Chương 4 đai so 10 song ngữ

Preparation date:
Teaching date:

Chapter IV:
INEQUALITIES AND INEQUATIONS
PERIOD: 28 §1 INEQUALITIES. (p1)

I. AIMS OF THE LESSON
-To realize concept and properties of inequalities
-To realize Cauchy’s inequalities and inequalities with absolute value bars
-Proof some simple inequalities
II. TEACHING AIDS
-Teacher:lesson plan,text book
-Student: To realize concept and methods in order to proof of inequalities in class 8
and 9

III. TEACHING PROCEDURE
1.Orginization:
2. Checking the previous lesson: NO
3.New lesson:
Activity 1: Review of inequalties
TEACHER’S ACTIVITIES
?1.Which of the following propositions are
true?: a/3,25<4,

1
b/-5>- 4 , c/- 2 ≤ 3
4

?2. Choose a suitale symbol(=; <; >) such
that we get a true propositions when we fill

in the blank box:
a/2 2

3

4
3
(1 + 2) 2

b/

2
3

c/3 + 2 2
d/ a2 +1 0
?3.State the concept of inequalities
?4. Proof :a < b ⇔ a - b < 0
.

STUDENTS’ACTIVITIES

I/ Review of inequalties
?1. proposition a.
a/2 2 < 3

b/

4 2
>

3 3

c/3+2 2 = (1 + 2) 2
d/ a2+1 > 0
1.Concept of inequalities
:textbook p74
proposition “a > b” or “a < b” are called
inequalties.
2.Resulting inequalities and equivalent
inequalities:textbook p74
a < b ⇒ a – b < 0 và a – b < 0 => a < b
3.Properties of inequalities: p75
Ex:x > y ⇔ x + 2 > y + 2
x > 2 ⇒ x2 > 4
Note:page76

Activity 2 :Cauchy’s inequality
TEACHER’S ACTIVITIES

STUDENTS’ACTIVITIES
II. Cauchy’s inequality

1


a+b
≥ ab , ∀a, b ≥ 0 .
2
Sign “=” occurs ⇔ a = b

1.Theorem:

Proof: textboob page 76
Express theorem cauchy
?Proof

2.Consequence:
Consequence 1:textbook page76

?When does Sign “=” occurs
? a > 0 ,proof:

1
a

Consequence 2: textbook page77

a+ ≥2

?Among all rectangles with the same
perimeter,which has the largest area?
Consequence 3: textbook page77
a + b ≥ 2 ab ,a,b are lengths 2 sides
when is ab max
If both x and y are positive and their sum is
invariablr then the sum x+y is the smallest if
and only if x=y

Activity 3: INEQUALITIES WITH ABSOLUTE VALUE BARS
TEACHER’S ACTIVITIES

STUDENTS’ACTIVITIES
?Compute absolute value of the following
numbers
a/ 0

b/1,25

c/ −

3
4

d/ −π

?State the definition

III. INEQUALITIES WITH ABSOLUTE
VALUE BARS
State the definition :
|A| = A if A ≥ 0
|A| = - A if A < 0
Properties: textbook page78
Ex: textbook page78

IV. Consolidate:
State the definition inequalities and Cauchy’s inequality
V.Homeworks:
Exercises in textbook page 79

..............................................................

Preparation date:
Teaching date:
PERIOD: 29

§1 INEQUALITIES. (p2)

2


I. AIMS OF THE LESSON
- Consolidate concept and properties of inequalities ,Cauchy’s inequalities and
inequalities with absolute value bars
- Consolidate methods in order to proof of inequalities
-Applying methods in order to proof of inequalities in doing exercises

II. TEACHING AIDS
-Teacher:lesson plan,text book
-Student: text book,notebook

III. TEACHING PROCEDURE
1.Orginization:
2. Checking the previous lesson:
-State concept and properties of inequalities
- Cauchy’s inequalities.Prove
a+b+c+d 4
≥ abcd ∀a, b, c, d ≥ 0
4

