xcxd.net
当前位置:首页 >> 已知xyz都是正数,则x2+y2+z2/2xy+yz的最小值 >>

已知xyz都是正数,则x2+y2+z2/2xy+yz的最小值

由于1=x2+y2+z2=(x2+23y2)+(13y2+z2)≥223xy+213yz=233(2xy+yz),∴2xy+yz≤1233=32,当且仅当x=23yz=13y时取等号,则2xy+yz的最大值为 32故答案为:32.

M=x^2+2y^2+z^2-2xy-8y+2z+17 =x^2-2xy+y^2+y^2-8y+16+z^2+2z+1 =(x-y)^2+(y-4)^2+(z+1)^2 由上述公式可知,M≥0. M+x^2≤7 x^2≤7-M 因M≥0,故0≤x^2≤7. 又因为x,y,z都为非负整数,故在0-7之中x只能有x只有三个可能的值,即0,1,2.而X为偶数,故x取 0,2两...

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

上面的等号应该是大于等于吧。将X视为未知数只需正判别式非负。我记得这个不等式等价于厄尔多斯定理

x^2+y^2+z^2-2xy+2yz+2zx =(x^2-2xy+y^2)+2(x+y)z+z^2 =(x-y)^2+2(x+y)z+z^2 =(x-y+z)^2

真快啊楼上,打这么多字就用了这么几秒...-_-## 楼主 你看懂了没? 我帮你翻译一下 若 x^2+y^2+z^2≥2yzcosA+2xzcosB+2xycosC 则 z^2 -(2ycosA + 2xcosB)z +x^2+y^2-2xycosC≥0 上式为关于z的一元二次不等式 令 f(z)=z^2 -(2ycosA + 2xcosB)z +x^2...

5x2+2y2+2z2+2xy+2yz-4xz-6y-4z+6=0?(x2+2y2+2xy)+(y2+z2+2yz)+(4x2+z2-4xz)-6z-4z+6=0?(x+y)2+(y+z)2+(2x-z)2-6y-4z+6=0设上式可分转化成x、y、z一次方因式平方和的形式,即(x+y+a)2+(y+z+b)2+(2x-z+c)2=0?(x+y)2+(y+z)2...

x2+y2+z2≥2xycosC+2yzcosA+2zxcosB x²+y²+z²-x(2ycosC+2zcosB)-2yzcosA≥0 (x-ycosC-zcosB)²+y²+z²-2yzcosA-(ycosC+zcosB)²≥0 (x-ycosC-zcosB)²+y²sin²C+z²sin²B-yz(2cosA+2cosCcos...

x+4y-3z=0 (1) 4x-5y+2z=0 (2) (1)*4-(2)得 21y=14z y=2z/3 代入(1)得 x=z/3 (3x²+2xy+z²)/(x²+y²) =z²(1/3+4/9+1)/[z²(1/9+4/9)] =16/5

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

网站首页 | 网站地图
All rights reserved Powered by www.xcxd.net
copyright ©right 2010-2021。
内容来自网络,如有侵犯请联系客服。zhit325@qq.com