z=1-√(x^2 y^2)的极值
来源:学生作业帮助网 编辑:作业帮 时间:2024/11/07 19:19:13
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
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+
|4x-4y+1|+1/3√2y+z+(z-1/2)^2=0由几个非负数的和为零,要求每一项为零故4x-4y+1=02y+z=0z-1/2=0x=-1/2y=-1/4z=1/2所求为(y+z)*x^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=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
题目是否应该是2(√x+√y-1+√z-2)=x+y+z?如果是的话,解法如下x-2√x+y-2√(y-1)+z-2√(z-2)=0x-2√x+1+(y-1)-2√(y-1)+1+(z-2)-2√(z
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-2y=02y+z=01-2z=0解出来x=-0.5y=-0.25z=0.5快采纳吧~~~再问:给你点个赞再答:O(∩_∩)O哈哈~
√x+√(y-1)+√(z-2)=1/4(x+y+z+9)4√x+4√(y-1)+4√(z-2)=x+y+z+9x-4√x+(y-1)-4√(y-1)+(z-2)-4√(z-2)+9=0看出来了吧,是
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
方法一:2√x+2√(y-1)+2√(z-2)=x+y+z移项,得x+y+z-2√x-2√(y-1)-2√(z-2)=0(x-2√x+1)+[(y-1)-2√(y-1)+1]+[(z-2)-2√(z-
由已知条件,可以移项合并得到(√x-1)^2+(√(y-1)-1)^2+(√(z-2)-1)^2=0所以√x-1=0=>x=1√(y-1)-1=0=>y=2√(z-2)-1=0=>z=3所以x+y+z
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