Question

初一数学题目，求简便运算，求过程，谢谢

-9又24分之23×48-(4分之1-6又3分之2)×12

Answer 1

解：由题意知：-9又23/24×48-（1/4-6又2/3）×12
=-（9×24+23）/24-1/4×12+（6×3+2）/3×12
=-216/24-3+80
=-9-3+80
=68
望楼主采纳！

Answer 2

-9又24分之23×48-(4分之1-6又3分之2)×12
=-(9×24+23)/24×48-[1/4×12-(3×6+2)/3×12]
=-239/24×48-(3-20/3×12)
=-478-(3-80)
=-478+77
=401