由原题可得：(x+2y)^2+(y+1)^2+√(z+1/2)=0
∴y= -1 z= -1/2 x= 2
∴（4x-10y）^z=(8+10)^(-1/2)=(√2)/6
x,y,z满足条件｜x^2+4xy+5y^2︱+√(z+1/2)=-2y-1 则（4x-10y）^z等于
1个回答
相关问题
-
实数x ,y ,z 满足条件∣x^2+4xy+5y^2∣+(z+1/2)^-1/2=-2y-1,则(4x-10y)^z的
-
实数x,y,z满足x=y+根号2,2xy+2*根号2*z*z+1=0,则x+y+z等于多少
-
已知x,y,z满足x+y=5及z^4=xy+y-9则x+2y+3z=______
-
实数x,y,z,满足条件√x+√y-1+√z-2=1/4(x+y+z+9),求xyz
-
若x,y,z满足xy/(x+y)=1,yz/(y+z)=1/2,zx/(z+x)=1/5,则xyz=?
-
xy=2x+2y yz=4y+4z zx=5z+5x y=?z=?
-
方程组(xy+x)/(x+y+1)=2 (xz+2x)/(z+y+2)=3 (y+1)(z+2)/(z+y+3)=4
-
4x+2y+z=5,4x-2y+z=1,z=-1
-
4x+2y+z=5 4x-2y+z=1 z=-1