select date_format(submit_time, '%Y%m') as month,
round((count(distinct uid, date_format(submit_time, '%y%m%d'))) / count(distinct uid), 2) as avg_active_days,
count(distinct uid) as mau
from exam_record
where year(start_time)=2021
and submit_time is not null
group by date_format(submit_time, '%Y%m')

count(distinct 参数1，参数2) 只要两个参数有一个不一样就会返回该条记录