Mathematics teacher, secondary school No. 23, Astrakhan

Novakova S.A.

LESSON TOPIC: RATIONAL INEQUALITIES

Grade 9

The purpose of the lesson: to consolidate and deepen the knowledge of students in the process of solving various exercises on a given topic; to promote the development of mutual assistance and mutual assistance, the ability to conduct a cultural discussion.

Lesson objectives:

1. consolidate the ability to solve rational inequalities by the interval method; consider rational inequalities of various levels of complexity; to test students' ability to solve rational inequalities;
2. create conditions for the development of skills and abilities to apply knowledge in new situations; for the development of the qualities of thinking: flexibility, purposefulness, rationality, criticality, taking into account individual characteristics.

Lesson type : general lesson; consolidation and improvement of knowledge and skills.

Forms of organizing activities in the lesson:

1. frontal
2. individual
3. collective

Lesson structure:

1. Organizing time;
2. motivational conversation;
3. updating knowledge;
4. individual or collective work with assignments;
5. summarizing.

Methods:

1. verbal;
2. visual;
3. practical.

Equipment:

1. computers;
2. multimedia projector;
3. personal cards.

Predicted result:strengthening the skills and abilities of solving rational inequalities; the formation of the ability to plan their work; achievement by each student of the level of skills that he needs:

I level - to solve the simplest rational inequalities; solve inequalities according to a given algorithm;

Level II - solve rational inequalities, independently choosing a solution method;

Level III - apply the acquired knowledge in a non-standard situation.

DURING THE CLASSES.

1. Organization. Setting goals.
2. Updating of basic knowledge. oral exercises.(Slide 2-4)

1) Are the following inequalities equivalent?

a) and (no)

b) and (yes)

2) Determine the method for solving the equation:

3) Determine the course of solving the inequality:

b) ﴾2х 2 +11х+6)﴾2х 2 +11х+13)

1. Repeat the algorithm for solving a rational inequality using the interval method:(Slide 5)

1. In each factor, the coefficient at the highest degree of the variable must be positive, for this it is necessary to take out the minus from all factors in which the coefficient at the highest degree is negative, and if there is still a minus sign in front of the expression, then the entire inequality must be multiplied by (-1).

Get the roots of the numeratorand discontinuity points of the denominator.

1. On the number line, we plot all the obtained values ​​​​and draw a curve of signs.

1. Problem solving.(Slide 6, 7)

1. Solve the inequality.

Answer:

2. Solve the inequality.
Answer:

3. Find the difference between the integer largest and smallest solutions of the inequality

Answer: 4.

4. Solve the inequality.
Answer:

5. Find the product of the largest negative integer and the smallest positive integer solution of the inequality

Answer: -42.

6. Find the smallest integer solution to the inequality.

7. How many prime numbers are solutions to the inequality?

Answer: 1.

1. Personal cards for verification work.

Card number 1.

1. Solve the inequality:

≤ .

a) [-4; -2) ∪ (0;5],

b) (–1, 0] ∪ ,

d) there are no solutions.

2. Find the largest integer x satisfying the inequality:

- > 1.

a) x ∈ (- ∞ ; -3.5),

B) -3,

at 4,

d) there are no solutions.

Card number 2.

1. Find the largest integer x satisfying the inequality:

- > -.

a)5,

b) -3,

at 4,

d) there are no solutions.

2. Solve the inequality:

a) (-9; -5) ∪ (0; 8),

B) (–8, -7) ∪ (1; 3),

B) (- ∞ ; -7) ∪ (1; 3),

D) there are no solutions.

Card number 3.

1. Solve the inequality:

a) (- ∞ ; -3) ∪ (0; 3,

B) (–3, 0) ∪ (0; ∞ ),

B) (5; 7),

D) there are no solutions.

2. Find integer solutions of inequalities:

a) 0, 1, 2,

B) 4, 5,

AT 7,

D) there are no solutions.

Card number 4.

1. Solve the inequality:

a) (- ∞ ; -3/25) ∪ (0; ∞ ),

b) (–12, 0) ∪ (7;9),

B) (- ∞ ;) ∪ (; 5),

D) there are no solutions.

2. Find the sum of integer solutions of the inequality

a) 2,

b) 4,

c) 0,

d) 1,

e) 3.

1. Summarizing.

During the lesson, students consolidated the ability to solve rational inequalities, considered the solution of rational inequalities of various levels of complexity. Students in practice showed the ability to apply the method of intervals in solving rational inequalities. Particular attention should be paid to solving non-strict rational inequalities.

