Quick calculation skills in primary school mathematics and calculation teaching are often associated with "abstract, boring and tasteless". How to make it easy to understand and be loved by students in teaching has always been a problem that many teachers think about. Let's take a look at the quick skills of primary school mathematics.
Quick skills in primary school mathematics 1 1. Clear mathematics.
Teaching students the key to solving problems, fast calculation requires students to master the commonly used simple operation methods, including the method of directly applying laws and properties to make the operation simple, and the indirect application of operation laws and properties after decomposition and combination. This is a simple operation method.
The former is more popular and acceptable. The latter is more difficult, but we should pay attention to cultivating students' thinking habit of looking before thinking. Once students can see and want to discover the relationship between data by themselves, and do indirect quick calculation through decomposition or combination, contact law and nature, it shows that students have mastered the "key" of quick calculation and have a high level of quick calculation.
In order to cultivate students' thinking habit of looking before thinking and their ability of decomposition or combination. For example, 70-70×3/5 can be converted into 70× (1-3/5), 125× 32× 25 can be converted into 125×8×4×25 and so on. Doing this exercise often can not only deepen students' understanding of arithmetic, but also be effective.
2. Memorize commonly used data
Improve the agility of fast computing. Practice has proved that memorizing common data can not only speed up calculation, but also develop students' thinking ability. Primary schools need to memorize a lot of data, such as the product of 125×8 and 25×4,1/41/8 ...1/20, etc.
Second, do a good job in comparative teaching and guide students to choose the best quick calculation method.
As far as a calculation problem is concerned, there are more than one calculation method, and one of them must be simple. In order to make the calculation fast, we should try to choose the simplest one that conforms to the calculation principle. Therefore, we should pay attention to the discussion of calculation methods in class, so that students can understand that the method is simple.
On this basis, students can write several methods for a problem, so that students can find the simplest one, or they can express similar problems in different ways, so that students can find the best quick calculation method for each problem, such as: 240 ÷ 6/ 15 ÷ 6/ 13 ÷ 6/.
These topics all have scores, and they are all divisions, but the quick calculation methods are different. Finally, teachers should help students guide and summarize some common types, methods, quick tricks and shortcuts, so that students can gradually form skills mastery methods.
Third, do all kinds of exercises to guide students to improve their calculation speed.
Students will use rules and arithmetic to calculate, which does not mean that they have the corresponding skills. This requires skills training through various channels to further speed up the calculation. Usually, teaching should pay attention to:
Fast calculation skills in primary school mathematics 2 1, fast calculation essentials
The formula of the "same head and same tail sum 10" algorithm is: the head is multiplied by 1 and then multiplied by the product of two tails (less than two digits make up 0). Indicates that the sum of single digits is 10. When multiplying two digits with the same ten digits, the number on the first two digits plus 1 is multiplied by the number on the next two digits.
Its product constitutes the first two digits of the product result of two digits; The product of two digits constitutes the last two digits of the product of two digits (if the product of two digits is less than 10, add 0 before the product result to form two digits), and then arrange the two digits formed by the two products in order to form the product result of two digits with the same head and the same tail.
2. Algorithm analysis
According to the quick calculation formula, it is transformed into a scientific counting method, which is expressed as: there are two digits multiplied by (10a+b) and (10a+d), and b+d= 10. The verification is: (10a+b) × (10). It is proved that according to the algebraic expression (10a+b)×( 10a+d), we can get: (10A+B )× (10A+D) =10A× 65438. +BD∶b+ d = 10∴ 10a( 10a+b+d)+BD = 10a( 10a+ 10)+BD = 65438。 The image expression of the result of = 100a(a+ 1)+bd is the basic formula of this algorithm: AB and AD are multiplied by two digits, and B+D= 10. The result is a four-digit EFGH, where EF=A(A+ 1) and GH=BD.
Second, the algorithm analysis of "same tail and same head 10"
Key points of fast calculation
The head is multiplied by the head and the tail, and the product of the two tails follows the tail (when the product of the two tails is less than 10, the decimal number is added with 0). When two two-digit numbers are multiplied, if the digits of the two numbers are the same and the sum of the ten digits is 10, the digits of the two digits are multiplied, and then the sum of the digits of any two digits is added.
