z=0由3x-2y+z=2x+

3xyz+2(x^2y+y^2z-xyz)-xyz+2z^2x原式=3xyz+2（x²y+y²z+z²x）-3xyz=2（x²y+y²z+z²

4x-y+3z=0(1)2x+y+6z=0(2)()+(2)6x+9z=06x=-9zz/x=-2/3(1)*2-(2)8x-2y-2x-y=06x-3y=06x=3yx/y=1/2z/x=-2/3x

x²+y³-xyz=0,z=（x²+y³）/（xy）=x/y+y²/x；故z/x=1/y+y²/x²z/y=x/y²+y

不妨用特殊代入法啊令a=b=c=0或者a=1,b=-1,c=0结果都是x^3+x^2z-xyz+y^3=0

因为：X+Y+Z=0得：Z+Y=-X------(1)X+Y=-Z------------(2)Z+Y=-X------------(3)X^3+X^2Z-XYZ+Y^2Z+Y^3=X^3+XZ(X+

=(x+y+z)^2+yz(y+z+x)=(x+y+z)(x+y+z+yz)

原式=2x^3-xyz-2x^3+2y^3-2xyz+xyz-2y^3=-2xyz=-2×(-1)×(-2)×(-3)=12

柯西【x^2/(y+z)+y^2/(x+z)+z^2/(x+y)】*（y+z＋x+z＋x+y）≥（x+y+z）^2即x^2/(y+z)+y^2/(x+z)+z^2/(x+y)≥（x+y+z）/2＝（3

1、隐函数对x求导得1+az/ax+yz+xy*az/ax=0,故az/ax=－(1+yz)/(1+xy)；F对x求导得aF/ax=e^x*y*z^2+e^x*y*2z*az/ax；当x＝0,y＝1时

由2x+3y-3z=0得：z-y=2x/3(2x+y-z)/(2x-y+z)=(2x-（z-y）)/(2x+（z-y）)将z-y=2x/3代入上式得：(2x+y-z)/(2x-y+z)=(2x-（2x

1、{x+y+z=301){3x+y-z=502){5x+4y+2z=403)1)+2)得：2x+y=404)3)-1)×2得：3x+2y=-205)4)×2-5)得：x=1006)6)代入5)得：y

是指所构造的方程存在实数解时,其判别式△不小于0.再问：：t^2-(y+z)t+yz=0这个是什么意思再答：题目抄错了，应当是证明x²≥3.利用韦达定理啊!依条件式知：yz＝x²,

4x-3y+z=0(1)x+2y-8z=0(2)(1)-(2)×4得-11y+33z=0∴y=3z把y=3z代入(2)得x=2z把x=2z,y=3z代入x+y-z/x-y+2z得原式=(2z+3z-z

25xy^2z^2(x+y-z)-30xyz(z-x-y)^2+5xyz^3(z-x-y)=25xy^2z^2(x+y-z)+30xyz(x+y-z)^2-5xyz^3(x+y-z)=5xyz(x+y

(x+1)^2+|y-1|+|z|=0(x+1)^2=0x+1=0x=-1y-1=0y=1z=0A=2x^3-xyz=2*（-1）^3-0=-2B=y^3-z^3+xyz=1^3-0+0=1C=-x^

x:y:z=2:3:4=4:6:8x+y+z=18x=4y=6z=8xyz=4x6x8=192

√x+y-2011+√2011-x-y要成立则x+y≥2011x+y≤2011最终：x+y=2011所以等号右边为0,则左边也为03x+y-z-2=02x+y-z=0则x=2,y=2011,z=201