题目内容
(
+
)×8+
÷[
×(
+
)]
×
-
999
+99
+9
+
.
| 5 |
| 8 |
| 2 |
| 27 |
| 11 |
| 27 |
| 1 |
| 2 |
| 2 |
| 3 |
| 1 |
| 6 |
| 1 |
| 3 |
| 2.5×3+0.8 |
| 16×1.5+0.9 |
| 9 |
| 11 |
| 7 |
| 33 |
999
| 8 |
| 9 |
| 8 |
| 9 |
| 8 |
| 9 |
| 1 |
| 3 |
分析:(1)运用乘法分配律简算;
(2)中括号内运用乘法分配律简算;
(3)把算式中的小数扩大10倍,化成整数计算较简便;
(4)前面的三个分数有一定特点,把它们拆成整数和分数相加的形式,通过进一步计算,出现了999+99+9+3,把3拆成三个1.
(2)中括号内运用乘法分配律简算;
(3)把算式中的小数扩大10倍,化成整数计算较简便;
(4)前面的三个分数有一定特点,把它们拆成整数和分数相加的形式,通过进一步计算,出现了999+99+9+3,把3拆成三个1.
解答:解:(1)(
+
)×8+
,
=
×8+
×8+
,
=5+1,
=6;
(2)
÷[
×(
+
)],
=
÷[
×
+
×
],
=
÷
,
=
×3,
=1
;
(3)
×
-
,
=
×
-
,
=
×
-
,
=
-
,
=
;
(4)999
+99
+9
+
,
=999+99+9+
×3+
,
=999+99+9+3,
=(999+1)+(99+1)+(9+1),
=1000+100+10,
=1110.
| 5 |
| 8 |
| 2 |
| 27 |
| 11 |
| 27 |
=
| 5 |
| 8 |
| 2 |
| 27 |
| 11 |
| 27 |
=5+1,
=6;
(2)
| 1 |
| 2 |
| 2 |
| 3 |
| 1 |
| 6 |
| 1 |
| 3 |
=
| 1 |
| 2 |
| 2 |
| 3 |
| 1 |
| 6 |
| 2 |
| 3 |
| 1 |
| 3 |
=
| 1 |
| 2 |
| 1 |
| 3 |
=
| 1 |
| 2 |
=1
| 1 |
| 2 |
(3)
| 2.5×3+0.8 |
| 16×1.5+0.9 |
| 9 |
| 11 |
| 7 |
| 33 |
=
| 25×3+8 |
| 16×15+9 |
| 9 |
| 11 |
| 7 |
| 33 |
=
| 1 |
| 3 |
| 9 |
| 11 |
| 7 |
| 33 |
=
| 3 |
| 11 |
| 7 |
| 33 |
=
| 2 |
| 33 |
(4)999
| 8 |
| 9 |
| 8 |
| 9 |
| 8 |
| 9 |
| 1 |
| 3 |
=999+99+9+
| 8 |
| 9 |
| 1 |
| 3 |
=999+99+9+3,
=(999+1)+(99+1)+(9+1),
=1000+100+10,
=1110.
点评:此题考查学生对分数和小数四则混合运算的计算能力,以及运用所学知识灵活简算的能力.
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