1, addition

There are many ways to train additive thinking within 20 minutes: counting, connecting, adding ten, oral calculation, deduction, subtraction and so on. According to different cultural environment, family background and their own thinking, it should be realized through students' own hands-on practice, independent exploration and cooperation and exchange. The emphasis here is on subtraction.

We stipulate that the two numbers that can make up 10 are complementary, 1 and 9, 2 and 8, 3 and 7, and so on. They complement each other.

The method is: subtract the complement of the second addend from the first addend and add 10. For example:

9+4= 13

Thinking method: the complement of the second addend is 6; The first addend 9 subtracts the complement 6 of 4 to get 3; Add 10 to 3 and you get 13. That is 9+4 = 9-6+10 = 3+10 =13.

This way of thinking is conducive to cultivating students' reverse thinking ability, but it can only satisfy students with strong thinking ability. Teachers can guide according to the situation.

Step 2 subtract

The subtraction of abdication within 20 is based on addition within 20, and the methods are: addition subtraction, division by ten, decomposition subtraction, decimal to large number, subtraction, addition, complement and so on. Here, we mainly introduce the supplementary methods:

The method is: add the complement of the minuend with the number in the unit, and remove the tenth "1", such as the minuend.

13 - 4 = 9

Thinking method: 3 in the minuend unit is not reduced enough; The complement of subtrahend 4 is 6; 6 plus 3 in the minuend gets 9, and the tenth "1" is removed.

Second, two-digit addition and subtraction oral calculation:

Two-digit addition and subtraction This paper focuses on the methods of subtraction and addition. First of all, we stipulate that two numbers whose sum is 100 are complementary to each other.

1, addition

There are four phenomena in the addition of two digits, that is, one digit and ten digits do not carry; Carry one digit, not ten digits; There is a decimal place and a single place without decimal place; All figures are rounded. The following are introduced respectively:

(1), two digits plus one digit plus ten digits do not carry, and they are added directly by the method of number synthesis.

For example: 34+52 = 30+50+4+2 = 86

(2) Two-digit addition, carry one digit, not carry ten digits. The way to think is:

One plus ten digits and another plus ten digits plus "1" get ten digits, and each digit subtracts the complement of 100 of the other digit and adds the digits to get the digits.

For example: 36+ 47 = 83

Oral calculation process: the number on the tenth digit is 3+4+ 1=8.

The number in the unit is 6-3(3 is the ten complement of 7) =3.

Or 7-4(4 is the ten complement of 6) =3.

So: 36+47 The decimal digit is 8, and the single digit is 3, which is equal to 83.

(3) Two digits plus decimal number, and one digit without decimal number. The way to think is:

Firstly, the number of digits in the "hundred digits" is determined as "1", then the digits are obtained by subtracting the complement of ten from the number of digits plus one, and the digits are added directly by the method of digit synthesis.

For example: 83+64 = 147

Oral calculation process: the hundred digits are "1".

The decimal number is 8-4 = 4 or 6-2 = 4.

The unit is 3 +4 = 7.

So: 83+64 hundred digits are 1, 10 digits are 4, and 7 digits are equal to 147.

(4) two-digit addition, all digits carry, thinking method is:

First, make sure that the digit of 100 is "1", and then subtract the complement of 100 of another addend from one addend to get ten digits and one digit.

For example: 86+59= 145

Oral calculation process: the hundred digits are "1".

The decimal and unit numbers are 86-4 1(59' s complement) =45.

Or 59- 14(86' s complement) =45.

So: 86+59 digits are 1, and the decimal digits and digits are 45, which is equal to 145.

2, abdication subtraction

Two-digit subtraction We focus on abdication subtraction.

(1) two digits minus two digits, thinking method is:

Subtract the negative ten digits from the negative ten digits first, then subtract "1", which is the difference of ten digits, and then add the negative ten's complement of the negative one digit to the negative one digit, which is the difference of one digit.

For example: 83-26 = 57

Oral calculation process: the ten digits are 8-2-1= 5.