Forming the first two digits of the product result of two digits; Multiply the product of two digits by two digits (if it is less than two digits, make up the last two digits of the product result of two digits), and then arrange the two digits formed by the two products in order to form the product result of two digits with the same tail and the head of 10.
2. Algorithm analysis is based on quick calculation formula, which is converted into scientific counting method: there are two digits (10b+a) and (10d+a), and b+d= 10. The verification is: (10b+a) × (65433).
Prove: According to the algebraic expression (10b+a)×( 10d+a), we can get:
( 10 b+a)×( 10d+a)= 10b× 10d+ 10b×a+a× 10d+aa = 10b 10d+ 10a(b+d)+aa
∫b+d = 10
∴ 10b 10d+ 10a(b+d)+aa= 100bd+ 100a+aa= 100×(bd+a)+aa
The image expression of the result is the basic formula of this algorithm: BA and DA are multiplied by two digits, and B+D= 10. The result is a four-digit EFGH, where EF=BD+A and GH=AA.
Third, the algorithm analysis of "tail 5, head even number"
1, fast calculation essentials "tail 5, even head" algorithm formula: the head is multiplied by the head plus the first half, and the product of the two tails is followed by the head. It means that when two numbers are multiplied, if one digit is 5 and the sum of ten digits is even, the sum of the products of ten digits and half of the sum of ten digits constitutes the first two digits of the product of two digits, while the product of two digits constitutes the last two digits of the product of two digits. After being combined in sequence, the product of two digits is formed.
2. Algorithm analysis
According to the quick calculation formula, it is converted into a scientific counting method: two digits with mantissa of 5 (10b+5) and (10d+5), and the sum of b and d is even. Proof: (10b+5) × (10d+5).
Prove: According to the algebraic expression (10b+5)×( 10d+5), we can get:
( 10b+5)×( 10d+5)= 10b× 10d+ 10b×5+5× 10d+5 = 10b 10d+50×(b+d)+5×5
∫b+d = even
∴ 10b 10d+50(b+d)+5×5= 100bd+ 100(b+d)/2+5×5
Syndrome: (10b+5) × (10d+5) =100 [BD+(b+d)/2]+5× 5.
The image expression of the result is the basic formula of this algorithm: two-digit B5 and mantissa D5, and B+D= even number. Its product is four-digit EFGH, where EF=BD+(B+D)/2 and GH=5×5.
Quick math skills in primary schools 3 100 Quick math skills in addition and subtraction.
1. Method 1: carry addition of two digits plus two digits.
Formula: plus 9 minus 1, plus 8 minus 2, plus 7 minus 3, plus 6 minus 4, plus 5 minus 6, plus 3 minus 7, plus 2 minus 8, plus/kloc-0 minus 9. (Note: The words added in the oral decision refer to the number in the unit. )
Example: 26+38=64 Solution: Add 8 and subtract 2. Who will subtract 2? 26 minus 2 equals 6. Three out of 38 will get four points.
(Note: How to round the last ten digits is 1, I enter 2, 2, 3, 4, and so on. Where are you taking it? It reaches the tenth place in the second two digits. If this is 3, I enter 4, and these two digits are 2+4=6. ) Here, 26+38=64 means that 6-2=4 is written with one digit, and 3 into 4 plus 2 means that 6 is written with ten digits.
Another example is 42+29=7 1. Just add 9 and subtract 1, 2- 1= 1, and write 1 in one place, which is 2. I enter 3,4+3 = 7, and write 7 in ten places to get 7 1.
If two numbers don't add up to carry, you can write the number directly, for example, 25+34=59, write one digit and one digit on the number after the equal sign, 5+4=9, and write ten digits and ten digits on the tenth digit, which is 2+3=5, which is 59. No vertical calculation is required.
2. Method 2: abdication subtraction of two digits minus two digits.
Oral decision:? Minus 9 plus 1, 8 plus 2, 7 plus 3, 6 plus 4, 5 plus 5, 4 plus 6, 3 plus 7, 2 plus 8, and 1 plus 9. (Note: orally decided to reduce a few words, that is, reduce one digit).
Example: 73-46=27, solution: subtract 6 and add 4, who will add 4? Three plus four equals seven. Written in one place, the decimal place of subtraction is four. I will retire. Who will retire? 7 returns 5, which is 27.