Two. Teaching objectives:
1. Knowing the definition of unary linear inequality will correctly distinguish unary linear inequality.
2. Master the general steps of linear inequality in one variable and express the solution set of inequality on the number axis.
3. By analogy with the definition and general steps of linear equation, master the solution and general steps of linear inequality.
Cultivate students' rational reasoning ability.
4. Inequalities can be abstracted according to the inequality relations in practical problems, and solutions or solution sets that conform to practical significance can be obtained.
Three. Teaching emphases and difficulties:
Teaching emphasis: the general steps to solve linear inequality of one variable.
Teaching difficulty: the solution of one-dimensional linear inequality. And accurately abstract inequalities by using inequality relations in practical problems.
Four. Classroom teaching:
(1) Key points of knowledge:
Knowledge point 1: one-dimensional linear inequality
Like 2x-1>; 5、3x+70 > 100、y+4 & lt； 0, etc. , (1) contains only one unknown, (2) and the highest degree of the unknown is 1, and (3) the coefficient is not equal to 0. This inequality is called unary linear inequality.
Inequalities that satisfy these three conditions are linear inequalities.
For example: 2x+y > 3, 2x2-3x-2 < 0, > x is not a linear inequality.
Knowledge point 2: the solution of one-dimensional linear inequality
The general steps of solving linear inequalities are very similar to those of solving linear equations.
(1) denominator; (2) remove the brackets; (3) moving items;
(4) merging similar projects; (5) The coefficient is 1.
Knowledge point 3: Compare the similarities and differences between the solutions of linear inequality and linear equation.
The steps of solving linear inequalities are similar to solving linear equations.
The difference is that when two sides of inequality are multiplied (or divided) by the same number that is not equal to zero, we must correctly apply the inequality property 2 according to whether the number is positive or negative, and pay special attention to changing the direction of inequality when both sides of inequality are multiplied (or divided) by the same negative number.
Knowledge point 4: Similarities and differences between finding integer solutions of one-dimensional linear inequalities and finding solution sets of one-dimensional linear inequalities.
The steps of (1) solution are similar: denominator, brackets, shifting terms, merging similar terms and converting them into 1.
(2) Finding the integer solution of one-dimensional linear inequality is one step more than finding the solution set of one-dimensional linear equation: finding the integer solution in the set.
Knowledge point 5: Solving practical problems by using one-dimensional linear inequality
In real life, inequality relations are everywhere, and many quantitative relations can be realized by establishing mathematical models, especially inequality, which requires us to abstract inequalities from the analysis of problems.
Knowledge point 6: the comparison between solving practical problems with linear equations and solving practical problems with linear inequalities.
Enumerating one-dimensional linear inequalities to solve practical problems
Enumerating one-dimensional linear equations to solve practical problems
Set an unknown number according to the meaning of the question.
Set an unknown number according to the meaning of the question.
Find out the inequality relation that can summarize the quantitative relation in the problem.
Find out the equality relation that can summarize the quantitative relation in the problem.
Representing inequalities with unknowns
Equal with unknown number
List inequalities and find their solution sets.
List the equations and find their solutions.
According to the requirements of the actual problem, test and write the solution or solution set that meets the meaning of the question and write the answer.
Test and write the answers.
Knowledge Point 7: Mathematical Thought
1. Analogy:
Analogy refers to comparing different objects or things according to their similarities in some aspects (such as characteristics, attributes and relationships). Through analogy, we can find the similarities and differences between old and new knowledge, which is helpful to use existing knowledge to understand new knowledge and deepen our understanding of new knowledge, such as learning the basic properties of inequality. To compare with the basic properties of equality, learn the solution of one-dimensional linear inequality and compare it with the solution of one-dimensional linear equation.
Comparison of basic attributes of (1):
equation
inequality
Add (or subtract) the same _ _ _ _ or the same _ _ _ _ on both sides, and the result is still an equation.
Add (or subtract) the same _ _ _ _ or the same _ _ _ _ _ on both sides, and the direction is _ _ _ _.
Multiply (or divide) the same _ _ _ _ _ (divisor _ _ _ _) on both sides, and the result is still an equation.
Both sides are multiplied by (or divided by) the same _ _ _ _ _ _, and the direction is not equal to _ _ _ _ _.
Both sides are multiplied by (or divided by) the same _ _ _ _ _ _, and the direction is not equal to _ _ _ _ _.
(2) Comparison of solving steps:
Solving a linear equation with one variable:
Solve one-dimensional linear inequality;
solve
law
step
suddenly
(1) Go to _ _ _ _ _ _ _;
(2) Go to _ _ _ _ _ _;
(3)____________;
(4)____________;
(5) The coefficient becomes 1.
