当x=2时,有最大值,为1过程y
=x+1/(x-3)
=x-3+1/(x-3)+3
=-[(3-x)+1/(3-x)]+3由于x<3，所以将x-3变成3-x，使之为正
≤-2√{(3-x)·[1/(3-x)]}+3
=-2+3
=1