3.New lesson
TEACHER’S ACTIVITIES

STUDENTS’ACTIVITIES

Call student answer Exercise 1

Exercise 1: d.
Exercise 2: c. explain:since x > 5

Exercise 2. To guide:
Given x > 5, compare

0<

5
x
and
x
5

;

x
>1
5

Exercise 3 text book page79
a. ( b-c)2 < a2 ⇔ (b – c - a)(b – c + a) < 0
we have:
(a - b)2 < c2 ;(c - a)2 < b2

Exercise 3 page79
When are a,b,c lengths of 3 sides of
1 triangle?
Are the following inequalities true?
(a - b)2 < c2 ;(c - a)2 < b2
b. Addition three inequalities sides
by sides

b. Addition three inequalities sides by sides

Exercise 4 We use equivalent
transformation

Exercise 4
We have:x3 + y3 - (x2y + xy2) =
= (x +y)(x2 + y2 - xy) – xy(x + y)
= (x + y)(x2 - 2xy + y2) = (x + y)(x - y)2 ≥ 0 true
Since x ≥ 0, y ≥ 0 , x + y ≥ 0, (x – y)2 ≥ 0
Equation occurs if and only if x = y.

Exercise 5
To guide students find solution
Set x = t
Consider two cases :
2
* 0 ≤ x <1
* x≥1
Exercise 6. Call H tangenital point
of AB and cirle
By applying Cauchy’s inequalities

5
5
5
< 1 ;1 < + 1; -1 < 0
x
x
x

y

Exercise 5
Set x = t we have t8 – t5 + t2 – t + 1 = f(t)
0 ≤ t < 1, f(t) = t8 + t2(1- t3) + (1 – t) > 0
t ≥ 1, f(t) = t5(t3 – 1) + t(t – 1) + 1 > 0.

B
H

O

Exercise
6
H
Call H tangenital point
of AB
A xand cirle
3

we can show that
AB=HA+HB ≥ 2 HA.HB .
AB min when equation occurs

⇒ OH ⊥ AB

AB = HA + HB ≥ 2 HA.HB .
= OH 2 = OH
⇒ ∆OAB right and isosceles at O ⇒ AB = 2OH = 2
AB
2
=
= 2.
2
2
Therefore,when A( 2;0), B(0; 2) ,AB min
⇒ OA = OB =

IV. CONSOLIDATE:
- Recall methods to solve of exercises

V.HOMEWORKS:
To read lesson:inequations and systems of inequations with one unknown
..............................................................
Preparation date:
Teaching date:

Period 30: Inequations and systems of inequations
with one unknown(p1)
I. AIMS OF THE LESSON

-Introduce concept of inequations and systems of inequations with one
unknown;conditions for an inequation;inequations containing parameters
-To understand some inequation transformations
-Find conditions for an inequation,solve the simple inequations and systems of inequations
with one unknown

II. TEACHING AIDS
-Teacher:lesson plan,text book
-Student: text book,notebook
III. TEACHING PROCEDURE
1.Orginization:
2. Checking the previous lesson:
State concept of inequations and systems of inequations with one
unknown;conditions for an inequation;inequations containing parameters
3.NEW LESSON
Activity 1: Inequations with one unknown and Conditions for an inequation
TEACHER’S ACTIVITIES

Given inequation:2x ≤ 3

Show left side and hand side of
inequation?

STUDENTS’ACTIVITIES
I. Concept of inequations with one unknown
1. Inequations with one unknown
Definition:text book page 80
H2: 2x is the left side, 3 is the right side
a.-2 is the solution to the above inequation
The other numbers are not the solution to the above

4


inequation

0

x
b. 2x ≤ 3 ⇔ x ≤

Call people to read definition
:
3 − x + x +1 ≤ x

2

3
2

2. Conditions for an inequation
Definition:text book page 81
Ex;Find conditions for an inequation
3 − x + x + 1 ≤ x2
3 − x ≥ 0
⇔ −1 ≤ x ≤ 3
are 
x +1 ≥ 0

Activity 2 Inequations containing parameters
TEACHER’S ACTIVITIES

STUDENTS’ACTIVITIES
State equations containing
parameters?To imply inequations
containing parameters
Call people to read definition
People do example

3. Inequations containing parameters
In an inequation,besides letters acting as
unknowns,there may be other letters that are
regarded as constants and called parameters
Ex: (2m – 1)x + 3 < 0
Solving and justify an inequality containing
parameters mean considering for which values of
parameters the inequality has no solutions or some
solutions and finding the solutions

