题目内容

(2011?盐城)用递等式计算:
①0.6-0.2×1.9÷7.6;
②5÷
5
7
-
5
7
×
1
5

③1.02÷(5+0.4×0.25);
7
12
÷(
1
2
+
2
3
)×
2
5

⑤3÷
3
4
×(
4
5
-
3
10
).
分析:这些题目按照先算乘除,再算加减,有括号的先算扩号里面的运算顺序计算.
解答:解:①0.6-0.2×1.9÷7.6,
=0.6-0.38÷7.6,
=0.6-0.05,
=0.55;

②5÷
5
7
-
5
7
×
1
5

=5×
7
5
-
5
7
×
1
5

=7-
1
7

=6
6
7


③1.02÷(5+0.4×0.25),
=1.02÷(5+0.1),
=1.02÷5.1,
=0.2;

7
12
÷(
1
2
+
2
3
)×
2
5

=
7
12
÷
7
6
×
2
5

=
7
12
×
6
7
×
2
5

=
1
2
×
2
5

=
1
5


⑤3÷
3
4
×(
4
5
-
3
10
),
=3÷
3
4
×
1
2

=3×
4
3
×
1
2

=4×
1
2

=2.
点评:本题考查了基本的四则混合运算,要找清楚运算顺序,按照运算法则,仔细计算.
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