(1) Go to _ _ _ _ _ _ _;
(2) Go to _ _ _ _ _ _;
(3)______;
(4)______;
(5)______ 1。
In the above steps (1) and (5), if the multiplier or divisor is _ _ _ _ _ _ _ _, the direction of the inequality sign is changed.
solve
situation
One-dimensional linear equation has only one solution.
The solution set of one-dimensional linear inequality contains an infinite number.
2. The idea of combining numbers and shapes:
Representing numbers on the number axis is the concrete embodiment of the idea of combining numbers with shapes. Representing the solution set on the number axis is a step forward than representing numbers on the number axis. In this chapter, the solution set of inequality is intuitively expressed on the number axis, so that we can intuitively see that there are countless solutions of inequality and easily determine the solution set of inequality group.
3. Overview of matters needing attention:
(1) When learning the properties of inequalities and solving linear inequalities with one variable, we should review the properties of comparative equations and the contents of solving linear equations with one variable, and compare the similarities and differences.
(2) Be careful when multiplying (or dividing) the numbers on both sides of the inequality, especially when the numbers are negative, and don't forget to change the direction of the inequality. If the quantity is not limited, there are three possibilities. Identity inequality 5 >; 3, for example in inequality 3 >; 2 When both sides are multiplied by the same number A, there are three situations:
3a & gt2a(a & gt； 0)3a = 2a(a = 0)3a & lt； 2a(a & lt； 0)
(3) the solution set of inequality x
Typical example
Example 1. Solve the following inequality and express the solution set on the number axis.
( 1)2x- 1 < 4x+ 13； 2(5x+3)≤x-3( 1-2x).
Solution: (1) Move the term to get 2x-4x.
Merge similar items to get -2x.
Divide both sides by -2 to get x & gt-7.
The solution set of this inequality is expressed on the number axis as follows
(2) Move the project to get 10x-7x ≤-9.
Merge similar items to get 3x ≤-9.
Divide both sides by 3 to get x ≤-3.
The solution set of this inequality is expressed on the number axis as follows
Example 2. Solve the unary linear inequality 2x- 1 > 4x+ 13 and express the solution set on the number axis;
2x- 1>4x+ 13，
2x-4x > 13+ 1, (shift term)
-2x > 14, (merging similar projects)
X >-7。 (The coefficient is 1)
Its solution set is shown on the number axis as follows:
Observe whether there are any mistakes in the above answers, and why?
A: There is a mistake. When the coefficient becomes 1, the direction of inequality sign should be changed.
Example 3. Finding positive integer solutions of inequalities
Solution: If you remove the denominator, you get 3 (2x+3) ≥ 8x-2.
Without parentheses, we get 6x+9 ≥ 8x-2.
Move items and merge -2x ≥- 1 1 similar items.
The coefficient is 1, and x≤
So the positive integer solution of the original inequality is 1, 2, 3, 4, 5.
Example 4. Solving inequality: -> 1
Analysis: Using the basic properties of fractions, multiply the numerator and denominator by 100 or 10 respectively, and then remove the denominator.
Solution: sort -> 1
Divide by the denominator to get 4 (8x+200)-3 (5x-20) > 12.
Without brackets, you get 32x+800-15x+60 >12.
Move projects and merge similar projects in 17x >-848.
The coefficient is 1, x >-
Example 5. When the value of x is taken, the difference between the sum of algebraic expressions is not greater than 1?
Solution: According to the meaning of the question, it is -≤ 1.
X≥0 for solving this inequality.
Therefore, when x≥, the difference between algebraic sum is not greater than 1.
Example 6. It is known that the solution of equation 3 (x-2a)+2 = x-a+ 1 is applicable to inequality 2 (x-5) ≥ 8a, and the value range of a is found.
Solution: firstly, solve equation 3 (X-2A)+2 = X-A+ 1 to get X =
Substitute x = into inequality 2 (x-5) ≥ 8a.
2( -5)≥8a
Then solve the inequality to get a≤
So the value range of a is a≤
Example 7. The mass of a carton is 1kg. When some apples are put in (the mass of each apple is 0.3kg), the total mass of the carton and apples does not exceed 10kg. How many apples can this carton hold at most?
Solution: Suppose this carton can hold up to X apples.
According to the meaning, it is 1+0.3x ≤ 10.
X≤30 for solving this inequality.
This carton can hold 30 apples.
Example 8. For flood fighting and emergency rescue, it takes 120km for materials to be transported to dangerous sections, and 1 hour for transportation. After 50km left in the first half hour, what is the speed of the second half hour to ensure timely delivery?
