1, knowledge and skills

(1), understand the problem of "chickens and rabbits in the same cage" and feel the interest of ancient mathematical problems.

(2) Try to solve the problem of "chickens and rabbits in the same cage" in different ways, so that students can realize the generality of algebraic methods.

2. Process and method

To solve the problem of "chickens and rabbits in the same cage", we can guess, enumerate, assume or solve the equation.

3. Emotions, attitudes and values

(1), cultivate students' logical reasoning ability.

(2) Let students realize the application of mathematical problems in daily life.

Unit weight and difficulty:

Try to solve the problem of "chickens and rabbits in the same cage" in different ways.

One kind: the problem of "chickens and rabbits in the same cage"

Teaching objectives:

1. By observing and thinking about some phenomena in daily life, students can find some special laws.

2. Solve the problem of "chickens and rabbits in the same cage" by guessing, listing, assuming or solving equations.

3. Through the study of this class, I know the history of mathematics related to "chickens and rabbits in the same cage", and edify and infect students with mathematical culture.

Teaching focus:

Try to solve the problem of "chickens and rabbits in the same cage" in different ways.

Teaching difficulties:

By observing and thinking about some phenomena in daily life, we can find some special laws.

Teaching preparation:

Story video, discussion form.

teaching process

First, a brief introduction to the story

Teacher: There are many interesting math problems in ancient China, and "chickens and rabbits in the same cage" is one of them. This problem was discussed as early as 1500 years ago.

Show me the topic: Today there are pheasants and rabbits in the same cage, with 35 heads on the top and 94 feet on the bottom. What are the geometric shapes of pheasant rabbits? There are some chickens and rabbits in the cage. From the top, there are 35 heads, and from the bottom, there are 94 feet. How many chickens and rabbits are there? )

Second, explore new knowledge.

1, teaching example 1: Some chickens and rabbits in cages. There are eight heads from the top and 26 feet from the bottom. How many chickens and rabbits are there?

Let the students discuss in pairs.

Report the results of the discussion.

(1), list:

Chicken 8 7 6 5 4 3

Rabbit 0 1 2 3 4 5

Feet 16 18 20 22 24 26

(2) Hypothesis method:

Assuming that the cage is full of chickens, it is 8× 2 = 16 feet, which is 26- 16 = 10 feet more than the topic.

Because the rabbit was just regarded as a chicken, one rabbit lost two feet, so the extra 10 feet became 10 ÷ 2 = 5 rabbits.

Therefore, chickens have: 8-5 = 3 (only)

(3), using the equation to solve:

Solution: Suppose there are X chickens, then there are (8-x) rabbits.

According to the fact that chickens and rabbits have 26 feet, make an equation.

2x+(8-x)×4=26

2x+8×4-4x=26

32-26=4x-2x

2x=6

x=3

8-3 = 5 (only)

2. Summarize the problem-solving methods:

Teacher: Which of the above three solutions is more convenient?

Summary: To solve the problem of "chickens and rabbits in the same cage", we can use the hypothesis method or the equation solution. It is more direct to solve with equations.

3. Solve the interesting problems in the book independently.

(1), equation solution:

Solution: Suppose there are x chickens, then there are (35-x) rabbits.

According to the fact that chickens and rabbits have 94 feet, make an equation.

2x+(35-x)×4=94

2x+35×4-4x=94

140-94=4x-2x

2x=46

x=23

35-23 = 12 (only)

A: There are 23 chickens and 0/2 rabbits.

(2) Arithmetic solution:

Suppose it's all chickens.

2× 35 = 70 (only)

94-70 = 24 (only)

24 ÷ (4-2) = 12 (only)

35- 12 = 23 (only)

A: There are 23 chickens and 0/2 rabbits.

Third, in-class evaluation

1, complete the title on page 1 15 of the textbook.

Students read the questions independently and then answer them in columns. Encourage solving with equations.

2. Complete the second question on page 1 15 of the textbook.

Question: What information can you learn from the pictures? (Six in the big boat and four in the small boat)

Please answer the questions independently. (When commenting, focus on the arithmetic of each step of the arithmetic solution)

6× 8 = 48 people

Suppose eight ships are big enough to hold 48 people.

48-38 = 10 (person)

Suppose the number of people is more than the actual number 10.

The reason why there are 10 people is that some ships are big ships, that is, each ship has two more people. The extra 10 people divided by the extra people in each boat is the number of boats.

10 ÷ (6-4) = 5 (article)

8-5 = 3 (bar)

This means there are three big ships.

Fourth, class summary.

Through this lesson, you can solve those problems in life.

Design intent:

The original data of 1 and "chickens and rabbits in the same cage" are relatively large, which is not conducive to students who are exposed to this kind of problem for the first time to explore. Therefore, the textbook first arranges an example of 1 to help students explore the general methods to solve this kind of problems, and then solve the original problem of large data in the calculation of Sun Tzu's Art of War. On the one hand, it can cultivate students' logical reasoning ability; On the other hand, students can understand the generality of algebraic methods.

2. The methods of guessing, enumerating, assuming or solving equations are all based on the actual situation of society.

3. Arrange some similar exercises in the exercises, such as the problem of "turtle crane" and some practical problems in life, so that students can further understand the application of such problems in daily life and consolidate the use of "hypothesis method" or equation method to solve such problems.