①令f(x)=0,得3x+x=0,化为3x=-x,分别作出函数y=3x,y=-x的图象,由图象可知函数f(x)的零点a<0;②令g(x)=log3x+2=0,解得x=3?2= 1 9 ,∴b= 1 9 ;③令h(x)=log3x+x=0,可知其零点c>0,而h( 1 9 )=log3 1 9 + 1 9 =-2+ 1 9 <0=h(c),又函数h(x)单调递增,∴ 1 9 <c.综上①②③可知:a<b<c.故答案为a<b<c.