16a²-12a≤0两边同时除以4得4a²-3a≤0a(4a-3)≤0当a≤0,那么4a-3>=0;即a>=3/4；当a>=0,那么4a-3≤0，即0=综上所述a<=0,不符合舍去；所以0=
16a²-12a≤04a(4a-3)≤0 0 ≤ a≤3/4
