Example solving equation:

( 1)2x+xx+3 = 1； (2) 15x = 2× 15 x+ 12；

(3)2( 1x+ 1x+3)+x-2x+3 = 1。

Multiply both sides by x(3+3) to solve the equation (1) and remove the denominator.

2(x+3)+x2=x2+3x, that is 2x-3x =-6.

So x=6.

Test: When x=6, x(x+3)=6(6+3)≠0, so x=6 is the root of the original fractional equation.

(2) multiply both sides of the equation by x(x+ 12), and remove the denominator to get.

15(x+ 12)=30x。

In order to solve this problem, you must

x= 12。

Test: when x= 12, x (x+12) =12 (12+12) ≠ 0, so x= 12 is the original score.

(3) sorting to obtain

2x+2x+3+x-2x+3 = 1, that is 2x+2+x-2x+3 = 1,

That is 2x+xx+3= 1.

Multiply both sides of the equation by x(x+3) and get it by removing the denominator.

2(x+3)+x2=x(x+3)，

Namely 2x+6+x2=x2+3x,

That is 2x-3x =-6.

Solve the whole equation and get x=6.

Test: When x=6, x(x+3)=6(6+3)≠0, so x=6 is the root of the original fractional equation.

Second, the new lesson

A group of students went to visit outside the school. After they have walked for 30 minutes, the school will send an urgent notice to the team leader, and send a student to start from the school by bike and catch up with the team by the same route. If the speed of cycling is twice that of the team, the distance from the school when students catch up with the team is 15 km. How long did it take the student from school to catch up with the team?

Please find out the equivalent relationship in the topic according to the meaning of the topic.

Answer: Riding distance = parade distance =15 (km);

Cycling speed = twice the walking speed;

Cycling time = walking time -0.5 hours.

Please list the equations according to the above equivalence relation.

Answer:

Method 1 suppose it takes x hours for the student to catch up with the team by bike, and the equation according to the meaning of the question is as follows

15x=2× 15 x+ 12。

Method 2 Let the walking speed be x km/h and the riding speed be 2x km/h, and the equation according to the meaning of the question is

15x- 15 2x= 12。

Solution method 1 listed equation, which has been solved in the review. The equation listed in Method 2 is solved as follows.

Multiply the two sides of the equation by 2x and divide by the denominator.

30- 15=x，

So x= 15.

Test: when x= 15, 2x=2× 15≠0, so x= 15 is the root of the original fractional equation, which is consistent with the meaning of the question.

So the time to catch up with the team by bike is 15km 30km/h = 12h.

A: It takes 30 minutes to catch up with the team by bike.

It is pointed out that in the example 1, we use two relationships, namely, time = distance speed and speed = distance time.

If the speed is unknown, then find the equation of equal relationship according to time; If the time is unknown, please press.

The equations of velocity equivalence relation are all fractional equations.

A project needs to be completed within the specified date. If Team A did it, it would be finished on schedule. If Team B does it, it will take three days to finish it. Now team A and team B work together for two days, and the rest of the projects are completed by team B alone, only on the specified date. How many days is the specified date?

Analysis; This is an engineering problem. There are three quantities in engineering problems. The workload is set to s, the working time is set to t, the working efficiency is set to m, and the relationship between the three quantities is

S=m=st, or t=sm, or m=st.

Please list the equations according to the equivalence relation in the question.

Answer:

Method 1 The specified date of the project is the number of days required for Party A to complete the project alone. If it is set to x days, then the number of days required for Party B to complete the project alone is (x+3) days. If the total project amount is 1, the work efficiency of Party A is x 1, and the work efficiency of Party B is 1x+3. According to the meaning of the question, the equation is

2( 1x+ 1x 3)+x2-xx+3 = 1。

It is pointed out that the meaning of work efficiency is the amount of work completed in unit time.

Method 2 sets the specified date to x days. After two days of cooperation between Party B and Party A, the remaining projects were completed by Party B alone, just on the specified date. Therefore, Party B's working time is X days, and the equation is formulated according to the meaning of the question.

2x+xx+3= 1。

Method 3 According to the equivalence relation, the total workload -A workload = B workload. If the specified date is X days, the equation can be listed.

1-2x=2x+3+x-2x+3。

We have solved the equations listed in method 1 ~ method 3 before the new lesson, so we don't understand the fractional equation here. The key is to find the equation of equality relationship.

Third, classroom exercises.

1.A takes time to process 180 pieces, and b can process 240 pieces. It is known that A processes 5 parts less than B per hour, so find the number of parts processed by two people per hour.

2. The distance between A and B is 135km. There are two cars, a big car and a small car, which drive from A to B. The big car leaves 5 hours earlier than the small car and the small car arrives 30 minutes later. As we all know, the speed ratio of big car to small car is 2: 5. Find the speed of two cars.

Answer:

1.a processes 15 pieces per hour, and b processes 20 pieces per hour.

