题目内容
列递等式计算[(2)、(4)简算]
(1)
×[(1.6+
÷0.84-1
)]
(2)
×22
+16×
+
×
(3)[2.25+(3
-1.21×
)]÷40%
(4)1999÷1999
+
.
(1)
| 15 |
| 7 |
| 1 |
| 9 |
| 7 |
| 18 |
(2)
| 4 |
| 9 |
| 8 |
| 15 |
| 1 |
| 9 |
| 2 |
| 9 |
| 4 |
| 15 |
(3)[2.25+(3
| 3 |
| 5 |
| 5 |
| 11 |
(4)1999÷1999
| 19999 |
| 2000 |
| 1 |
| 2001 |
分析:(1)(3)依据四则运算计算方法,先算第二级运算,再算第一级运算,如果只含有同一级运算,按照从左到右顺序计算,有括号先算括号里面的解答,
(2)化16×
=4×
,
×
=
×
,再运用乘法分配律解答,
(4)化1999
=1999×(1+
),在运用除法性质即可解答.
(2)化16×
| 1 |
| 9 |
| 4 |
| 9 |
| 2 |
| 9 |
| 4 |
| 15 |
| 4 |
| 9 |
| 2 |
| 15 |
(4)化1999
| 1999 |
| 2000 |
| 1 |
| 2000 |
解答:解:(1)
×[(1.6+
÷0.84-1
)],
=
×[1.6+
-1
],
=
×[
-1
],
=
×
,
=
;
(2)
×22
+16×
+
×
,
=
×22
+4×
+
×
,
=(22
+4+
)×
,
=26
×
,
=11
;
(3)[2.25+(3
-1.21×
)]÷40%,
=[2.25+(3
-0.55)]÷40%,
=[2.25+3.05]÷40%,
=5.3÷40%,
=13.25;
(4)1999÷1999
+
,
=1999÷[1999×(1+
)]+
,
=1999÷1999÷
+
,
=
+
,
=1.
| 15 |
| 7 |
| 1 |
| 9 |
| 7 |
| 18 |
=
| 15 |
| 7 |
| 25 |
| 189 |
| 7 |
| 18 |
=
| 15 |
| 7 |
| 1637 |
| 945 |
| 7 |
| 18 |
=
| 15 |
| 7 |
| 3841 |
| 17010 |
=
| 3841 |
| 23814 |
(2)
| 4 |
| 9 |
| 8 |
| 15 |
| 1 |
| 9 |
| 2 |
| 9 |
| 4 |
| 15 |
=
| 4 |
| 9 |
| 8 |
| 15 |
| 4 |
| 9 |
| 4 |
| 9 |
| 2 |
| 15 |
=(22
| 8 |
| 15 |
| 2 |
| 15 |
| 4 |
| 9 |
=26
| 2 |
| 3 |
| 4 |
| 9 |
=11
| 23 |
| 27 |
(3)[2.25+(3
| 3 |
| 5 |
| 5 |
| 11 |
=[2.25+(3
| 3 |
| 5 |
=[2.25+3.05]÷40%,
=5.3÷40%,
=13.25;
(4)1999÷1999
| 1999 |
| 2000 |
| 1 |
| 2001 |
=1999÷[1999×(1+
| 1 |
| 2000 |
| 1 |
| 2001 |
=1999÷1999÷
| 2001 |
| 2000 |
| 1 |
| 2001 |
=
| 2000 |
| 2001 |
| 1 |
| 2001 |
=1.
点评:本题考查知识点:(1)四则运算计算方法,(2)简便算法的正确运用.
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