The number of units is 3+4(4 is the ten complement of 6) =7.

So the decimal digit of 83-26 is 5, and the single digit is 7, which is equal to 57.

(2) The minuend is more than one hundred abdication subtraction, and the thinking method is:

First, make sure that the hundred digits are 1- 1=0, that is, the difference of this number is dozens, and then the number of decimals and digits plus the complement of decimals and digits is the difference.

Example 132-67 = 65

Oral calculation process: 32+33(33 is the hundreds' complement of 67) =65.

Third, two-digit multiplication oral calculation

One-digit multiplication oral calculation is a formula table, which needs to be recited on the basis of Qing theory. This paper mainly introduces several special algorithms of two-digit multiplication.

1, two-port algorithm of the same factor product; (Square mouth algorithm)

(1), sum port algorithm of basic number and difference number:

Basic number: this number is squared separately to form a new number called basic number. The square of ten digits is a number above hundred digits, the square of one digit is a number of ten digits and single digits, and ten digits are occupied by zeros.

Difference: Multiplying the product of these dozens of bits and one bit by 20 is called difference.

Cardinality+Difference = the product of these two same factors.

For example, 1,13x13.

Basic number: Hundred: 1× 1= 1.

Ten places: use 0 to occupy one place.

Unit: 3×3=9

So the cardinality is 109.

Difference: 1×3×20=60

Base number+difference number = 109+60 = 169

So 13× 13= 169.

Example 2, 67×67

Basic number: 100 digits or more is 6×6=36.

Ten digits and single digits are 7×7=49.

So the cardinal number is 3649.

Difference: 6×7×20=840

Basic number+difference number =3649+840=4489

So: 67×67 = 4489

(2) Three-step method

Thinking process:

Step 1: Square this number. The number obtained, one digit is the product, and ten digits are reserved.

Step 2: Multiply these numbers by ten digits, then multiply them by two, and then add up the digits reserved in step 1. The number of digits obtained is the ten digits of the product, which are reserved.

Step 3: Square these tens of digits and add the number reserved in step 2 to get the hundreds and thousands of digits of the product.

Example 1, 24×24

Step 1: 4× 4 =16 "1"is reserved, and "6" is the single digit of the product.

Step 2: 4× 2× 2+1=17 "1"is reserved, and "7" is the ten digits of the product.

Step 3: 2× 2+1= 5 "5" is one percent of the product.

So 24×24=576

Example 2: 37×37

Step 1: Keep 7× 7 = 49 "4", and "9" is the single digit of the product.

Step 2: 3× 7× 2+4 = 46 "4" is reserved, and "6" is the ten digits of the product.

Step 3: 3× 3+4 =13 "13" is the hundredth and thousandth bit of the product.

So: 37×37= 1369

(3) Oral calculation of two products with the same factor close to 50.

Thinking method: the product of two identical numbers greater than 50 is equal to 5 times 5 plus one digit, and then the square of one digit (two digits must be occupied, and ten digits should be occupied by zero): the product of two identical numbers less than 50 is equal to 5 times 5 minus one digit's ten complement, and then the square of one digit's ten complement (two digits must be occupied, and ten digits should be occupied by zero).

Example 1, 53×53

5×5+3=28 plus 3×3=9 (there must be two digits 09) equals 2809.

So: 53×53=2809

Example 2, 58×58

5×5+8=33 plus 8×8=64 equals 3364.

So: 58×58=3364

Example 3, 47×47

5×5-3(3 is the decimal complement of 7) =22, plus 3×3=9 (there must be two digits 09).

Equal to 2209

So: 47×47=2209

(4) Oral calculation of the product of two identical factors with the last digit of 5.

Thinking method: Let the decimal place of this number be k, then the product of these two same factors is: K×(K+ 1) plus 5×5=25 or K×(K+ 1)× 100+25.

For example, 1, 35× 35 = 3× (4+1)×100+25 =1225.

Example 2: 75× 75 = 7× (7+1)×100+25 = 5625.

