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Inequalities and groups of inequalities (inequalities and their solution sets) in the second book of mathematics in senior one.
Chapter VIII One-dimensional One-time Inequality

-Thematic review

Summary of this chapter

1. In this chapter, we know this inequality and study its properties. We learned how to solve a linear inequality (group) by using the properties of inequality, and expressed the solution set of linear inequality on the number axis, which can be used to get the solution set of linear inequality group intuitively.

2. The understanding of inequality comes from the reality of life. We should learn to analyze the unequal relationship between quantity and quantity in practical problems, abstract inequalities (groups), and use the obtained inequalities (groups) to solve practical problems.

3. The process of solving one-dimensional linear inequality is similar to solving one-dimensional linear equation. It includes: (1) denominator; (2) remove the brackets; (3) moving items; (4) merging similar projects; (5) the step of converting the coefficient into 1. When solving inequalities, we should flexibly arrange and reasonably choose the solving steps according to the requirements of practical problems. It should be noted that when the coefficient is 1, if both sides of the inequality are multiplied or divided by the same positive number, the unsigned direction will not be changed; But when both sides of inequality are multiplied or divided by the same negative number, the direction of inequality must be changed.

4. When solving a set of one-dimensional linear inequalities, first find the solution set of each inequality, and then find their common parts. The latter usually uses four basic situations, such as number axis or memory, and adopts the method of "taking the big from the big, taking the small from the small, taking the middle from the small, and there is no solution between the big and the small".

5. Representing the solution set of one-dimensional linear inequality on the number axis can not only deepen our understanding of the solution set of one-dimensional linear inequality (group), but also facilitate us to get the positive equivalent solution set of one-dimensional linear inequality and the special solution set of one-dimensional linear equations more intuitively.

Comprehensive interpretation of theme

Topic 1: Transforming Inequalities by Using their Properties

Example 1 multiple choice questions

(1) If -a < 2, then the correct one in the following categories is ().

a、a 2C 、- a+ 1 < 3D 、-a- 1> 1

(2) if a > b, the following inequality must hold ()

a、b、c 、- a >-bD、a-b>0

(3) (Sui Zhou, 2003) If a < 0, the solution set of the inequality AX+ 1 > 0 about x is ().

a、x > B、x < C、x > D、x 2xB、3 x2 > 2x2c、3x > 2xB、3+x2>2

Solution: (1) c (2) d (3) d (4) d.

Comments: (1) The key to solve this problem is to understand and master the basic properties of inequality. After solving it by using three basic properties of inequality, it is screened.

(2) For some multiple-choice questions, if they are difficult or too complicated to be solved directly, special values can be used to help screen and reduce the time for answering questions. For example, (4) X =- 1, 0 can choose to eliminate A, C and B respectively, so choose D.

Example 2 judges whether the deformation of the following inequality is correct.

(1) AC < BC from A < B (2) from X > Y and m≠0.

(3) xz2 > yz2 comes from x > y (4) x > y comes from xz2 > yz2.

Solution: (1) is incorrect. C may be zero or negative, and the dimensional relationship cannot be determined after deformation.

(2) incorrect. -m is not necessarily a negative number, and the direction of inequality cannot be determined after deformation.

(3) Incorrect. Z can be 0.

(4) correct. According to the conditions, z2 > 0.

Comments: Accurately understanding the essence of inequality is the key to solving the problem. Pay attention to consider the problem comprehensively. Pay special attention to the application of attribute 3.

Special double solution inequality or inequality group

Example 1 inequality

Solution: change the decimal into a fraction and get,

Divide by the denominator to get 4 (2x-1)-6 (3x-5)-2 (x+1)+3x5 > 0,

Remove the brackets to get 8x-4-18x+30-2x-2+15 > 0.

Combining with similar projects, we get-12x+39 > 0,

Move the item and get-12x+39 > 0.

The coefficient is 1, x.

Comments: Inequalities with both denominators and decimals can be converted into fractions or fractions, but the latter may have infinite decimals, which will make the operation answer incorrect. Often, all decimals are converted into fractions before solving.

Example 2 Solving Inequality System

Solution: Solve the inequality (1) to get X.

The solution set of the inequality group is-4 ≤ X.

