Judging from the psychological characteristics of primary school students of different ages, the basic requirements of oral arithmetic are different. The middle and low grades are mainly in the addition of one or two digits. It is best for senior students to take one-digit multiplied by two-digit oral calculation as the basic training. The specific requirement of oral calculation is to multiply one digit by the number in the tenth place of two digits, immediately add the product of one digit and the number in the first digit of two digits to the three digits, and quickly say the result. This kind of oral arithmetic training, including the practice of the spatial concept of numbers, the comparison of numbers and the memory training, can be said to be the sublimation training of digital abstract thinking in primary school, which is very beneficial to promoting the development of thinking and intelligence. This exercise can be arranged in two periods. One is an early reading class, and the other is an assignment group. Each group is divided as follows: select a digit, and one or ten digits in the corresponding two digits all contain a certain number. Each group has 18 rows. Let the students write the formula first, and then write the numbers directly after several oral calculations. After this lasts for a period of time (usually 2 ~ 3 months), the speed and accuracy of oral calculation will be greatly improved.

Second, targeted training.

The main form of elementary school senior grade series has changed from integer to fraction. In the operation of numbers, fractional addition with different denominators is the most time-consuming and error-prone place for students, and it is also the key and difficult point in teaching and learning. How to break this key and difficult point? Through research, comparison and teaching practice, it is proved that it is correct to add the scores of different denominators in the oral operation of fractions. Through analysis and induction, there are only three cases of addition (subtraction) of different denominator fractions, and each case has its own oral calculation rules. As long as students master it, the problem will be solved.

1. Two fractions, where the big number in the denominator is a multiple of the decimal.

For example, "112+1/3", in this case, oral calculation is relatively easy. The method is: the big denominator is the common denominator of the two denominators. As long as the small denominator is multiplied by the multiple until it is the same as the big number, the denominator is multiplied by several times, and the numerator is multiplied by the same multiple, you can perform oral calculation by adding the scores of the same denominator:

2. Two fractions, the denominator is a prime number. This situation is more difficult in form, and it is also the biggest headache for students, but it can be turned into an easy thing: except for the future, the common denominator is the product of two denominators, and the numerator is the sum of the products of the numerator of each fraction and another denominator (if it is subtraction, it is the difference between the two products), such as 2/7+3/ 13, and the oral calculation process is: the common denominator is 7 × 650.

If the numerator of both fractions is 1, oral calculation is faster. For example, "1/7+ 1/9", the common denominator is the product of two denominators (63), and the numerator is the sum of two denominators (16).

3. Two fractions and two denominators are neither prime numbers nor multiples of decimals. In this case, the mother of centimeters is usually found by short division. In fact, you can also calculate the total score directly in the formula and get the result quickly. The common denominator can be obtained by multiplying the large numbers in the denominator. The specific method is: multiply the big denominator (large number) by the expansion until it is a multiple of the decimal of another denominator. For example, 1/8+3/ 10 expands a large number 10, 2 times, 3 times and 4 times, and each expansion is compared with the decimal 8 to see if it is a multiple of 8. When expanded to 4 times, it is a multiple of 8 (5 times), then the common denominator is 40, and the numerator is also expanded accordingly.

The above three situations are also applicable to the oral calculation method in addition and subtraction with fractions.

Third, memory training.

The content of advanced computing is extensive, comprehensive and comprehensive. Some common operations are often encountered in real life. Some of these operations have no specific rules for oral calculation, and they must be solved by strengthening memory training. The main contents are:

The square result of 1 0 ~ 24 in1.natural number;

2. Approximate value of pi 3. Product of14 with a digit and several common numbers, such as 12, 15, 16 and 25;

3. The decimal values of the simplest fractions with denominators of 2, 4, 5, 8, 10, 16, 20, 25, that is, the reciprocity of these fractions and decimals.

The results of these figures are often used in daily work and real life. After mastering and memorizing skillfully, it can be converted into energy, resulting in high efficiency in calculation.

Fourth, regular training.

1. Master the operation rules. There are five laws in this respect: the commutative law and associative law of addition; Commutative law, associative law and multiplicative distribution law. Among them, multiplication and division have a wide range of uses and forms, including positive and negative use, integers, decimals and fractions. When a fraction is multiplied by an integer, students often ignore the application of the law of multiplication and distribution, which makes the calculation complicated. For example, 2000/ 16×8, the result can be directly calculated by using the multiplication distribution law, that is, 100 1.5, but it is time-consuming and error-prone to use the general method of forging scores. In addition, there are applications of subtraction and quotient invariance.

2. Regular training. Mainly the oral calculation method of the result that the number in the unit is the square of the two digits of 5 (the method is abbreviated).

3. Master some special circumstances. For example, fractional subtraction, generally, the molecules are not reduced enough after the fraction, and the molecules reduced are often larger than those reduced by 1, 2, 3, etc. No matter how big the denominator is, it can be calculated directly. For example, 12/7-6/7, its numerator is only 1, and its difference numerator must be smaller than the denominator 1, and the result is 6/7 without calculation. Another example is: 194/99-97/99, where the difference between numerator and denominator is 2 and the result is 97/99. When the reduced molecule is 3, 4 and 5 larger than the reduced molecule, the result can be calculated quickly. Another example is the oral calculation of the product of any two digits and 1.5, which is two digits plus one of it.