Activity 3 : A systems of inequations with one unknown
TEACHER’S ACTIVITIES
Call people to read definition
People do example

STUDENTS’ACTIVITIES
II. A systems of inequations
with one unknown
Definition:text book page 81
3 − x ≥ 0
x + 2 ≥ 0

Ex:Solve a systems of inequations: 

3 − x ≥ 0
⇔ −2 ≤ x ≤ 3

x + 2 ≥ 0

Hence,the solution to the system is: [- 2; 3]

IV. CONSOLIDATE:
- Find conditions for inequations:

5


x−2
≤ 5x + 2
3x + 1
- Find values of x satisfying conditions for each of the following inequations
a.

a)

2 − 5x + 3x − 4 > 0

1
≥2
x +1

b.

b)

2
x
>
2
( x − 1)( x + 3)
1− x

V.HOMEWORKS:
Remember concept of inequations and systems of inequations with one unknown
Do the exercise 1,2 page 88
..............................................................
Preparation date:
Teaching date:

Period 31:Review end of term I
I. AIMS OF THE LESSON
-Review propositions,sets,linear and quadratic funtions,equations and system of
equations,inequations;inequalities
II. TEACHING AIDS
Teacher:syllabus,textbook,exercise
Students:Review exercises from chapter I to from chapter IV
III. TEACHING PROCEDURE
1.Orginization:
2. Checking the previous lesson:
State concept of inequations and systems of inequations with one
unknown;conditions for an inequation;inequations containing parameters
3.NEW LESSON
Activity 1:Solve equation contain root

Exercise 4: Solve equation
Let students recognize the form of the a) 2 x − 9 = 1
equation and state the method to slove
9
Condition : x ≥
the equation.
2
Ask student to solve equation.
2x − 9 = 1 ⇒ 2x − 9 = 1
⇒ 2 x = 10 ⇒ x = 5 ( satisfy )
Ask 2 students to present on the board
Thus,equation has 1 solution x = 5
b) 2 x + 9 = 1
Folow and help student with difficulty.
9
Condition x ≥ −
2
Ask student to comment.
2x + 9 = 1 ⇒ 2x + 9 = 1
⇒ 2 x = −8 ⇒ x = −4 ( no satisfy)
Assess and grade.
Thus,equation has no solution

Activity 2:Solve biquadratic equation:
Let students recognize the form of the
equation and state the method to slove the

Exercise 5: Solve equation

6

equation.
Ask student to solve equation.
Ask 3 students to present on the board
(satisfy)
Folow and help student(satisfy)
with difficulty.
Remind student to compare conditions to
find the roots.
Ask student to comment.
Assess and grade.

(satisfy)
(no satisfy)

(satisfy)
(no satisfy)

a) x4 – 5x2 + 6 = 0
Given x2 = t ( t ≥ 0)
We have
t2 – 5t + 6 = 0 (a = 1; b = - 5 ; c = 6 )
∆ = (−5) 2 − 4.1.6 = 1 > 0
t = 2
⇒ ∆ = 1 =1⇒ 
t = 3
With t = 2, we have: x2 = 2 ⇒ x = ± 2
With t = 3, ta có: x2 = 3 ⇒ x = ± 3
Thus S = { − 3; − 2; 2; 3 }

b) –x4 – 5x2 + 6 = 0
Given x2 = t ( t ≥ 0)
We have:
–t2 – 5t + 6 = 0 ( a = –1; b = –5; c = 6)
We have: a + b + c = –1–5 + 6 = 0
t = 1
⇒
 t = −6
With t = 1, we have: x2 = 1 ⇒ x = ±1
Thus S = {–1 ; 1}
c) –x4 + 8x2 + 9 = 0
Given x2 = t ( t ≥ 0)
We have:
–t2 + 8t + 9 = 0 ( a = –1; b = 8; c = 9)
We have: a – b + c = –1– 8 + 9 = 0
t = 9
⇒
 t = −1
With t = 9 : x2 = 9 ⇒ x = ±3
Thus S = {–3 ; 3}

Activity 3: Inequalities.
Let student read the question carefully.
Guide student to prove using (A – B )2 ≥ 0
Ask student to present the proof.

