Four plus five
Publish: 2021-04-14 07:14:25
1. Computer. (calculation)
attack 25 hit base 0
attack + [
attack main) equipment
attack this / 1 attack = material auxiliary attack
object
defense D assistant
Magic
through attack] according to x
attack object
reason has no essence
its essence
knowledge
Ben +
pieces
equipped with weapon (middle =
Wu
attack quantity base plate n yushiben + F principle (attached
x
+ attack base according to attack)
attack 25 hit base 0
attack + [
attack main) equipment
attack this / 1 attack = material auxiliary attack
object
defense D assistant
Magic
through attack] according to x
attack object
reason has no essence
its essence
knowledge
Ben +
pieces
equipped with weapon (middle =
Wu
attack quantity base plate n yushiben + F principle (attached
x
+ attack base according to attack)
2. 1、 Basic training:
from the psychological characteristics of different ages of primary school students, the basic requirements of oral arithmetic are different. Low and middle grade students mainly add one or two digits. It is better for senior students to take the one digit by two digit mental arithmetic as the basic training. The specific requirement of oral arithmetic is to multiply the number of one digit and the number of ten digits of two digits, and then add the proct of multiplying the number of one digit and the number of one digit of two digits to the three digits, and quickly say the result. In primary school, this training is a sublimation training of abstract thinking of numbers. It is very beneficial to promote the development of thinking and intelligence
Second, targeted training: the main form of the number of senior primary school students has changed from integer to score. In the operation of numbers, the addition of different denominators is the most time-consuming and error prone place for students, and it is also the key and difficult point of teaching and learning
two fractions, the large number in the denominator is the multiple of the decimal. For example, "1 / 12 + 1 / 3", in this case, oral calculation is relatively easy, the method is: the big denominator is the common denominator of two denominators, as long as the small denominator is expanded by multiple until it is the same as the big number, and the denominator is expanded by several times until it is the multiple of another denominator decimal< Third, memory training
the content of senior calculation is extensive, comprehensive and comprehensive. Some common operations are often encountered in real life. Some of these operations have no specific rules of oral arithmetic and must be solved by strengthening memory training. The main contents are as follows:
1
2. The proct of the approximate value of PI 3.14 with one digit and the proct with several common numbers 12, 15, 16 and 25
3. The denominator is the decimal value of the simplest fraction of 2, 4, 5, 8, 10, 16, 20 and 25, that is, the interaction of these fractions and decimals.
from the psychological characteristics of different ages of primary school students, the basic requirements of oral arithmetic are different. Low and middle grade students mainly add one or two digits. It is better for senior students to take the one digit by two digit mental arithmetic as the basic training. The specific requirement of oral arithmetic is to multiply the number of one digit and the number of ten digits of two digits, and then add the proct of multiplying the number of one digit and the number of one digit of two digits to the three digits, and quickly say the result. In primary school, this training is a sublimation training of abstract thinking of numbers. It is very beneficial to promote the development of thinking and intelligence
Second, targeted training: the main form of the number of senior primary school students has changed from integer to score. In the operation of numbers, the addition of different denominators is the most time-consuming and error prone place for students, and it is also the key and difficult point of teaching and learning
two fractions, the large number in the denominator is the multiple of the decimal. For example, "1 / 12 + 1 / 3", in this case, oral calculation is relatively easy, the method is: the big denominator is the common denominator of two denominators, as long as the small denominator is expanded by multiple until it is the same as the big number, and the denominator is expanded by several times until it is the multiple of another denominator decimal< Third, memory training
the content of senior calculation is extensive, comprehensive and comprehensive. Some common operations are often encountered in real life. Some of these operations have no specific rules of oral arithmetic and must be solved by strengthening memory training. The main contents are as follows:
1
2. The proct of the approximate value of PI 3.14 with one digit and the proct with several common numbers 12, 15, 16 and 25
3. The denominator is the decimal value of the simplest fraction of 2, 4, 5, 8, 10, 16, 20 and 25, that is, the interaction of these fractions and decimals.