There are many oral calculation methods for two products of the same factor, which will not be introduced here. We can use the oral calculation method of the product of two identical factors to calculate many similar products of two numbers. Examples are as follows:

For example, 1, 13× 14.

Because: 13x13 =169 plus13 gives 182, so:13x14 =182.

Or 14× 14 because: 14 = 196, if you subtract 14, you get 182.

Example 2, 35×37

Because: 35×35= 1225 plus 70(2×35) is 1295.

So 35×37= 1295

2. Oral calculation of numbers that start and end regularly

(1) Folding of head and tail (head and tail mending)

Thinking method: multiply the first number by "1" and add the product of mantissa (two digits) to the right. If the product is a single digit, the ten digits will be occupied by zeros.

For example: 76× 74 = (7+1)× 7×100+6× 4 = 5624.

(2) Tail with head (tail with head complement)

Thinking method: multiply the first number by the mantissa and add the square of the mantissa (two digits) to the right. If the product is a single digit, the ten digits will be occupied by zeros.

For example: 76× 36 = (7× 3+6 )×100+6× 6 = 2736.

(3) One is the same, folded (one number with two digits is the same, and one number with two digits is complementary)

Thinking method: multiply the ten digits of two numbers, add the same number, and add the product of two mantissas to the right. If the product is a number, ten numbers are occupied by zeros.

For example: 33× 64 = (3× 6+3 )×100+3× 4 = 2112.

The above three methods can be calculated by one formula:

(head × head+same) × 100+ tail × tail

3. Use special numbers for multiplication and oral calculation.

Some sizes are special, and their products are regular.

(1)7 times a multiple of 3 or 3 times a multiple of 7.

Let's look at the following formula:

7×3=2 1 7×6=42 7×9=63

7× 12=84 7× 15= 105 7× 18= 126......7×27= 189

We observe that the multiplicand of these formulas is 7 and the multiplier is a multiple of 3. It is a multiple of 3, and what is the unit of the product? The product of ten digits or more is always twice the unit.

Therefore, we can say that the multiple of 7 times 3 equals the multiple plus 20 times.

If we set this multiple as n, it can be expressed by the formula: 7× 3n = n+20n (where n > 0 is a positive integer).

Example: 1, 7×27=7×3×9=9+20×9= 189.

Example 2, 7× 57 = 7× 3×19 =19+20×19 = 398.

This conclusion also applies to multiples of 3 times 7. With this conclusion, we can orally calculate the multiplication of two numbers that are multiples of 3 and 7.

Example 3:14×15 = 7× 2× 3× 5 = 7× 3×10 =10+20×10 = 210.

Example 4: 28× 36 = 7× 4× 3×12 = 7× 3× 48 = 48+20× 48 =1008.

(2) 17 times 3 or 3 times 17.

17 times 3 is equal to this multiple plus 50 times this multiple. (A multiple of 3 times 17 is also applicable. )

If we set this multiple to n, it is expressed by the formula: 17× 3n = n+50n (n > 0 is a positive integer).

For example, 1,17× 21=17× 3× 7 = 7+50× 7 = 357.

Example 2:17× 84 =17× 3× 28 = 28+50× 28 =1428.

Example 3, 34× 24 =17× 2× 3× 8 =17× 3×16 =16+50×16 = 816.

(3) multiply 17 by 13 or multiply 13 by 17.

17 times 13 is equal to this multiple plus 20 times this multiple, plus 200 times.

If we set this multiple to n, it is expressed by the formula:17×13n = n+20n+200n (n > 0 is a positive integer).

For example, 1,17× 78 =17×13× 6 = 6+20× 6+200× 6 =1326.

Example 2: 34× 65 =17× 2×13× 5 =17×13×/kloc-0 =10+20×/kloc-0.

=22 10

Example 3: 34× 78 =17× 2×13× 6 =17×13×12 =12+20×/kloc-0.

=2652

(4)43 times a multiple of 7 or 7 times a multiple of 43.

The multiple of 43 times 7 is equal to this multiple plus 300 times this multiple.

