保留1位小数：round(..., 1)

select tag, difficulty, 
round((sum(score) - max(score) - min(score)) / ((count(score) - 2)), 1) as clip_avg_score 
from exam_record as er left join examination_info as ei
on er.exam_id = ei.exam_id
where difficulty = 'hard' and tag = 'SQL';