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用递等式计算.①[1-(18+23)]×36②(29+712)×36③7.5×4.7+4.7

2022-08-14 02:46:09 问答库 阅读 200 次

问题详情

用递等式计算.
①[1-(1

8

+2

3

)]×36
②(2

9

+7

12

)×36
③7.5×4.7+4.7×2.5
④25.3×12-33.46÷0.35
⑤5

6

×[2

3

÷(1-13

15

)]
⑥1

4

+(5

8

-3

10

)÷1

2

参考答案

①[1-(18+23)]×36,
=[1-(324+1624)]×36,
=[1-1924]×36,
=524×36,
=152;

②(29+712)×36,
=36×29+712×36,
=8+21,
=29;

③7.5×4.7+4.7×2.5,
=4.7×(7.5+2.5),
=4.7×10,
=47;

④25.3×12-33.46÷0.35,
=303.6-95.6,
=208;

⑤56×[23÷(1-1315)],
=56×[23÷215],
=56×[23×152],
=56×5,
=416;

⑥14+(58-310)÷12,
=14+(2540-1240)×2,
=14+1340×2,
=520+1320,
=910.

考点:等式