Folow and help student with difficulty.
.
Ask student to comment.
Assess and grade.

Exercise 6:Prove
a 2 + b 2 + c 2 ≥ ab + bc + ca
∀a, b, c ≥ 0
Solution:
We have : a 2 + b2 ≥ 2ab
b 2 + c 2 ≥ 2bc
c 2 + a 2 ≥ 2ca
Imply :
a 2 + b2 + b2 + c2 + c2 + a2 ≥
2ab + 2bc + 2ca
2
⇒ 2a + 2b 2 + 2c 2 ≥ 2ab + 2bc + 2ca
⇒ 2 ( a 2 + b 2 + c 2 ) ≥ 2 ( ab + bc + ca )
⇒ a 2 + b 2 + c 2 ≥ ab + bc + ca

7


IV- Consolidation:
Let student review important knowledge..
V- Home work:
Review the forms of above problem.
Prepare for final exam.
..................................................................................................................................................
Preparation date:
Teaching date:

Period 32: Semester 1 Exam.
Question and solution are given by educational council of Vinh Phuc province.

...................................................................................................................................................
.
Preparation date:
Teaching date:

Period 33: Correct semester 1 exam’s questions.
Question and solution are given by educational council of Vinh Phuc province.
...................................................................................................................................................
.
Preparation date:
Teaching date:

Period 34: Inequations and systems of inequations
with one unknown(p2)
I. AIMS OF THE LESSON:
-Introduce concept of inequations and systems of inequations with one
unknown;conditions for an inequation;inequations containing parameters
-To understand some inequation transformations
-Find conditions for an inequation,solve the simple inequations and systems of inequations
with one unknown

II. TEACHING AIDS
-Teacher:lesson plan,text book
-Student: text book,notebook
III. TEACHING PROCEDURE
1.Orginization:

8

2. Checking the previous lesson:
3.New lesson:state concept of inequations, conditions for an inequation
3.NEW LESSON
Activity 1: Equivalent inequations ;Addition/subtracttion
TEACHER’S ACTIVITIES
STUDENTS’ACTIVITIES
Call a student to read definition
equivalent inequations
Compare it with equivalent
equations?
Call people to read definition
equivalent transformations?
Compare it with equivalent
transformations
Call people to read definition
addition

I.
Some inequation transformations
1.Equivalent inequations
Definition in text book 82
2. Equivalent transformations
Definition in text book 82
3. Addition/subtracttion
Text book page 83
P(x) < Q(x) ⇔ P(x) + f(x) < Q(x) + f(x)
Ex 2 solve inequation
(x + 2)(2x - 1) - 2 ≤ x2 + (x - 1)(x + 3)
⇔ 2x2 +3x – 4 ≤ x2 + x2 + 2x – 3 ⇔ x ≤ 1
Thus the solution set to the inequation ( −∞;1]

Activity 2: Multiplication/ division
TEACHER’S ACTIVITIES

STUDENTS’ACTIVITIES
4.Multiplication/ division

Call people to read definition
multiplication

Text book page 84
P(x) < Q(x) ⇔ P(x).f(x) < Q(x).f(x) if f(x) > 0
P(x) < Q(x) ⇔ P(x).f(x) > Q(x).f(x) if f(x) < 0
Ex solve inequation:

When we square 2 sides of
inequation,What do we must note?

x2 + x + 1 x2 + x
> 2
x +1
x2 + 2

Activity 3: Squaring
TEACHER’S ACTIVITIES
When we square 2 sides of
inequation,What do we must note?

