z=1 (2-x*x-y*y)的间断点为

根据柯西不等式(x^2+y^2+z^2)(1+4+16)≥(x+2y+4z)^2=1(x^2+y^2+z^2)*21≥1x^2+y^2+z^2≥1/21所以最小值为1/21

左-右,以xyz为分母进行通分,化简合并后,得分子：z(x-y)^2+x(y-z)^2+y(z-x)^2分母：xyz除成3个式子:(x-y)^2/xy+(y-z)^2/yz+(z-x)^2/xz利用x

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-1)+√(z-2)=1/2(x+y+z)变形后得[x-2√x+1]+[(y-1)-2√(y-1)+1]+[(z-2)-2√(z-2)+1=0即（√x-1）^2+[√(y-1)+1]^2+

因为x+2y+4z=1所以2x+4y+8z=2所以x^2+y^2+z^2=x^2+y^2+z^2+2-2=x^2+y^2+z^2+2x+4y+8z-2=x^2+2x+1-1+y^2+4y+4-4+z^

x/(y+z)=y/(x+z)=z/(x+y)当x+y+z=0时,x+y=-z(x+y)/z=-z/z=-1当x+y+z≠0时,由x/(y+z)=y/(x+z)=z/(x+y)根据等比性质可得(x+y

x+y-z=6y+z-x=2z+x-y=0三式相加得x+y+z=8-得2z=2z=1-得2x=6x=3-得2y=8y=4x=3y=4z=1

2x-y+4z=8.1x-2y-z=7.21+2,得：3x-3y+3z=15即：x-y+z=5

x=2,y=3z=3

因为x/y+z+y/z+x+z/x+y=1所以x/y+z=1-y/z+x-z/x+y,两边同乘以x得x^2/y+z=x-xy/z+x-xz/x+y同理y^2/x+z=y-xy/z+y-yz/x+y,z

因为x:y:z=3：4：5所以设x=3k,y=4k,z=5k(k≠0)(1)z/(x+y)=5k/(3k+4k)=5k/7k=5/7(2)x+y+z=63k+4k+5k=612k=6k=1/2x=3k

由|X-3|+|Y-2|+|Z-1|=0可知X-3=0,Y-2=0,Z-1=0得X=3,Y=2,Z=1所以3x+2y+Z=3×3+2×2+1=14

(x+y+z)²+(x-y-z)(x-y+z)-2·z(x+y)=(x+y)²+2z(x+y)+z²+(x-y)²-z²-2z(x+y)=(x+y)&

x：y：z=1：2：3,x=k,y=2k,z=3kx+y+z=k+2k+3k=6k=12k=2x=2,y=4,z=6

x+y大于等于2倍根号下xy同理x+z大于等于2倍根号下xzz+y大于等于2倍根号下zy所以(x+y)(y+z)(z+x)大于等于8xyz当取到8xyz时分数值最大为1/8此时x=1/3y=1/3z=

这是线性规划的题目根据x-y≥-1,x+y≥1,2x-y≤1画图可以发现满足的点位于三条直线所围成的三角形内.x-y=-1与2x-y=1交点为x=2,y=3即交点为(2,3)Z=(x-2y)/(x+y

x+2y=32y=3zx-y=-1x+2y=3①2y=3z②x-y=-1③①-③得3y=4,得y=4/3代入③,得x=y-1=1/3代入②,得z=2/3y=8/9x+y+z=36x-y=12x+z-y

2x+y-3z=1,①x-2y+z=6,②3x-y+2z=9③①+③得：5x-z=10④①×2+②得：5x-5z=8⑤④-⑤得：4z=2∴z=1/2x=21/10=2.1y=-1.7

2z=1-x-yx＾2＋y＾2＋3－3x－3y＋1．5＝0（x－1．5）＾2＋（y－1．5）＾2＝0x＝y＝1．5z＝－1

2x-y+4z=8①x-2y-z=7②①+②得3x-3y+3z=15∴x-y+z=5因为小明得到正确的解x=4y=-2所以4a-2b=26因为小刚写错c得到x=7y=3说明此解满足第一个方程式所以7a