f(x)>=1,即有:ax^2+2x-3a≥2 即a(x^2-3)≥2-2x
x^2-3>0在[2,3]上恒成立
所以有:a≥(2-2x)/(x^2-3)=-2/[(x-1)-2/(x-1)+2]
x=3时不等号右边取得最大-2/3
所以有:a≥-2/3a≥(2-2x)/(x^2-3)=-2/[(x-1)-2/(x-1)+2]怎么出来的(2-2x)/(x^2-3),(分子分母同除以x-1)=-2/[(x^2-3)/(x-1)]=-2/{[(x-1)^2+2(x-1)-2]/(x-1)}=-2/[(x-1)-2/(x-1)+2]