1. Homework.(Slide 8)

1. Find the smallest integer negative solution to the inequality

2. Solve the inequality.
3. Find the sum of the largest and smallest integer solutions of the inequality

.

1. Bibliography:

1. Algebra: Proc. For 9 cells. general education institutions. / S.M. Nikolsky, M.K. Potapov, N.N. Reshetnikov, A.V. Shevkin. - 2nd ed. – M.: Enlightenment, 2003. – 255 p.
2. Algebra 8th grade. Tasks for the training and development of students. / Belenkova E.Yu., Lebedintseva E.A. - M.: Intellect - center, 2003. - 176 p.
3. "Small USE" in mathematics: Grade 9: Preparing students for final certification / M.N. Kochagin, V.V. Kochagin. – M.: Eksmo, 2008. – 192 p.

Lesson No.: 16 Date:_________

Lesson topic: "Systems of rational inequalities".

Lesson Objectives:

educational: promote the development of skills for solving systems of inequalities; learn to find common decision systems of inequalities; teach to solve a system containing quadratic inequalities; repeat the interval method;

developing: to teach to express thoughts in oral speech, to draw conclusions, to summarize, to form self-control skills;

educational: to teach to listen and accept the point of view of others, to cultivate a sense of patriotism, to instill love for the subject.

Lesson type: combined.

During the classes

Organizing time:

greetings;

checking the readiness of students for the lesson;

announcement of the topic and formulation of the objectives of the lesson;

examination homework.

Verbally check homework. Discuss assignments that are difficult for students.

II. Doing exercises.

1. Recall the formula for factoring a square trinomial.

2. Repeat what the interval method is when solving square inequalities.

3. Solve No. 4.9 (d). The solution is explained by the teacher.

1) Solve inequality 3 X – 10 5X – 5; 3X – 5X – 5 + 10; – 2X 5;
X

2) Solve the inequality X 2 + 5X + 6 X 2 + 5X + 6 = 0; D = 1; X 1 = – 3;
X 2 = - 2; then ( X + 3)(X + 2)

We have - 3 X

3) Find a solution to the system of inequalities

Ans: – 3 X

4. Solve No. 4.9 (c) independently with verification.

Ans: No solutions.

5. Solve No. 4.10 (d). The teacher explains. First, repeat the square trinomial theorem with negative discriminant.

G)

1) Solve the inequality - 2 X 2 + 3X – 2 X 2 + 3X – 2 = 0; D = 9 – 16 = = – 7 X.

2) Solve the inequality –3(6 X – 1) – 2X X; – 18X + 3 – 2X X; – 20X – X X X Solution of this system of inequalities X

Answer: X

6. Solve No. 4.10 (c) on the board and in notebooks.

in)

Let's solve inequality 5 X 2 – 2X + 1 ≤ 0. 5X 2 –2X + 1 = 0; D = 4 – 20 = –16

By the theorem, the inequality has no solutions, which means that the given system has no solutions.

Ans: No solutions.

7. Solve No. 4.11 (c) on your own. One student solves on the board, others in notebooks, then the solution is checked.

in)

1) Solve inequality 2 X 2 + 5X + 10 0. 2X 2 + 5X + 10 = 0; D = –55

By theorem, the inequality is true for all values X.

2) Solve the inequality X 2 ≥ 16; X 2 – 16 ≥ 0; (X – 4)(X + 4) ≥ 0; X = 4;
X = – 4.

Solution X≤ –4 and X ≥ 4.

3) Solving the system of inequalities

Answer: X ≤ – 4; X ≥ 4.

8. Solve number 4.32 (b) on the board and in notebooks.

Solution

The smallest integer is -2; the largest integer is 6.

A n e t: -2; 6.

9. Repetition of previously studied material.

1) Solve No. 4.11 (a; b) on p. 12 orally.

2) Solve No. 4.12 (b) by plotting function graphs (p. 12).

b)

We build graphs of functions
And y = –1 – x.

Answer: -2.

III. Lesson results.

1. In the 9th grade algebra course, we will consider only systems of two inequalities.

2. If in a system of several inequalities with one variable one inequality has no solutions, then the system has no solutions.

3. If in a system of two inequalities with one variable one inequality holds for any values ​​of the variable, then the solution of the system is the solution of the second inequality of the system.

Homework: solve Nos.: 4.9 (a; b), No. 4.10 (a; b), No. 4.11 (a; b), No. 4.32 (a).

