已知三个数x,x,z满足x y分之xy=-3
来源:学生作业帮助网 编辑:作业帮 时间:2024/10/02 06:33:55
实数x,y,z,满足那么x+y=6,z^2=xy-9,∴xy=z^+9,(x-y)^=(x+y)^-4xy=-4z^>=0,∴z=0,(x+y)^z=6^0=1.
XY/X+Y=-2,-->(x+y)/(xy)=-1/2,-->1/x+1/y=-1/2YZ/Y+Z=4/3,-->(y+z)/(yz)=3/4,-->1/y+1/z=3/4&
倒数解方程组..(x+y)/xy=1/x+1/y=-1/2类推~~~
XY/X+Y=-2,-->(x+y)/(xy)=-1/2,-->1/x+1/y=-1/2YZ/Y+Z=4/3,-->(y+z)/(yz)=3/4,-->1/y+1/z=3/4ZX/Z+X=-4/3,-
由对称性,求z的取值情况.x+y+z=5,===>x+y=5-z,由xy+yz+zx=3.===>xy=3-z(x+y)=3-z(5-z)=z²-5z+3.∴x+y=5-z,xy=z&sup
xy/(x+y)=-2(x+y)/(xy)=-1/2,1/x+1/y=-1/2yz/(y+z)=4/3,(y+z)/(yz)=3/4,1/y+1/z=3/4zx/(z+x)=-4/3,(z+x)/(z
xy/(x+y)=-2,yz/(y+z)=4/3,zx/(z+x)=-4/3(x+y)/xy=-1/2,即:1/x+1/y=-1/2同理:1/y+1/z=3/4,1/z+1/x=-3/4三式相加得:1
将x=5-y-z代入xy+yz+zx=3,整理成关于y的一元二次方程y²+(z-5)y+z²-5z+3=0由于y为实数,所以△≥0.即(z-5)²-4(z²-5
xyz=x+y+z<3z∴xy<3由于x<y,故xy=2,x=1,y=2∴z=3
∵xyx+y=-2,yzy+z=43,zxz+x=-43,∴1x+1y=-12,1y+1z=34,1z+1x=-34,∴2(1x+1y+1z)=-12,即1x+1y+1z=-14,则xyzxy+yz+
题目有问题,yx/(y+x)=4/3应该是yz/(y+z)=4/3xy/(x+y)=-2(x+y)/(xy)=-1/21/x+1/y=-1/2(1)yz/(y+z)=4/3(y+z)/(yz)=3/4
-4再问:请问第三步是怎么算出-1/4的可以写一下过程么再答:1/x+1/y+1/y+1/z+1/z+1/x=2(1/x+1/y+1/z)=-1/2∴1/x+1/y+1/z=-1/2/2=-1/4
【x+y】分之xy=-2,xy分之【x+y】=-1/21/x+1/y=-1/2(1)【y+z】分之yz=3分之4,yz分之【y+z】=3/41/y+1/z=3/4(2)【z+x】分之zx=-3分之4,
xy/x+y=1/3可得:1/x+1/y=31式yz/y+z=1/4可得:1/y+1/z=42式xz/z+x=1/5可得:1/x+1/z=53式由1、2、3式可解得:x=1/2,y=1,z=1/3
因为X-Y=8所以X.Y的值会出现4种情况1.X为正,Y为负X+|Y|=82.X为负,Y为负|Y|-X=83.X为正,Y为正X-Y=84.X为负,Y为正Y-X=8又有条件了个条件:XY+Z2=-16那
可知:1/x+1/y=-2……①1/y+1/z=4/3……②1/z+1/x=-4/3……③①+②+③得:1/x+1/y+1/z=-1……④④-①得:1/z=1④-②得:1/x=-7/3④-③得:1/y
xy/(x+y)=-2(x+y)/xy=-1/21/y+1/x=-1/2yz/(y+z)=4/3(y+z)/yz=3/41/z+1/y=3/4zx/(z+x)=-4/3(z+x)/zx=-3/41/x
xy/x+y=-2,取倒数就得1/x+1/y=-1/2①yz/y+z=3/4取倒数就得1/y+1/z=4/3②zx/z+x=-3/4取倒数就得1/x+1/z=-4/3③①+②+③就得2(1/x+1/y
解题思路:本题的关键是将三个方程两边取倒数,化简后分别将方程等号左边和右边相加,得到1/x+1/y+1/z的值,最后将要求的分式化简,把1/x+1/y+1/z的值带入即可。解题过程: