Question

帮我解决一道数学题1.如果lx-1/2l+ly+1/3l+lz-1/6l=0,求x+y+z的值. 注：l l 是绝对值符

Answer 1

1.∵lx-1/2l+ly+1/3l+lz-1/6l=0,绝对值为非负数.

∴x-1/2=0,y+1/3=0,z-1/6=0

x=1/2,y=-(1/3),z=1/6

x+y+z=(1/2)-(1/3)+(1/6)=1/3

2.∵la-1l+(b-3)的平方+l3c-1l=0

∴a-1=0,b-3=0,3c-1=0

a=1,b=3,c=1/3

(abc)^1000=[1x3x(1/3)]^1000=1^1000=1