Analysis: the quantitative relationship in the topic is that the sum of the distances traveled in the first half hour and the second half hour should be at least 120km, and the inequality can be established by grasping this quantitative relationship.
Solution: Let the speed of the last half hour be x kilometers per hour.
50+0.5x≥ 120
X≥ 140 for solving this inequality.
Answer: The speed of the second half hour140km ensures timely delivery.
Example 9. At the weekend, a company decided to organize 48 employees to take a boat trip to a nearby water park. First, the company sent a person to find out the rent of the ship. The rent list that this person saw is as follows:
ship form
Number of people loaded each time (person)
Rent (yuan)
boat
five
three
boat
three
2
So, how to design a charter scheme to minimize the rent paid? (Overloading is prohibited)
Solution: If there are X people in the big boat and (48-X) people in the small boat, the rent is W yuan.
So w = =
Because x is an integer of 1≤x≤48, W = 29 yuan when x takes 45.
Therefore, when renting 45÷5=9 large ships and 3÷3= 1 small boats, the rent is the least.
Example 10. A school held a cultural performance to celebrate New Year's Day, and awarded five first prizes, five second prizes 15 and five third prizes 15. The school decided to award prizes to the winning students. The prizes in the same order are the same, and only one of the items listed in the table below can be selected:
Title of article
violin
Athletic wear
flute
dance shoes
mouth organ
photo album
notebook
pen
Unit price/yuan
120
80
24
22
16
six
five
four
(1) If the higher the prize order, the higher the unit price of the prize, then how much should the school spend on the prize at least?
(2) The school requires that the unit price of the first prize is 5 times that of the second prize, and the unit price of the second prize is 4 times that of the third prize. On the premise that the total cost does not exceed 1 1,000 yuan, how many purchase schemes are there? How much is the most expensive plan?
Solution: (1) 6× 5+5×10+4×15 =140 (yuan)
(2) If the third prize is X yuan, the second prize is 4x yuan and the first prize is 20x yuan.
100x+40x+15x ≤1000 according to the meaning of the question.
Solve this inequality x≤6, so X takes 4, 5 and 6, so there are two schemes: the third prize is 4 yuan, the second prize is 16 yuan, and the first prize is 80 yuan;
Third prize 6 yuan, second prize 24 yuan, first prize 120 yuan.
So the most expensive scheme needs 155× 6 = 930 yuan.
Example 1 1. Type A refrigerators sold in shopping malls consume 2 190 yuan each, and the daily power consumption is 1 kwh, while the price of type B energy-saving refrigerators is 10% higher than that of type A refrigerators, but the daily power consumption is 0.55 kwh. In order to reduce inventory, the mall decided to sell Type A refrigerators at a discount (after a 10% discount). Please answer the following questions:
(1) It is known that the purchase price of type A refrigerator is 1700 yuan. In order to ensure that the profit is not less than 3%, the shopping mall tries to determine the price reduction range of type A refrigerator.
(2) If only the price and power consumption are considered, then the shopping mall will give at least a few discounts, and consumers will buy the same two kinds of refrigerators.
Economy (calculated by service life 10 year, 365 days per year, 0.40 yuan per kilowatt hour)?
Solution: (1) Set up a shopping mall and reduce the price of Type A refrigerator by X yuan, which can guarantee a profit of not less than 3%.
The solution is x≤439.
A: The mall will reduce the price of Type A refrigerator by 439 yuan, which can guarantee a profit of not less than 3%.
(2) When a shopping mall is set up and the price of Type A refrigerator is discounted by Y, consumers can buy two kinds of refrigerators as cost-effectively.
=2 190( 1+ 10%)+0.4 10 365 0.55
y=8
A: Shopping malls offer a 20% discount on the price of Type A refrigerators, which is as cost-effective for consumers to buy two kinds of refrigerators.
Simulated test questions (answer time: 30 minutes)
(1) multiple choice questions
The value of 1. X is not greater than 3, and the value range of X expressed by inequality is ().
A.x & gt3 B. x & lt3 C. x≠3 D. x≤3
2. Among the four numbers given below, the inequality is 3-2x >; The solution of 7 is ()
A.-2 b .–2.5 c .+3d .– 1.5
3. The following statement is wrong ()
A.x<2 has countless negative integer solutions. There are countless integer solutions of 2.
The positive integer solutions of C.x<2 are 1 and 2d.x.
4. A two-digit number, in which ten digits are exchanged with single digits, and the difference between the obtained two digits and the original two digits is equal to 27.
Then this two-digit number is ()
A.36 B. 57 C. 64 D. 79
5. It is known that b is a positive number, so the value range of n is ().
A.n x ，a=，a