Methodical development

algebra lesson in grade 9 (2).

Teacher R.I. Maslyuk

Topic: Solving fractional rational inequalities by the interval method

Goals:

Strengthen the skills of solving square inequalities

To form the ability to solve fractional-rational inequalities using the interval method.

Form the concept of a set of solutions; to develop in students a culture of designing a geometric interpretation to solve inequalities.

Update knowledge about methods for solving quadratic inequalities based on visual-geometric interpretations;

Develop the ability to independently apply knowledge in a complex in new conditions.

Tasks:

Educational: in-depth study of the topic based on existing knowledge, consolidation of practical skills and skills in solving problems of increased complexity as a result of independent work of students and lecture and advisory activities of the most prepared of them.

Educational: development cognitive interest, independence of thinking, memory, initiative of students through the use of communicative-activity methods and elements problem learning.

Educational: formation of communication skills, culture of communication, cooperation.

Conducting methods:

Lecture with elements of conversation and problem-based learning;

Lecture and advisory activities of a group of students with a high level of skill in solving problems of increased complexity;

Independent work students;

Development of a culture of registration of the solution of quadratic inequalities.

Key competencies:

Information and educational: the ability to work with abstracts, the ability to listen to the solution presented by a classmate, to choose the main thing in the solution, to draw conclusions and generalize.

Communicative: the ability to conduct a dialogue, to prove their point of view.

Subject: the ability to explore a quadratic function on a segment, using the constancy of the function on a certain interval; use the graph-analytical method in solving equations and inequalities.

By the time of the lesson students should be able to:

Using the number line to find the intersection and union of number sets

Using the discriminant formula and Vieta's theorem, find the roots of a square trinomial

Transform square trinomial into the product of linear factors

During the classes

Organizational moment.

Knowledge check:

1) Checking homework No. 333; 334; (verification of answers with a discussion of the points that caused difficulties in doing homework)

2)Updating of basic knowledge .

oral work

(slides) with discussion and geometric interpretation on the board:

Yes

Not

Yes

Not

Factorize

Solve the inequality

Find a solution to the inequality

Answers: 1) (х+3) 2 ;2) (-∞;-3) U (-3;+ ∞); 3)(-∞;-1) U (1;+ ∞);4)(0;2);5)(-4;-2)

3. Motivation for applying the solution algorithm

fractional rational inequalities.

Solution of fractional rational inequalities

Answers

but)

(-∞ ;-3)U(5;+∞)

b )

(-∞ ;-4)U(-1; 1)U

c) x

(-2;1]

2) a) x

(-∞;-2)U U (2;+ ∞)

b) x

(-∞;-1]U (0;+1]U (2;+ ∞)

in)

[-4 ;-2)U (1 ;3 ]

3) a)

[-3;-1) U U U(-2 ;1)U U (2 ;+ ∞)

in)

(-∞;-8)U(-1 ;8)U (8 ;+ ∞)

G )

(-∞;-2]U(-1 ;2]U (3 ;+ ∞)

Group work is carried out by levels. Each group presents their solution at the board. The remaining groups act as opponents. Estimates for the work are set collectively by voting.

Generalization of the topic

Solving inequalities and systems of inequalities by the interval method.

Who were you interested in working with?

What would you like to praise yourself for in class?

What did you like the most about the lesson?

Who would you like to thank for the lesson?

Homework ChapterIII , point 6

Level I - No. 334 (a, c), 339 (a)

II level - №№335,339 (b)

III level - №№ 336, 339,379

This lesson is taught in the ninth grade and is the first lesson in which the solution of inequalities other than linear is proposed. The volume is designed for one lyceum lesson (80 minutes). It is given before the lesson, where they show how to expand the module. In the textbooks for the 8th grade (Alimov) and the 9th grade (Makarychev), this material is presented insufficiently, and the analysis of errors indicates a poor understanding of the use of this method by students in the future.

Practice shows that experienced teachers try to expand the concept of the interval method in grades 10-11, but this takes extra time. The presented approach allows students of the 9th grade to form the ability to solve complex inequalities and, on this basis, use the possibilities of the method without additional explanations. In grades 10-11, it remains to show the method of intervals for solving inequalities containing exponential, logarithmic function etc.

Lesson outline

"Solution of Rational Inequalities".

Methods: explanatory-illustrative, reproductive, research.

Type of lesson: formation and consolidation of knowledge.

Form: lecture-conversation.

