用递等式计算.2005+1911÷4924÷1.5-2322÷(23-17)×20%25÷[1÷(110-3100)].-数学

题目简介

用递等式计算.2005+1911÷4924÷1.5-2322÷(23-17)×20%25÷[1÷(110-3100)].-数学

题目详情

用递等式计算.
2005+1911÷49
24÷1.5-
2
3

22÷(
2
3
-
1
7
)×20%
25÷[1÷(
1
10
-
3
100
)].
题型:解答题难度:中档来源:不详

答案


(1)2005+1911÷49,
=2005+39,
=2044;

(2)24÷1.5-class="stub"2
3

=16-class="stub"2
3

=15class="stub"1
3


(3)22÷(class="stub"2
3
-class="stub"1
7
)×20%,
=22÷class="stub"11
21
×0.2,
=42×0.2,
=8.4;

(4)25÷[1÷(class="stub"1
10
-class="stub"3
100
)],
=25÷[1÷class="stub"7
100
],
=25÷class="stub"100
7

=class="stub"7
4

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