Question

已知［x+y-3］+［y+z-4］+［z+x-2］=0,求ｘ,ｙ,ｚ的值

Answer 1

∵［x+y-3］+［y+z-4］+［z+x-2］=0 ∴x+y-3=0..(1) y+z-4=0..(2) z+x-2=0..(3) (1)-(2)得:x-z+1=0..(4) (3)+(4)得2x-1=0,∴x=1/2,y=3-x=5/2,z=2-x=3/2