A-A+

用递等式计算:(1)113×[21÷(4112-2.625)-412]÷335(2)(258

2022-08-14 02:36:37 问答库 阅读 200 次

问题详情

用递等式计算:
(1)11

3

×[21÷(41

12

-2.625)-41

2

]÷33

5



(2)(25

8

-14

7

)÷[(31

12

+4.375)÷198

9

]

(3)10

13

÷219

22

-1.4×11

13

+7+22

63

×20%.

参考答案

(1)113×[21÷(4112-2.625)-412]÷335,
=113×[21÷3524-412]÷335,
=113×[725-412]÷335,
=113×9910×518,
=113;

(2)(258-147)÷[(3112+4.375)÷1989],
=1356÷[71124÷1989],
=5956÷38,
=5921;


(3)1013÷21922-1.4×1113+7+2263×20%,
=1013×2263-75×1113+7+2263×15,
=7+(1013+15)×2263-75×1113,
=7+6365×2263-75×1113,
=7+2265-7765,
=72265-11265,
=6213.

考点:等式