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Quick oral calculation skills in grade two of junior high school
1. Practice makes perfect.

"Rome wasn't built in a day", and verbal ability is a basic skill for children. It is necessary to plan long and arrange short, carry out teaching and training purposefully, planned and step by step, and embody the basic principle of teaching step by step and according to the new curriculum standard.

In daily life, I insist on oral arithmetic training for 3-5 minutes every day and practice 1 crossing calculation every day. At the beginning, it is completed under the supervision of parents, and children can gradually let go to form a habit and voluntarily complete it.

2. Quality and quantity training

Each exercise should record the time taken to complete 1 page, correct right and wrong immediately after completion and analyze the cause of the error. Every time you do training, you should compare it with the last time to see if there is any progress. Parents should praise their children appropriately and say, "Great, progress!"

Children need timely encouragement at this time, just as players who struggle on the court need teammates to be cheerleaders, they can also post their excellent exercises in eye-catching places at home as incentives, and sometimes they can give a small prize.

Parents should never be impatient at the sight of their children's slowness and say, "What a fool!" " If parents will have a bad influence on children, it is something that needs patience and love to do well.

For children who regress, we can analyze the reasons for retrogression with them, then gently touch their heads and put forward what to do in the future. If their grades improve next time, they will be praised in time to encourage them to continue their efforts and build up their confidence.

3. Diversified training forms

Doing more and practicing more is the premise, but children are used to being interested in new things, especially learning in games to increase their knowledge. If they practice alone for a long time, they will easily get bored.

Therefore, oral arithmetic practice should be lively and varied, and the supplementary methods that can be taken in practice include: playing poker (24 o'clock), listening to arithmetic, driving a train, looking up passwords, winning red flags, sending letters, finding friends, fighting for the championship, regular tests and so on. (practical, simple)

At the same time, through some mathematical practice activities, let children realize the importance of cultivating language ability in our daily life (such as buying food and visiting supermarkets). ).

4. Know arithmetic and master clever calculation methods.

The improvement of verbal ability depends on children's understanding of arithmetic. Only on the basis of understanding can we get the effect of giving inferences by analogy, greatly improve the speed and accuracy of oral calculation and form oral calculation ability. To this end, we should pay attention to strengthening children's understanding of arithmetic.

For example, rounding method, rounding method, decomposition method and memorizing some commonly used data are commonly used in oral calculation. Look at 25×4= 100, 125×8= 1000 to remind children that they can often be used as crutches for oral calculation, and sometimes they can be converted into 25×4= 100,125× 00 by decomposition method.

Let the children write down the clever calculation methods they usually find and share them with their classmates. At the same time, it also cultivates children's interest in verbal arithmetic.

5. Develop good computing habits

Developing good computing habits is an effective way to improve children's computing ability. Help children develop the following good calculation habits, and do the serious calculation habits of "seeing, thinking and calculating".

Calculation is a very serious matter, and we can't be careless, but just some children don't have good study habits. After getting the calculation problem, they can't see the numbers clearly, and they can't see the operation order clearly, so they calculate blindly.

For example, when calculating a simple calculation problem like 6+4÷2, the child is careless and the result is 5. If you calculate carefully, it is easy to see that the operation order of this problem is to divide first and then add, and the correct result should be 8.

Fast calculation skills of primary school mathematics

It is said that this is the fastest mathematical calculation method in the world! For the sake of children, parents must put their children away!

1, ten times ten:

Formula: head joint, tail to tail, tail to tail.

For example: 12× 14=?

Solution: 1× 1= 1.

2+4=6

2×4=8

12× 14= 168

Note: Numbers are multiplied. If two digits are not enough, use 0 to occupy the space.

2. The heads are the same and the tails are complementary (the sum of the tails is equal to 10):

Formula: After a head is added with 1, the head is multiplied by the head and the tail is multiplied by the tail.

For example: 23×27=?

Solution: 2+ 1 = 3

2×3=6

3×7=2 1

23×27=62 1

Note: Numbers are multiplied. If two digits are not enough, use 0 to occupy the space.

3. The first multiplier is complementary and the other multiplier has the same number:

Formula: After a head is added with 1, the head is multiplied by the head and the tail is multiplied by the tail.

For example: 37×44=?

Solution: 3+ 1=4

4×4= 16

7×4=28

37×44= 1628

Note: Numbers are multiplied. If two digits are not enough, use 0 to occupy the space.

4, dozens of eleven times dozens of eleven:

Formula: head joint, head joint, tail to tail.

For example: 2 1×4 1=?

Answer: 2×4=8

2+4=6

1× 1= 1

2 1×4 1=86 1

5. Multiply 1 1 by any number:

Formula: head and tail do not move down, middle and pull down.

For example: 1 1×23 125=?

Answer: 2+3=5

3+ 1=4

1+2=3

2+5=7

2 and 5 are at the beginning and end respectively.

1 1×23 125=254375

Note: If you add up to ten, you will get one.

6, more than ten times of any number:

Formula: The first digit of the second multiplier does not drop, the single digit of the first factor multiplies each digit after the second factor, and then drops.

For example: 13×326=?

Solution: 13 bit is 3.

3×3+2= 1 1

3×2+6= 12

3×6= 18

13×326=4238

Note: If you add up to ten, you will get one.