x^2+y^2+z^2=(x+y+z)^2-2(xy+yz+zx)=5^2-2*7=25-14=11

来源：蜘蛛抓取(WebSpider) 时间：2015-01-19 22:56 标签： xy x y 1 yz y z 2
已知x+y+z=2,xy+yz+zx=-5,求x²+y²+z²的值_百度作业帮
已知x+y+z=2,xy+yz+zx=-5,求x²+y²+z²的值
(x+y+z)^2=4x^2+y^2+z^2+2xy+2xz+2yz=4x^2+y^2+z^2+2(-5)=4x^2+y^2+z^2=14
(x＋y＋z)²=x²＋y²＋z²＋2xy＋2yz＋2xz4=(x²＋y²＋z²)－10x²＋y²＋z²=14
(x+y+z)^2=x^2+y^2+z^2+2xy+2yz+2xzx²+y²+z²=(x+y+z)^2-2(xy+yz+zx)=4+10=14=(x+y+z）/3">
X,y,z是实数,求证x^+Y^十Z^>=(x+y+z）/3_百度作业帮
X,y,z是实数,求证x^+Y^十Z^>=(x+y+z）/3
[[[[[[[[注:
大哥,题目错了,当x=y=z=0.1时,左边=0.03,
哈哈.]]]]题目:求证x²+y²+z²≥(x+y+z)²/3证明:由基本不等式可得x²+y²≥2xyy²+z²≥2yzz²+x²≥2zx三式相加,整理可得x²+y²+z²≥xy+yz+zx上面结果的两边同乘以2, 再同加x²+y²+z².可得3(x²+y²+z²)≥x²+y²+z²+2xy+2yz+2zx=(x+y+z)²∴x²+y²+z²≥(x+y+z)²/3
证：∵(x-y)^2+(y-z)^2+(z-x)^2+4(x^2+y^2+z^2)>=0∴6x^2+6y^2+6z^2-2（x+y+z)≥0∴x^2+y^2+z^2≥(x+y+z)/3解方程组：2+y^2+z^2=xy+yz+zx x+y+z=2004】_百度作业帮
解方程组：2+y^2+z^2=xy+yz+zx x+y+z=2004】
x2+y2+z2=xy+yz+xz2x2+2y2+2z2=2xy+2yz+2zx(x-y)^2+(y-z)^2+(z-x)^2=0x=y=z3x=2004x=y=z=668
您可能关注的推广回答者：xyz=1 x+y+z=2,x^2+y^2+z^2=3,1/(xy+z-1)+1/(xz+y-1)+1/(yz+x-1)=_百度作业帮
xyz=1 x+y+z=2,x^2+y^2+z^2=3,1/(xy+z-1)+1/(xz+y-1)+1/(yz+x-1)=
由已知条件：x+y+z=2x^2+y^2+z^2=3所以xy+yz+zx=(1/2)[(x+y+z)^2-(x^2+y^2+z^2)]=1/2又因为左式第一项1/(xy+z-1)=1/[xy+(2-x-y)-1]=1/[(x-1)(y-1)]同理1/(yz+x-1)=1/[(y-1)(z-1)]1/(zx+y-1)=1/[(z-1)(x-1)]三式相加（此时通分便很简单）得：(3-x-y-z)/[(1-x)(1-y)(1-z)]1/[(1-x)(1-y)(1-z)]=1/(1-x-y-z+xy+yz+zx-xyz)=1/(1-2+1/2-1)=-2/3

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