1. Educational: define rational inequalities and teach how to solve inequalities using the interval method; work out the concepts of “special” cases and take them into account when solving inequalities.
2. Developing: prepare students for lecture forms of classes, accustoming them to perceive information in large blocks; develop logical thinking, independence, self-control; formation of mental operations (analysis, synthesis, selection of the main); vision of connection with subsequent material.

Educational tasks: development of rational communication; development of personal qualities (care, support, independence, helping others, empathy).

During the classes

I. Organizational moment.

II. Updating students' knowledge.

Oral counting is carried out in order to prepare students for the purpose of perceiving new material.

Examples are considered that allow us to draw conclusions about expressions that do not affect the inequality sign, but significantly affect the solution of the inequality.

Students conclude:

an expression that is to an even power does not affect the inequality sign, but it does affect the solution and cannot be discarded without additional restrictions.

2) Consider the solution of the inequality.

Emphasis is placed on the fact that the expression (x +3) also does not affect the sign of inequality, but it cannot be ignored, otherwise the solution will be incorrect.

These two cases (expressions to an even degree; expressions that have been abbreviated) will be classified as special occasions and this will be taken into account when describing the algorithm.

3) Students are given two expressions:

And av Consider the sign of expressions in the following cases:

a B C D)

Conclusion: what students do: sign quotient coincides with the sign of the product.

This allows in the future not to move from the particular to the product. Usually at this transition also there is a loss of a denominator in general.

4) We proceed to work with the graph of functions.

BUT)
	Y=f(x) When does a function change sign?

Output:when the function passes through zero. This is confirmed by figure B)

Output: this function belongs to the category of special cases, because even degree function does not affect the inequality sign, there is no sign change.

Conclusion: This means that those dots, that vanish denominator(break points) should also be taken into account as points, when passing through which the function changes its sign.

III. Formation of new knowledge

After the oral work done, an algorithm of the interval method is written, which allows even students with insufficient mathematical training to solve rather complex inequalities. In parallel with writing the algorithm, an example is analyzed, and when explaining it is not necessary to go from simple to complex, but on the contrary, you can safely move from complex to solving the simplest inequalities, making the remark that we have analyzed the algorithm that works in all cases, sometimes (depending on the example) . Some items will not work.

There are many different methods for solving rational inequalities, but the most common, most convenient, method that simplifies the solution of inequalities is the method of intervals.

We first make a few remarks that we will use in practice and introduce the definition of rational inequalities.

Definition: Rational are inequalities that contain only entire rational and fractional rational functions.

Rational inequalities can be solved by the interval method, based on a simple observation: the sign of the product (quotient) depends only on the signs of each of the factors (divisible and divisors).

The idea is as follows: the number line is divided by the zero of the function into a finite number of intervals, in each of which the function retains its sign. To determine this sign, you need to calculate the value of the function at any one point from each such interval.

It can be simplified if we stipulate the concept of special cases that affect the sign of the interval.

We will refer to them:

1. The linear multiplier is in an even power.
2. An expression that can be shortened.

In addition, you need to bring all the factors to the form (x-µ), because when the function has the form F(x)=(x-µ)(x-µ)….(x-µ) you can alternate the signs of the intervals, without defining the sign of each interval, because this is sometimes inconvenient (fractional values ​​that are close to each other).

Consider the algorithm using an example that provides for the remarks that we have stipulated.

Giving a general algorithm, it should be noted that not all points in some examples work, so it can be significantly reduced.

1. Arrange the expression in the numerator and in the denominator into linear factors.

> 0

2. Consider special cases (multipliers with an even exponent and those multipliers that will be reduced).

3. We rewrite the inequality, excluding those factors that fell into a number of special cases:

4. Equate to zero each factor of the numerator and denominator and find everything X from these equalities.

5. On the coordinate line, we mark those values X, which were obtained in step 4, taking into account the sign ( < ; >).

6. Check the sign of the function in one of the intervals. In the remaining intervals, the signs will strictly alternate

I

7. Taking into account special cases, write down the answer

After studying the algorithm, consider examples:

x 2 - 4 x + 6 > 0 at x

Homework:

Textbook examples

but. (x - 2) 3 (x+1)(x - 1) 2 (x 2 + 2x + 5)< 0

b.

Tasks for independent solution:

In preparing the lesson, materials from the IPKRO retraining courses were used.