3. In the primary school mathematics test questions, the questions involving calculation content account for more than 85% in a test paper. Therefore, it is a very important task for students to strengthen the calculation training and effectively improve the accuracy of calculation. The actual situation shows that the accuracy of a student's calculation is directly proportional to his oral calculation ability. So how to improve the computing power? Please take a look at the following training methods< Basic training
different age of primary school students, the basic requirements of oral arithmetic are also different. Low and middle grade students mainly add one or two digits. It is better for senior students to take the one digit by two digit mental arithmetic as the basic training. The specific requirement of oral arithmetic is to multiply the number of one digit and the number of ten digits of two digits, and then add the proct of multiplying the number of one digit and the number of one digit of two digits to the three digits, and quickly say the result. This mental arithmetic training includes the practice of several spatial concepts, digital comparison and memory training. In primary school, it can be said that it is a sublimation training of abstract thinking of numbers. It is very beneficial to promote the development of people's thinking and intelligence. You can arrange this exercise in two periods. One is to read early, the other is to arrange a group after homework. Each group is divided as follows: one digit is optional, corresponding to the number of one digit or ten digit in two digits. There are 18 questions in each group. Write the formula first, then write the number directly after several times of oral calculation. After a period of time, you will find that your speed and accuracy will be greatly improved< 2. Targeted training
the main form of the number of senior grades in primary school has changed from integer to score. In the number of operations, I believe we do not like the different denominator fractional addition, right? Because it's too error prone. Now, please think for yourself, is the addition (subtraction) method of different denominators only in the following three cases
1. For two fractions, the large number in the denominator is the multiple of the decimal
for example, "1 / 12 + 1 / 3", in this case, oral arithmetic is relatively easy. The method is: the big denominator is the common denominator of two denominators. As long as the small denominator is expanded by multiple, until it is the same as the big number, the denominator is expanded by several times, and the numerator is also expanded by the same multiple, It can be calculated by adding the same denominator fraction: 1 / 12 + 1 / 3 = 1 / 12 + 4 / 12 = 5 / 12
2. The denominator of two fractions is coprime
this kind of situation is more difficult in form, and I believe you are also the most headache, but it can be changed into easy: after it is divided, the common denominator is the proct of the two denominators, and the numerator is the sum of the proct of the numerator of each fraction and the other denominator (if it is subtraction, it is the difference of the two procts), such as 2 / 7 + 3 / 13. The oral calculation process is: the common denominator is 7 × 13 = 91, molecule 26 (2 × 13)＋21(7 × 3) = 47, the result is 47 / 91
if the molecules of both fractions are 1, the oral calculation is faster. For example, "1 / 7 + 1 / 9", the denominator is the proct of two denominators (63), and the numerator is the sum of two denominators (16)
3. Two fractions and two denominators are neither coprime numbers nor multiples of decimals
in this case, we usually use the short division method to get the common denominator. In fact, we can also directly calculate the general score in the formula and get the result quickly. The common denominator can be obtained by enlarging the large number in the denominator. The specific method is: to double the large denominator (large number) until it is a multiple of another denominator decimal. For example, 1 / 8 + 3 / 10 expands the large number 10, 2 times, 3 times and 4 times, and compares it with the decimal 8 every time to see if it is a multiple of 8. When it is expanded to 4 times, it is a multiple of 8 (5 times), then the common denominator is 40, and the numerator is expanded by the corresponding multiple and then added (5 + 12 = 17), and the number is 17 / 40
after reading the above, have you found the rule of mental arithmetic in every situation? So as long as you practice more and master it, the problem will be solved< (3) memory training
do senior students feel that sometimes the calculation content in the topic is very extensive? Some of these operations have no specific rules of oral arithmetic, so I have to solve them through memory training. The main contents are as follows:
1
2. The proct of the approximate value of PI 3.14 with one digit and with several common numbers 12, 15, 16 and 25
3. The denominator is the decimal value of the simplest fraction of 2, 4, 5, 8, 10, 16, 20 and 25, that is, the interaction between these fractions and decimals
