Question

设x,y,z是非零实数，且x^2+4y^2+z^2-3xy=2z根号(xy),则x+y+z/2z-x的值等于？

Answer 1

解：
算数平方根有意义，xy同号。
x²+4y²+z²-3xy=2z√(xy)
x²+4y²+z²-2z√(xy)-3xy=0
x²-4xy+4y²+z²-2z√(xy)+xy=0
(x-2y)²+[z-√(xy)]=0
平方项恒非负，两平方项之和=0，两平方项均=0
x-2y=0
z-√(xy)=0
解得x=2y z=√2|y|

y>0时，z=√2 y
(x+y+z)/(2z-x)
=(2y+y+√2y)/[2(√2)y -2y]
=[(3+√2)y]/[(2√2-2)y]
=(3+√2)/(2√2-2)
=(3+√2)(2√2+2)/4
=(10+8√2)/4
=(5+4√2)/2

y<0时，z=-√2 y
(x+y+z)/(2z-x)
=(2y+y-√2y)/[2(-√2)y -2y]
=[(3-√2)y]/[(-2√2-2)y]
=-(3-√2)/(2√2+2)
=-(3-√2)(2√2-2)/4
=-(-10+8√2)/4
=(5-4√2)/2

Answer 2

(x-2y)^2+(z-根号xy）^2=0
x=2y, z=根号xy
z=根号2y,x=2y
则x+y+z/2z-x=(3+根号2)y/（2根号2-1）y
=(8+7根号2)/7