设x,y,z为实数,且3乘以的绝对值++^2=0,求x^2+y^2+z^2+2xy-2xz-2yz的值?

题目

设x,y,z为实数,且3乘以<2x-1>的绝对值+<2倍根号9y^2-6y+1>+^2=0,求x^2+y^2+z^2+2xy-2xz-2yz的值?

答案

2x-1＝0
9y^2-6y+1＝0,(3y-1)^2=0
z-1=0
x=0.5
y=1/3
z=1
x^2+y^2+z^2+2xy-2xz-2yz=(z-x-y)^2=1/36