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Mathematical calculation exercises in primary schools
Simple operation-this is the most common math problem in primary schools. From the beginning of students' contact with calculation, the idea of simple operation has been infiltrated from different angles. In the fourth grade, simple operation appeared as an independent problem in the calculation problem. It is one of the most flexible calculation problems, which can fully exercise the flexibility of students' thinking and play a very important role in improving students' computing ability.

What is simple operation? This is a very simple question, but to understand it correctly, we must never do simple operations in pursuit of simple forms. In this regard, my understanding is that simple operation should be flexible, correct and reasonable use of various definitions, theorems, laws, properties, rules and so on. , change the original operation order to calculate, through simple operations, the calculation speed and accuracy should be greatly improved, making complex calculations simple. In other words, it is easier to complicate things, it is simpler to complicate things, and it is faster to slow down. The most important thing is to use all kinds of definitions, theorems, laws, properties and rules flexibly and reasonably. Especially emphasize "flexibility" and "rationality". Let's talk about my views on the situation I encountered in teaching.

1, "4.9+0. 1-4.9+0. 1" This is a very simple and common operation problem in Exercise 27 in Book 8 of primary school mathematics. When I assigned this problem to the students, I thought that the students would not hesitate to use the additive exchange rate and the combined exchange rate to successfully complete this problem, but when I corrected the students' homework, I found the following three situations:

①、4.9+0. 1-4.9+0. 1=(4.9-4.9)+(0. 1+0. 1);

②、4.9+0. 1-4.9+0. 1=4.9-4.9+0. 1+0. 1;

③、4.9+0. 1-4.9+0. 1=(4.9+0. 1)-(4.9+0. 1)。

Obviously, the third simplification is wrong, because it violates the order of the four operations, and its simplified result is definitely not equal to the original result. The problem lies in the first and second solutions. The simple calculation process of the first solution is very standard and impeccable. The second scheme seems nonstandard, but it makes sense. So, I organized students to discuss, and as a result, the students were divided into two opposing factions. One party thinks that the first scheme is absolutely correct, while the second scheme is not standardized and does not clearly indicate the simple operation process, so it cannot be regarded as correct. On the other hand, the first solution is very standard and definitely correct, but the second solution is also correct, because from left to right, if you calculate 4.9-4.9 first, you actually get 0, but you don't need to calculate it, just calculate 0. 1+0. 1 directly, and the simple calculation process is actually very clear.

Facing the different views of my classmates, I made a summary. I first affirmed the students' learning spirit, and then expounded my point of view: the first solution is absolutely correct, no doubt, but the second solution is also reasonable, reasonable and simple to operate, because they have grasped the key point of this problem, and there is no essential difference between the two. Simple operation can not only pursue form, but also grasp the essence of simplicity. As those students said, 4.9-4.9 can get 0 without calculation, just calculate 0. 1+0. 1. Since the same effect can be achieved without brackets, there is no need to emphasize adding brackets. The final result of simple operation is "simplicity".

2. "88× 25" This is a simple operation problem about multiplication. At that time, I had just finished learning the multiplication distribution rate, and there was such a problem in the exercise (80+8) × 25. After the students finished speaking, I immediately changed the question to "88×25" for students to consider. The next day, the students reported two answers:

①、88×25=80×25+8×25=2000+200=2200;

②、88×25= 1 1×(8×25)= 1 1×200=2200。

Then, I asked my classmates to introduce their ideas respectively, and their ideas were all very good. They say this: the first is to divide 88 into 80+8, and then multiply it by 25 respectively with the multiplication distribution rate; Second, 88 is divided by 8× 1 1, then 8 and 25 are multiplied by the multiplication exchange rate and the combined exchange rate, and then multiplied by 1 1.

After listening to the introduction of my classmates, I made a summary. First of all, I affirmed the correctness of the two answers, and then I made an analysis: the similarity between the two answers is that they both found it easy to multiply 8 by 25, so I tried my best to decompose 88, so I grasped the key of this question, so they were both right; The difference between the two schemes lies in the different decomposition methods. The first solution is additive decomposition, so we use multiplicative distribution rate, and the second solution is multiplicative decomposition, so we use multiplicative exchange rate and combination rate. Different methods have the same effect.

Finally, once again, there are many ideas for simple operation, but it should be correct as long as we grasp the key to solving the problem and use the laws and rules correctly and reasonably.

3. "5436 ÷18" This is a simple operation problem about division in the fifth question of Exercise 27 in Book 8. It is precisely because the requirement of the topic is "How to calculate the following questions as simply as possible", so the students' answers can be described as varied, and I summarized them, mainly including the following four items:

(1), direct calculation is very simple;

②、5436÷ 18=5400÷ 18+36÷ 18=300+2=302;

③、5436÷ 18=5436÷9÷2=604÷2=302;

④、5436÷ 18=5436÷6÷3=906÷3=302。

After careful analysis, the other three schemes are reasonable except that the first scheme does not conform to the simple operation rules. The second solution successfully applies the multiplication allocation rate to division. The third and fourth solutions successfully decompose the divisor 18 into the product of two one-digit factors, and then divide the divisor into two-digit numbers by using the property of "a ÷ (b× c) = a ÷ b ÷ c", which makes the calculation simple and convenient. So, I posted all four solutions on the blackboard in class to guide students to analyze them one by one, so that students can have a further understanding of the nature of simple operations.

4. Many students have a headache about the problem of "calculating the following questions, and simplifying what can be simplified". This problem is really difficult, because it not only requires students to make clear the operation order and calculate correctly, but also requires students to have certain observation ability, even intuition, and be able to conduct reasonable analysis, find out the parts that can be simplified, and perform simple operations reasonably. In order to successfully complete this kind of problems, students must have a thorough understanding of simple calculation and grasp the essence of simple calculation. They can neither miss any problem that can be simplified nor mistakenly simplify the problem that cannot be simplified.

In the process of teaching, I handled it like this: First of all, I didn't directly ask students to do such problems, but I did many simple problems and listed various ideas. Just like the above, through practice, I guide students to sum up some common simple arithmetic objects, such as "25 and 4", "125 and 8", "5 and any even number", and then review the operation order of mixed operations, so that students can further understand the operation order and basically complete it in normal order without thinking. Finally, do this kind of problem. At this point, students have already had the foundation of simple calculation, have an intuition about simple calculation, and at the same time firmly grasp the mixed operation under normal circumstances, and no longer think this kind of problem is difficult. Some students even think that this kind of problem is easy to calculate, and unconsciously apply this method to other places, such as other calculations, the calculation of applied problems, real life and so on, thus greatly improving students' computing ability.

Through these exercises, students not only learned the simple and clear operation, but more importantly, they initially understood the truth of applying what they have learned and truly understood the truth that the knowledge in books must be applied to practice.

Simple operation is a kind of advanced mixed operation and a skill of mixed operation. Learning simple operations well can not only improve the calculation ability and speed, but also make the definitions, theorems, laws, rules, properties and laws learned reach a comprehensive level, which is a kind of question that can best train students' thinking ability and broaden their thinking. Therefore, in the teaching of calculation problems, we must pay attention to simple operation and its flexibility. /browse /5 1 133.aspx