Comments: Pay attention to the correct use of denominator brackets when solving inequality (2), such as 0.2 (x-3)-0.5 (x+4) ≤-1.4; In this problem, the decimal coefficient can also be changed into an integer coefficient, such as ≤- 14.

Special solution of inequality (group) in special topic three seeking

Example 1 Find the positive integer solution of inequality.

Solution: Remove the denominator and get 2 (y+1)-3 (y-1) ≥ y-1(be careful not to forget to add brackets).

Remove the brackets 2y+2-3y+3 ≥ y- 1 (note the sign change).

Move and merge -2y ≥-6

The coefficient is 1 and y≤3 (note that the direction of inequality should be changed in this step).

Because there are three positive integers not greater than 3: 1, 2, 3,

So the positive integer solution of inequality is 1, 2, 3.

Comments: To determine the special solution of an inequality, we must first determine the range of inequality solution set, and then find out the qualified number within this range.

Example 2 Find the non-negative integer solution of inequality group.

Solution: x >-2 is obtained from the inequality 2x+ 1 < 3x+3; X≤5 is obtained from inequality, so the solution set of the original inequality group is -2 < x ≤ 5, and its non-negative integer solution is 6 numbers: 0, 1, 2, 3, 4, 5.

Comments: The solution set of solved inequalities (groups) can be expressed on the number axis, which can completely solve the phenomenon of missing solutions. As shown in this example, if the solution set of the obtained inequality group is represented as a graph on the number axis, obviously its non-negative integer solution is 0, 1, 2, 3, 4, 5.

-3 -2 - 1 0 1 2 3 4 5 6

The fourth question uses the concept of inequality solution set to solve related problems.

Example 1 It is known that the solution of inequality group is the same as that of, so the value of a is found.

Solution: solving the inequality group can be simplified to -2 < x < 1, and the solutions of the two inequality groups are the same, so -2 < x < a-4. So a-4 = 1, so a = 5.

Example 2 (Chongqing, 2003) knows that there is no solution to the inequality group about X, so the range of a is.

Solution: The original inequality group can be simplified as because the inequality group has no solution, x≤3 and X > A have no common parts, that is, a≥3.

Example 3 If inequality (ax-5) > x-a's solutions to X are all solutions of inequality 1-2x < 3, find the range of A..

Solution: from the inequality (ax-5) > x-a, we get (a-2) x > 5-2a;

From the inequality 1-2x < 3, we get x >-1; 2 < a ≤ 3 is derived from the meaning of the question.

Topic 5 inequality (group) combines calculation, estimation and equation to solve practical problems.

The comprehensive application of equations and inequalities is a common problem in the senior high school entrance examination in recent years. The key to solve this kind of problem is to find out the relationship between the quantities in the problem, list the equations and inequalities, and then solve them.

Example 1 (Heilongjiang, 2003) A middle school won 4 first prizes, 6 second prizes and 20 third prizes in the knowledge contest on SARS prevention and control. The school decided to give prizes to all the winners, and the prizes of the same first prize were the same.

(1) If the prizes of the first prize, the second prize and the third prize are watering can, mask and thermometer respectively, then it will cost 1 13 yuan to buy these three prizes, in which the total amount of money spent on watering can is more than the total amount of money spent on masks, and the unit price of masks is more than that of thermometers by 2 yuan, so the unit price of watering can, mask and thermometer.

(2) If the unit price of the third prize is an integer, and the unit price of the first prize is required to be twice that of the second prize, and the unit price of the second prize is twice that of the third prize, what are the unit prices of the first prize, the second prize and the third prize respectively when the total cost is not less than that of 90 yuan but less than 150 yuan? Calculate the unit price of the first, second and third prizes in each case.

Analysis: This question is based on the knowledge contest of SARS prevention in a middle school, and an application problem with perfect combination of equation and inequality is compiled.

Solution: (1) The unit price of watering can and mask is Y yuan and Z yuan respectively.

Then solve.

∴z-2=2.5。

A: The unit prices of watering can, mask and thermometer are 9 yuan, 4.5 yuan and 2.5 yuan respectively.

(2) If the unit price of the third prize is X yuan, the second prize is 2x yuan and the first prize is 4x yuan, then 90 ≤ 4× 4x+6× 2x+20x < 150,

∴≤x