If we set this multiple to n, it is expressed by the formula: 43× 7N = n+300N (n > 0 is a positive integer).

Example: 1, 43× 28 = 43× 7× 4 = 4+300× 4 =1204.

Example 2: 43× 84 = 43× 7×12 =12+300×12 = 3612.

4. Oral calculation of multiplication of two numbers close to 100

(1) Multiply two numbers by 100.

Thinking method: First add the difference between one factor and another factor with 100, and then add the difference between the two factors with the product of 100 respectively after the obtained results.

For example, 1,103x104 = (103+4) ×100+3x4 =10712.

Example 2:112×107 = (112+7 )×100+12× 7 =1.

(2) Multiply two numbers less than 100.

Thinking method: first subtract the difference between another factor and 100 from one factor, and then add the product of the difference between the two factors and 100 after the obtained result.

For example: 1, 92× 94 = (92-6 )×100+8× 6 = 8648.

Or: 92× 94 = (94-8 )×100+8× 6 = 8648.

(3) Multiply two numbers greater than 100 and less than 100.

Thinking method: the number exceeds 100, and the difference is less than100; If the magnification is 65,438+000 times, the product of the difference between two factors and 65,438+000 is subtracted.

For example, 1,104× 97 = (104-3 )×100-4× 3 =10/00-12.

There are too many skills in verbal calculation. The above is just to introduce some special oral arithmetic skills, which can be performed by using arithmetic rules and operational properties; Rounding method can be used for oral calculation and so on. There is only one key for our teachers to memorize and master these methods: finally, calculate the results quickly and accurately.

Proficient in basic verbal calculation. The addition and subtraction of carry within 20, the subtraction of abdication, and the multiplication and division in the table should all reach the proficiency of "blurting out". Because any one of the four arithmetic problems is the synthesis of a series of oral calculations, if one of them is wrong, it will be scrapped. The accuracy and proficiency of oral calculation directly restrict the cultivation and improvement of computing ability.

Remember common data. If we can remember the commonly used data in calculation on the basis of understanding, the accuracy and speed of calculation can be greatly improved. Such as 4×25= 100, 4×75=300, 8× 125= 1000, 1÷2=0.5,1÷ 4 =

Simple verbal calculation should be conscious. Making use of numerical characteristics and operational relations, and applying operational rules or properties to consciously carry out simple calculations are conducive to cultivating the flexibility and agility of students' thinking. For example, 389+298 and 654-496 can be simplified by the law of sum and difference. 389+298=389+300-2=689-2=687, 654-496 = 654-500+4 =154+4 =158, and then subtract a few; Lose more and add a few more. 312× 25,2700 ÷125 can be simplified by using the changing law of product sum quotient. 3 12×25=(3 12÷4)×(25×4)=78× 100=7800,2700÷ 125=(2700×8)÷( 125×8)=2 1600÷ 1000=2 1.6

Practice oral arithmetic often. Oral arithmetic practice should run through the whole process of teaching activities, focus on the teaching content and be targeted. Lower your posture purposefully. Before Protestantism, practice oral arithmetic, and "review the old and learn the new" plays a transfer role. Practicing oral arithmetic in new teaching has the stability of applying new knowledge. Practicing oral arithmetic after Protestantism is conducive to forming a good cognitive structure, making students consciously apply the laws or properties of operation, changing the original operation order and making calculation simple.

Mental arithmetic skills should be cultivated. Mastering the method of oral calculation on the basis of understanding arithmetic is the first and very important step in learning oral calculation, but to some extent, it is necessary to simplify and compress the thinking process and form the skills and techniques of oral calculation. For example, some formula problems of the same level calculation, 36÷7× 14, 72× 18÷24, cannot be calculated orally on the surface. According to the operation law or the nature of the budget, after reasonable adjustment, oral calculation can be made. 36 ÷ 7×14 = 36× (14 ÷ 7) = 36× 2 = 72,72×18 ÷ 24 = 72 ÷ 24×/kloc-0.