[(29/5+1/2)-(3/5+12/5 / 2/3)]/7/10

来源:学生作业帮助网 编辑:六六作业网 时间:2024/12/26 23:44:29
[(29/5+1/2)-(3/5+12/5/2/3)]/7/10[(29/5+1/2)-(3/5+12/5/2/3)]/7/10[(29/5+1/2)-(3/5+12/5/2/3)]/7/10原式=[

[(29/5+1/2)-(3/5+12/5 / 2/3)]/7/10
[(29/5+1/2)-(3/5+12/5 / 2/3)]/7/10

[(29/5+1/2)-(3/5+12/5 / 2/3)]/7/10
原式=[(29/5+1/2)-(3/5+18/5)]/(7/10)
=(29/5+1/2-21/5)*10/7
=(8/5+1/2)*10/7
=(16/10+5/10)*10/7
=(21/10)*(10/7)
=3

这题打错了吧。。。。哪来这么多除法。。。

53/700

700分之53。

[(29/5+1/2)-(3/5+12/5 / 2/3)]/7/10
=[(29/5+1/2)-(3/5+18/5]/7/10
=[63/10-21/5]/7/10
=21/10 / 7/10
3

[(29/5+1/2)-(3/5+12/5 / 2/3)]/7/10
=[(58/10+5/10)-(3/5+18/5)]/7/10
=(63/10-42/10)/7/10
=21/10/7/10
=3

等于3
3/5+12/5 / 2/3=3/5+12/5×3/2=21/5
原式=[29/5+1/2-21/5]×10/7
=[8/5+1/2]×10/7
=21/10×10/7
=3

[(29/5+1/2)-(3/5+12/5 / 2/3)]/(7/10 )
=[(29/5+1/2)-(3/5+18/5)]/(7/10)
=[29/5+21/5-1/2]/(7/10)
=(19/2)*(10/7)
=95/7