In this lesson, we will recall all the material covered on the topic and will solve examples with different types of inequalities. Let us first repeat the method of intervals and the operations of intersection and union of sets. Next, we will solve examples using standard solution techniques.

Topic: Rational inequalities and their systems

Lesson: Overview lesson on the topic: "Rational inequalities and their systems"

We gradually increased the complexity of systems of inequalities: first we solved linear systems, then added quadratic inequalities, rational inequalities, themselves constituted systems, and thus we developed a methodology for solving systems of inequalities.

It includes important elements:

1.Spacing method as a method for solving individual inequalities.

2. The operation of intersection and union of numerical sets.

Let's take a look at these elements. Recall the interval method in an example:

Consider the function

Find the roots of a square trinomial

Find the roots using Vieta's theorem

Let's single out intervals of sign constancy.

When passing through m.-1, the function does not change sign, because parenthesis to an even degree.

We made a mistake by not providing an isolated solution.

Answer:

Let's draw a sketch of the graph of the function.

The method of intervals is the most important element in solving rational inequalities and systems.

The meaning of the operations of intersection and union of sets, including numerical ones, helps to understand the following picture:

Intersection of many.

We have a set A of some elements and a set B. Some of these elements simultaneously fall into both set A and set B, and it is called the intersection of A and B (Fig. 3).

For example:

2.

Their intersection gives the following set:

Union of sets.

There are elements that are included only in set A, there are elements that are included only in set B. There are those that are included both there and there - these elements form the intersection of sets.

And all the elements from A and the missing elements from B form a union of sets (Fig. 5).

For example:

(Rice. 6).

The solution to the inequality is the union of two sets:

One more example.

Find the intersection and union of sets.

Intersection of many:

Union of sets:

The solution is any number

5.

Solve a system of simple inequalities.

Answer:

We repeated the method of intervals, the operations of union and intersection of sets. Now consider the inverse problem, which will allow us to better understand the meaning of solving inequalities.

Given a solution to an inequality, you need to come up with at least one inequality for which it is true.

6. Find an inequality whose solution is the given union of sets.

This could be the solution square inequality. schedule of the respective quadratic function is a parabola passing through points 2 and 4.

Consider tasks with a module.

Consider the first inequality. What's happened ? This is the distance from the point with coordinates x to point 3. A means that the distance between these points is no more than 2. Let's graph it:

Let's solve the second inequality.

Consider the function

The graph is a parabola, the branches are directed upwards.

Let's get back to the system.

Answer:

related tasks.

Find the smallest solution. Answer: There is no smallest solution to this system.

To find greatest solution. Answer:

We have reviewed the solution of systems of rational inequalities. We have considered the main elements that ensure the success of the technique of solving inequalities. What does it take to solve the inequality? interval method. What is needed to obtain a solution of typical systems? You need to imagine the operations of intersection and union.

We will need inequalities in what follows.

1. Mordkovich A.G. and others. Algebra 9th grade: Proc. For general education Institutions. - 4th ed. - M.: Mnemosyne, 2002.-192 p.: ill.

2. Mordkovich A.G. and others. Algebra Grade 9: Task book for students educational institutions/ A. G. Mordkovich, T. N. Mishustina and others - 4th ed. — M.: Mnemosyne, 2002.-143 p.: ill.

3. Yu. N. Makarychev, Algebra. Grade 9: textbook. for general education students. institutions / Yu. N. Makarychev, N. G. Mindyuk, K. I. Neshkov, I. E. Feoktistov. - 7th ed., Rev. and additional - M .: Mnemosyne, 2008.

4. Alimov Sh.A., Kolyagin Yu.M., Sidorov Yu.V. Algebra. Grade 9 16th ed. - M., 2011. - 287 p.

5. Mordkovich A. G. Algebra. Grade 9 At 2 pm Part 1. A textbook for students of educational institutions / A. G. Mordkovich, P. V. Semenov. - 12th ed., erased. — M.: 2010. — 224 p.: ill.

6. Algebra. Grade 9 At 2 hours. Part 2. Task book for students of educational institutions / A. G. Mordkovich, L. A. Aleksandrova, T. N. Mishustina and others; Ed. A. G. Mordkovich. - 12th ed., Rev. — M.: 2010.-223 p.: ill.

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1. Mordkovich A.G. et al. Algebra Grade 9: Taskbook for students of educational institutions / A. G. Mordkovich, T. N. Mishustina et al. - 4th ed. - M .: Mnemosyne, 2002.-143 p.: ill. No. 82 - 84; Home test № 1.