the results of the above numbers, whether in daily work or in real life, are used very frequently. After mastering and remembering them, they can be transformed into abilities and proce high efficiency in calculation< 4. Regular training
1. Mastering the law of operation. There are mainly five laws in this aspect: commutative law and associative law of addition; Commutative law, associative law and distributive law of multiplication. Among them, the multiplication distribution law is widely used and has many forms, including positive use and negative use, and the forms of integer, decimal and fraction. In the multiplication of fractions and integers, we often ignore the application of the law of distribution of multiplication, which makes the calculation complicated. Such as 2000 / 16 × 8. If we use the law of multiplicative distribution, we can calculate the result of 1000 by mouth directly. But if we use the general method of recing false fraction, it is time-consuming and easy to make mistakes. In addition, there are subtraction properties and quotient invariant properties< 2. Regular training. It is mainly the oral calculation method of the square result of the two digit number of 5
3. Master some special cases. For example, in fractional subtraction, if the numerator is not enough to be subtracted after general division, and the numerator subtracted is usually larger than the numerator subtracted by 1, 2, 3 and other smaller numbers, no matter how big the denominator is, it can be directly calculated orally. For example, the difference between 12 / 7 and 6 / 7 is only 1. The difference between 12 / 7 and 6 / 7 must be 1 less than the denominator. The result is 6 / 7 without calculation. Another example is: 194 / 99-97 / 99, if the difference between the numerator and denominator is 2, the difference between the numerator and denominator is 2, and the result is 97 / 99. When the subtracted molecule is larger than the subtracted molecule by 3, 4, 5 and other smaller numbers, the result can be quickly calculated orally. Another example is the mental calculation of the proct of any two digit number and 1.5, which is two digits plus half of it< 5. Comprehensive training
1
2< 3. Comprehensive training of four mixed operation sequences
comprehensive training is concive to the improvement of judgment ability, reaction speed and the consolidation of oral arithmetic
of course, we need to persevere in the above situations. Otherwise, it is difficult to achieve the expected effect if we fish for three days and dry the net for two days
the above five kinds of training should be carried out step by step, but also persevere. It will take some time to improve your math scores. Don't be too eager to succeed.
different age of primary school students, the basic requirements of oral arithmetic are also different. Low and middle grade students mainly add one or two digits. It is better for senior students to take the one digit by two digit mental arithmetic as the basic training. The specific requirement of oral arithmetic is to multiply the number of one digit and the number of ten digits of two digits, and then add the proct of multiplying the number of one digit and the number of one digit of two digits to the three digits, and quickly say the result. This mental arithmetic training includes the practice of several spatial concepts, digital comparison and memory training. In primary school, it can be said that it is a sublimation training of abstract thinking of numbers. It is very beneficial to promote the development of people's thinking and intelligence. You can arrange this exercise in two periods. One is to read early, the other is to arrange a group after homework. Each group is divided as follows: one digit is optional, corresponding to the number of one digit or ten digit in two digits. There are 18 questions in each group. Write the formula first, then write the number directly after several times of oral calculation. After a period of time, you will find that your speed and accuracy will be greatly improved< 2. Targeted training
the main form of the number of senior grades in primary school has changed from integer to score. In the number of operations, I believe we do not like the different denominator fractional addition, right? Because it's too error prone. Now, please think for yourself, is the addition (subtraction) method of different denominators only in the following three cases
1. For two fractions, the large number in the denominator is the multiple of the decimal
for example, "1 / 12 + 1 / 3", in this case, oral arithmetic is relatively easy. The method is: the big denominator is the common denominator of two denominators. As long as the small denominator is expanded by multiple, until it is the same as the big number, the denominator is expanded by several times, and the numerator is also expanded by the same multiple, It can be calculated by adding the same denominator fraction: 1 / 12 + 1 / 3 = 1 / 12 + 4 / 12 = 5 / 12