2. The speed of bus and car are 18km/h and 45km/h respectively.

Four. abstract

1. The method and steps of solving application problems by column fractional equation are basically the same as those by column linear equation. The difference is that the root must be tested to solve the fractional equation. On the one hand, it depends on whether the original equation has roots, on the other hand, it depends on whether the roots solved conform to the meaning of the problem. The added roots of the original equation and the roots that do not conform to the meaning of the problem should be discarded.

2. When solving application problems with fractional equations, it is generally assumed that the required quantity is unknown. This method of setting unknowns is called setting direct unknowns. However, sometimes the required quantity cannot be directly set as an unknown quantity according to the characteristics of the problem, but another quantity should be set as an unknown quantity. This method of setting unknowns is called setting indirect unknowns. When solving application problems with fractional equations, setting indirect unknowns can sometimes make the solution simple. For example, in the first type of exercise, the condition of the topic remains unchanged, and the problem is changed to find the time for two cars to travel from A to B. If the direct unknown is set, it is assumed that it takes x hours for a car to travel from A to B and (X+5- 12) hours for a bus to travel from A to B. According to the meaning of the topic, the formula is as follows.

135 x+5- 12: 135 x = 2:5。

Solving this fractional equation is a complicated operation. If the setting is indirect unknown, that is, the speed is set to unknown, you can calculate the speed of the cart and the car first, and then calculate the time from A to B separately, so the operation is much simpler.

Verb (short for verb) homework

1. Fill in the blanks:

(1) It takes m hours to complete a work alone, and n hours to complete it alone. If two people work together, the time to finish the work is _ _ _ _ _ _ _ _;

(2) A canteen has m kilograms of rice, and it was originally planned to use one kilogram of grain every day. Now save B kilograms of grain every day, so the number of days that can be used more than originally planned is _ _ _ _ _;

(3) Dissolve one kilogram of salt in b kilograms of water, so the salt content in this m kilograms of brine is _ _ _ _ _ kilograms.

2. Solve application problems with column equations.

(1) A worker's master processed 1500 parts twice. In the second processing, he innovated tools and improved the operation method. As a result, it took 18 hours less than the first time. It is known that the efficiency of his second processing is 2.5 times that of the first one. How many parts can he process per hour in the second processing?

(2) If someone walks 8 kilometers more per hour by bike than by walking, and if the time for him to walk 12 kilometers is equal to the time for him to ride 36 kilometers, how many hours will it take him to walk 40 kilometers?

As we all know, ships travel 20 kilometers per hour in still water. If the time it takes a ship to sail 72 kilometers downstream in a river is the same as the time it takes to sail 48 kilometers upstream, what is the speed of the river?

(4) The distance between A and B is 135km. Two cars drive from A to B. The bus leaves 5 hours earlier than the car, and the car arrives 30 minutes later than the bus. It is known that the speed ratio of two cars is 5: 2. Find out the speed of each car.

Answer:

1.( 1)Mn m+n； (2)m a-b-ma； (3) ma a+B.

2.( 1) During the second processing, 125 parts are processed every hour.

(2) It takes 40 4 =10 hour to walk 40 kilometers. It takes 10 hour to walk 40 kilometers.

(3) The velocity of the river is 4 km/h. 。

Description of classroom teaching design

1. In teaching design, for example, 1, guide students to find three equal relations according to the meaning of the question and list the equations in two different ways; Example 2, guide students to list equations in three different ways according to the meaning of the question. This arrangement is intended to inspire students to think from different angles and directions, and encourage students to develop flexible thinking habits when solving problems. This provides a broad space for cultivating students' divergent thinking in the teaching of solving application problems with fractional equations.

2. Teaching design embodies the mode function of giving full play to examples. Example 1 is a travel problem, in which the distance is a known quantity, and the speed (or time) is found; Example 2 is an engineering problem, the total workload is a known quantity, and the time (or work efficiency) to complete the workload is found. These are typical problems solved by fractional equations. In teaching, students are guided to deeply analyze the equivalence relationship between known quantities and unknown quantities and topics, as well as the idea of solving equations, so as to promote students to deepen their understanding and understanding of the main characteristics of the model. More importantly, let students know what kinds of problems can be solved by fractional equation and what the thinking is. After completing classroom exercises and homework, students can identify the types of problems, connect the problems they face with the patterns they have mastered in their minds, and explore ways to solve them.

3. Using fractional equation to solve applied mathematics permeates the thinking method of equation, from which students realize that the thinking method of equation is a sharp weapon to solve mathematical problems. The thinking method of the equation can be described in two sentences: "take the fake seriously" and "make the fake come true". How to assume that the required quantity is X by setting a direct unknown quantity or an indirect unknown quantity, and then treat it as a real quantity. By finding an equal relationship, the equation is listed. At this time, treating the known quantity and the assumed unknown quantity equally is "confusing the fake with the real". By solving the equation to get the solution of the problem, the previously assumed unknown quantity X becomes a definite quantity, which is "turning the false into the true".