Question

9×(1-2/5÷(9/20)÷(3/20))-1又3/7这个脱式计算怎样算？

Answer 1

9×(1-2/5÷(9/20)÷(3/20))-1又3/7这个脱式计算怎样算？

9×(1-2/5÷(9/20)÷(3/20))-1又3/7
=9x（1-2/5x20/9x20/3）-10/7
=9x（1-160/27）-10/7
=9-160/3-10/7
=9-1120/21-30/21
=9-1150/21
=-961/21
=-45又21分之16

Answer 2

四则运算有个规定：先算括号内再算括号外，先算乘除后算加减。
9×（1-2/5÷（9/20）÷（3/20））-1又3/7
=9×（1-2/5×20/9×20/3）-10/7
=9×（1-160/27）-10/7
=9×（-133/27）-10/7
=-133/3-10/7
=-931/21-30/21
=-961/21

Answer 3

脱式计算是一个数学学科术语,即递等式计算,把计算过程完整写出来的运算,也就是脱离竖式的计算。
9×(1-2/5÷(9/20)÷(3/20))-1又3/7
=9x(1-2/5x20/9x20/3)-10/7
=9x(1-160/27)-10/7
=9x(-133/27)-10/7
=-133/3-10/7
=-931/21-30/21
=-961/21。

Answer 4

解:
原式=9×(1-2/5÷(9/20 ×20/3)-10/7
=9×(1-2/5÷3)-10/7
=9×(1-2/5 ×1/3)-10/7
=9×(15-2)/15-10/7
(3×13×7-5×10)/35
=223/35
即为所求。

Answer 5

9×【1-2/5÷（9/20÷3/20）】-1又3/7
=9×【1-2/5÷（9/20×20/3）】-1又3/7
=9×【1-2/5÷3】-1又3/7
=9×【1-2/5×1/3】-1又3/7
=9×【1-2/15】-1又3/7
=9×13/15-1又3/7
=39/5-1又3/7
=7又4/5-1又3/7
=7又28/35-1又15/35
=6又13/35