题目内容
用递等式计算:
(
+
)×15
200-180÷15×6
(3.8-1.8÷2)×0.3
÷9+
×
×[(
-
)÷
].
(
1 |
3 |
2 |
5 |
200-180÷15×6
(3.8-1.8÷2)×0.3
7 |
12 |
1 |
9 |
5 |
12 |
8 |
15 |
3 |
4 |
1 |
2 |
3 |
5 |
①(
+
)×15
=
×15+
×15,
=5+6,
=11;
②200-180÷15×6
=200-12×6,
=200-72,
=128;
③(3.8-1.8÷2)×0.3
=(3.8-0.9)×0.3,
=2.9×0.3,
=0.87;
④
÷9+
×
=(
+
)×
,
=1×
,
=
;
⑤
×(
-
)
=
×
-
×
,
=
-
,
=
.
1 |
3 |
2 |
5 |
=
1 |
3 |
2 |
5 |
=5+6,
=11;
②200-180÷15×6
=200-12×6,
=200-72,
=128;
③(3.8-1.8÷2)×0.3
=(3.8-0.9)×0.3,
=2.9×0.3,
=0.87;
④
7 |
12 |
1 |
9 |
5 |
12 |
=(
7 |
12 |
5 |
12 |
1 |
9 |
=1×
1 |
9 |
=
1 |
9 |
⑤
8 |
15 |
3 |
4 |
1 |
2 |
=
8 |
15 |
3 |
4 |
8 |
15 |
1 |
2 |
=
6 |
15 |
4 |
15 |
=
2 |
15 |
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