2. The denominator of two fractions is coprime
this kind of situation is more difficult in form, and I believe you are also the most headache, but it can be changed into easy: after it is divided, the common denominator is the proct of the two denominators, and the numerator is the sum of the proct of the numerator of each fraction and the other denominator (if it is subtraction, it is the difference of the two procts), such as 2 / 7 + 3 / 13. The oral calculation process is: the common denominator is 7 × 13 = 91, molecule 26 (2 × 13)＋21(7 × 3) = 47, the result is 47 / 91
if the molecules of both fractions are 1, the oral calculation is faster. For example, "1 / 7 + 1 / 9", the denominator is the proct of two denominators (63), and the numerator is the sum of two denominators (16)
3. Two fractions and two denominators are neither coprime numbers nor multiples of decimals
in this case, we usually use the short division method to get the common denominator. In fact, we can also directly calculate the general score in the formula and get the result quickly. The common denominator can be obtained by enlarging the large number in the denominator. The specific method is: to double the large denominator (large number) until it is a multiple of another denominator decimal. For example, 1 / 8 + 3 / 10 expands the large number 10, 2 times, 3 times and 4 times, and compares it with the decimal 8 every time to see if it is a multiple of 8. When it is expanded to 4 times, it is a multiple of 8 (5 times), then the common denominator is 40, and the numerator is expanded by the corresponding multiple and then added (5 + 12 = 17), and the number is 17 / 40
after reading the above, have you found the rule of mental arithmetic in every situation? So as long as you practice more and master it, the problem will be solved< (3) memory training
do senior students feel that sometimes the calculation content in the topic is very extensive? Some of these operations have no specific rules of oral arithmetic, so I have to solve them through memory training. The main contents are as follows:
1
2. The proct of the approximate value of PI 3.14 with one digit and with several common numbers 12, 15, 16 and 25
3. The denominator is the decimal value of the simplest fraction of 2, 4, 5, 8, 10, 16, 20 and 25, that is, the interaction between these fractions and decimals
the results of the above numbers, whether in daily work or in real life, are used very frequently. After mastering and remembering them, they can be transformed into abilities and proce high efficiency in calculation< 4. Regular training
1. Mastering the law of operation. There are mainly five laws in this aspect: commutative law and associative law of addition; Commutative law, associative law and distributive law of multiplication. Among them, the multiplication distribution law is widely used and has many forms, including positive use and negative use, and the forms of integer, decimal and fraction. In the multiplication of fractions and integers, we often ignore the application of the law of distribution of multiplication, which makes the calculation complicated. Such as 2000 / 16 × 8. If we use the law of multiplicative distribution, we can calculate the result of 1000 by mouth directly. But if we use the general method of recing false fraction, it is time-consuming and easy to make mistakes. In addition, there are subtraction properties and quotient invariant properties< 2. Regular training. It is mainly the oral calculation method of the square result of the two digit number of 5
3. Master some special cases. For example, in fractional subtraction, if the numerator is not enough to be subtracted after general division, and the numerator subtracted is usually larger than the numerator subtracted by 1, 2, 3 and other smaller numbers, no matter how big the denominator is, it can be directly calculated orally. For example, the difference between 12 / 7 and 6 / 7 is only 1. The difference between 12 / 7 and 6 / 7 must be 1 less than the denominator. The result is 6 / 7 without calculation. Another example is: 194 / 99-97 / 99, if the difference between the numerator and denominator is 2, the difference between the numerator and denominator is 2, and the result is 97 / 99. When the subtracted molecule is larger than the subtracted molecule by 3, 4, 5 and other smaller numbers, the result can be quickly calculated orally. Another example is the mental calculation of the proct of any two digit number and 1.5, which is two digits plus half of it< 5. Comprehensive training
1
2< 3. Comprehensive training of four mixed operation sequences
comprehensive training is concive to the improvement of judgment ability, reaction speed and the consolidation of oral arithmetic
of course, we need to persevere in the above situations. Otherwise, it is difficult to achieve the expected effect if we fish for three days and dry the net for two days
the above five kinds of training should be carried out step by step, but also persevere. It will take some time to improve your math scores. Don't be too eager to succeed.
