Training to improve mathematical computing ability I. Basic training
The elementary school students' age is different, and the basic requirements of oral calculation are also 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, the practice of digital space concept, the comparison and memory training of numbers can be said to be the sublimation training of digital abstract thinking in primary school, which is very beneficial to promoting the development of everyone's thinking and intelligence. You can arrange this exercise in two time periods. One is when reading in the morning, and the other is to arrange a group after finishing homework. 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 channels. Write the formula first, and write the numbers directly after a few oral calculations. After this continues for a period of time, you will find that the speed and accuracy of your oral calculation will be greatly improved.
Training to improve mathematical calculation ability II. target training
The main form of elementary school senior grade series has changed from integer to fraction. In the operation of numbers, I believe everyone hates fractional addition with different denominators, right? Because it's too easy to make mistakes. Now, please think for yourself, are there only three ways to add (subtract) different denominator fractions?
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 until it is the same as the large number, the denominator is expanded several times, and the numerator is also expanded by the same multiple, you can do 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 I believe it is the biggest headache for everyone, 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 × 60.
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.
After reading the above, have you found the rules of oral calculation in each situation? Then as long as you practice more and master it, the problem will be solved.
Training to improve mathematical calculation ability. Memory training
Do senior students think that sometimes the calculation content in the topic is very extensive? Some of these operations have no specific rules for oral calculation, and I must solve them through 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 the above figures are frequently used in daily work and real life. After mastering and memorizing skillfully, it can be transformed into ability and produce high efficiency in calculation.
Training to improve mathematical calculation ability. Routine 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, people often ignore that the application of the law of multiplication and distribution makes the calculation complicated. For example, 2000/ 16×8, the result can be calculated directly by the law of multiplication distribution, 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.
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 half of it.
Training to improve mathematical computing ability V. Comprehensive training
1. The comprehensive performance of the above situation;
2. The comprehensive performance of integers, decimals and fractions;
3. Comprehensive training of four mixed operation sequences.
Comprehensive training is conducive to the improvement of judgment ability, reaction speed and the consolidation of oral calculation methods.
Of course, the above situation requires persistent training, otherwise it is difficult to achieve the expected results by fishing for three days and drying the net for two days.
The above five kinds of training should be gradual, and more importantly, persistent training. It will take some time to improve your math scores, so don't be too eager for success.