题目内容

(2012?郑州模拟)3
3
8
×9-3-0.375

999×222+333×334
12
1
4
-(3
2
5
+2.25-3.6)

[19+19×(1.9-1
9
10
)÷0.95]÷0.38
分析:(1)3
3
8
×9-3-0.375,把0.375化成
3
8
,把3
3
8
拆成(3+
3
8
),在计算过程中简算;
(2)999×222+333×334,因为999是333的倍数,所以把999写成333×3,用乘法分配律简算;
(3)12
1
4
-(3
2
5
+2.25-3.6),把分数化成小数的形式,运用加法结合律和减法的性质简算;
(4)[19+19×(1.9-1
9
10
)÷0.95]÷0.38,按运算顺序进行计算,先算小括号内的,因为小括号内的得数为0,所以19×(1.9-1
9
10
)÷0.95的结果直接写成0,然后再计算即可.
解答:解:(1)3
3
8
×9-3-0.375;
=(3+
3
8
)×9-3-
3
8

=3×9+
3
8
×9-3-
3
8

=(27-3)+(
3
8
×9-
3
8
),
=24+3;
=27; 

(2)999×222+333×334,
=333×3×222+333×334,
=333×(3×222+334),
=333×(666+334),
=333×1000,
=333000;

(3)12
1
4
-(3
2
5
+2.25-3.6),
=12.25-(3.4+2.25-3.6),
=(12.25-2.25)+(3.6-3.4),
=10+0.2,
=10.2;

(4)[19+19×(1.9-1
9
10
)÷0.95]÷0.38,
=[19+19×(1.9-1.9)÷0.95]÷0.38,
=19÷0.38,
=50.
点评:此题属于数的四则混合运算的题目,考查学生对运算顺序的掌握,以及巧妙解题的能力.
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