4. Compared with the old textbook, the amount of calculation practice in the new textbook has been greatly reced, and the difficulty has also been reced a lot. However, I feel that the students' computational ability has dropped a lot compared with the previous students, the computational ability is poor, the computational speed is slow, and the computational error rate is high! To this end, I think the following measures should be taken: first, adhere to the practice of oral arithmetic, often unremitting“ In order to cultivate students' computational ability, we should pay attention to the basic training of oral arithmetic, insist on regular practice, and graally achieve proficiency. " This is the requirement of mathematics curriculum standard. In my teaching, I organized teaching and training in a planned, step-by-step and purposeful way, such as listening to arithmetic, reading cards and oral arithmetic, reading questions and calculating silently, and repeatedly emphasized that the first is correct and then fast. 2、 Strengthen the basic knowledge and improve the ability of written calculation. In the teaching of calculation, oral calculation is the basis of written calculation, and written calculation is the key. Written calculation plays an extremely important role in primary school mathematics teaching. No matter how advanced the technology is in the future, the ability of written calculation is always a necessary ability for primary school students. 1. To enable students to understand the basic knowledge of mathematics and master skills is the primary condition for the formation of computing ability. Each calculation is based on the corresponding concepts, rules, properties, formulas and other basic knowledge. If students don't understand these basic knowledge correctly and master them thoroughly, they can't calculate. Only after students understand and master the relevant operation properties, laws and skills, can they apply these knowledge in specific calculation to seek simple and reasonable methods, improve the accuracy of calculation and speed up calculation. 2. Strengthening practice and skill training is the key to the formation of students' computing ability. For example, in the calculation of four items of scores, we often encounter the situation that some students' calculation rules are correct but the calculation results are wrong. The reason for the error lies in the basic skills such as approximate score, general score or mutualization; Another example is the division of decimals, especially when the divisor is a decimal. The mistake is that when the divisor becomes an integer, the divisor does not expand the same multiple at the same time, resulting in a wrong quotient. So we should strengthen the training of basic calculation skills in calculation practice. 3. Memorize common data, improve the calculation speed. In the four operations, if students memorize some commonly used data, it will not only help students to achieve the "correct and rapid" requirements, but also help students to better master the skills of calculation. For example: special data with sum and proct of whole hundred and thousand (for example: 75 + 25 = 100 + 25) × 4=100 125 × 8=1000 The proct of the approximate value of PI 3.14 with one digit and the proct with several common numbers 12, 15, 16, 25 and 36; The denominator is the decimal value of the simplest fraction of 2, 4, 5, 8, 10, 16, 20 and 25, that is, the interaction of these fractions and decimals. The results of the above numbers, whether in daily work or in real life, are used very frequently. After mastering and remembering them, they can be transformed into abilities and proce high efficiency in calculation. 3、 Carry out computing competition activities. For the boring calculation, students often deal with it casually after mastering the calculation method, resulting in more calculation errors. At this time, the proper development of some computing competition activities, often can better mobilize the students' learning initiative, improve the interest in computing, and achieve the purpose of improving the accuracy of computing. This semester, I plan to carry out two calculation competitions in the class, in order to improve the students' interest in calculation and improve the accuracy of calculation. After the competition, pay attention to sort out the students' mistakes, analyze and classify them, and then make targeted remedies, so as to clear the obstacles for improving the students' computing ability. 4、 We should encourage students to have a pleasant learning experience. Teachers should encourage students to make mistakes e to carelessness. In the usual homework correction, we should affirm its strengths, enhance self-confidence, put forward ardent hope, and urge it to correct its shortcomings.
