已知|x-8y|+2(4y-1)^2+3|8z-3x|=0,求x+y+z的值

因为|x-8y|≥0,
(4y-1)^2≥0,
|8z-3x|≥0.
又因为|x-8y|+2(4y-1)^2+3|8z-3x|=0.
所以|x-8y|=0,
(4y-1)^2=0,
|8z-3x|=0.
即x-8y=0,
4y-1=0,
8z-3x=0
解得x=2
y=1/4
当 x=2
y=1/4 z=3/4时
x+y+z=2+1/4+3/4
=3