5. 10 yuan to buy mini Miner: 0.18 yuan / day, 180 days
50 yuan to buy small Miner: 1.03 yuan / day, 170 days
250 yuan to buy medium Miner: 5.47 yuan / day, 160 days
1250 yuan to buy large Miner: 33.34 yuan / day, 150 days
6250 yuan to buy giant Miner: 180.58 yuan / day, 140 days
computing power
every miner invested under the umbrella provides computing power for leaders. It must be a mining machine in operation
micro: 1 computing power
small: 5 computing power
medium: 25 computing power
large: 125 computing power
giant: 625 computing power
Dynamic award
1:1 burn: if you proce 3 coins on the same day and a single member under the umbrella proces 5 coins, Then take this member leadership award according to the output of 3 coins
ordinary Miner: need a mining machine in normal operation
enjoy 5% of the output of the first generation on the same day
first level Miner: the computing power reaches 100, directly push three miners
give a small mining machine as a gift< The output of the second generation is 2%
on the same day of starting the second generation
the second level Miner: the calculation power reaches 500, the direct push three first level miners
give a medium-sized miner
3% of the third generation members
the third level Miner: the calculation power reaches 2500, the direct push three second-level miners
give a large miner
4% of the fourth generation members
the fourth level Miner: the calculation power reaches 12500, Direct push three third-class miners
present a giant miner
open 5% of the five generation members
the currency is 0.1 yuan, the issue price
the total amount of coins that can be mined is 10 million
after the total amount of coins mined reaches 5 million, it will go to the exchange.
50 yuan to buy small Miner: 1.03 yuan / day, 170 days
250 yuan to buy medium Miner: 5.47 yuan / day, 160 days
1250 yuan to buy large Miner: 33.34 yuan / day, 150 days
6250 yuan to buy giant Miner: 180.58 yuan / day, 140 days
computing power
every miner invested under the umbrella provides computing power for leaders. It must be a mining machine in operation
micro: 1 computing power
small: 5 computing power
medium: 25 computing power
large: 125 computing power
giant: 625 computing power
Dynamic award
1:1 burn: if you proce 3 coins on the same day and a single member under the umbrella proces 5 coins, Then take this member leadership award according to the output of 3 coins
ordinary Miner: need a mining machine in normal operation
enjoy 5% of the output of the first generation on the same day
first level Miner: the computing power reaches 100, directly push three miners
give a small mining machine as a gift< The output of the second generation is 2%
on the same day of starting the second generation
the second level Miner: the calculation power reaches 500, the direct push three first level miners
give a medium-sized miner
3% of the third generation members
the third level Miner: the calculation power reaches 2500, the direct push three second-level miners
give a large miner
4% of the fourth generation members
the fourth level Miner: the calculation power reaches 12500, Direct push three third-class miners
present a giant miner
open 5% of the five generation members
the currency is 0.1 yuan, the issue price
the total amount of coins that can be mined is 10 million
after the total amount of coins mined reaches 5 million, it will go to the exchange.
6. Find professional books, this problem is too professional, it is estimated that experts in the instry do not really understand
7. The algorithm multiplies and divides first, then adds and subtracts. You see the front as a whole, and the following minus 50 and plus 50 can offset. Then 250 * 4 equals 1000, and then multiply by 5, so the answer is 5000
8. Other diagonal braces only play a stabilizing role, and four 50 * 5 angle steels are responsible for the tensile force. Then, as long as you check the material manual, check the tensile parameters of angle steels according to the material number, and then multiply by the total area of the four, you can get the total tensile force. Pay attention to the margin. It can't be used to design tension without tension test.
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