STUDENTS’ACTIVITIES
5. Squaring

P(x) < Q(x) ⇔ P2(x) < Q2(x) if P(x) ≥ 0, Q(x) ≥ 0, ∀x
9


Ex solve inequation:
State method to solve inequation
: f (x) > g (x)

f (x) >

Do the example in notebook
Call other student comment and
repair

x2 + x + 2 >

x 2 − 2x + 3

 f ( x) > g ( x )
 f ( x) > g ( x)

⇒
g (x) ⇔  f ( x) ≥ 0
 g ( x) ≥ 0
 g ( x) ≥ 0


Ex5 solve inequation

5x + 2 3 − x x 4 − 3 3 − x

> −
4
4
6

Note text book page 85
1
≥1.
x −1
17
1
> x+
Ex7 solve inequation : x 2 +
4
2

Ex6 solve inequation:

IV. CONSOLIDATE:
- are the following pairs of inquations equivalent?Explain
a) 2x - 3 > 0 và - 2x+3 < 0
b) x2 + 1 < 2x2 -3 và - x2+4 < 0
c)

1
≥ 1 và 1 ≥ x + 1
x +1

V.HOMEWORKS:

-Remember some inequation transformations
- Do the exercises 1, 2, 3, 4, 5 page 88.

..............................................................
Preparation date:
Teaching date:

Period 35: Practise
I. AIMS OF THE LESSON:
Condisolate concept of inequations and systems of inequations with one
unknown;conditions for an inequation;inequations containing parameters
-To understand some inequation transformations
-Find conditions for an inequation,solve the simple inequations and systems of
inequations with one unknown
- Củng cố khái nêm bất phương trình một ẩn, các phép biến đổi tương đương, phép
biến đổi hệ quả bất phương trình..

II. TEACHING AIDS
-Teacher:lesson plan,text book
-Student: text book,notebook
III. TEACHING PROCEDURE
1.Orginization:
2. Checking the previous lesson:
State some inequation transformations
10


.

3.NEW LESSON

Activity 1: Exercise 1 and Exercise 2
TEACHER’S ACTIVITIES
1.
* Condition of fraction funtion ?

* Condition of funtions contain
root ?
2. a.Compare left side and 0
b. Compute min of left side,compare
with hand side
c. Compare left side with 1

STUDENTS’ACTIVITIES

Exercise1
x ≠ 0
x ≠ 0
⇔
x +1 ≠ 0
x ≠ 1

a. Condition: 

 x ≠ ±2
 x 2 − 4 ≠ 0

⇔ x ≠ 1
b. Condition:  2
 x − 4 x + 3 ≠ 0
x ≠ 3


c. Condition : x ≠ −1
1 − x ≥ 0
x ≤ 1
⇔
d. Condition : 
x + 4 ≠ 0
 x ≠ −4

Exercise 2:
a. Condition x + 8 ≥ 0 ⇒ VT ≥ 0 , VP ≤ 0
thus, inequation no solution
2
2
b. inequation ⇔ 1 + 2( x − 3) + 1 + ( x − 2) <

3
2

We have VT ≥ 2 VP < 2. thus, inequation no
solution
c. VT <1, VP = 1. thus, inequation no solution

Activity 2: Exercise 3 and Exercise 4
TEACHER’S ACTIVITIES
3. Show method to change from
inequation (1) to inequation(2)

4.
a. Combining fraction using

common denominator

Simplifying algebraic fractions

STUDENTS’ACTIVITIES

Exercise 3:
a. - 4x + 1 > 0 ⇔ 4x – 1 < 0.
b. 2x2 + 5 ≤ 2x – 1 ⇔ 2x2 – 2x + 6 ≤ 0
c. Since we addition 2 sides with a expression
d. Condition of inequation (1) x ≥ 1 .
2 inequation are equivalent since we multiply 2
sides with a number >0
Exercise 4:

3x + 1 x − 2 1 − 2 x
−
<
⇔ 18 x + 6 − 4 x + 8 < 3 − 6 x
2
3
4
11
⇔ 20 x < −11 ⇔ x < − .
20
11
Thus,solution of inequation x < −
20
b. (2x – 1)(x + 3) – 3x + 1 ≤ (x – 1)(x + 3) + x2 – 5
⇔ 2x2 + 2 x − 2 ≤ 2 x2 + 2x − 8 ⇔ - 2 ≤ - 8

a.

Thus, inequation no solution

11


b.
Combining fraction using common
denominator

Simplifying algebraic fractions

Activity 3: Exercise 5.
TEACHER’S ACTIVITIES
5.
a. Combining fraction using
common denominator

Simplifying algebraic fractions

b.
Combining fraction using common
denominator

STUDENTS’ACTIVITIES
Exercise 5.
5


6 x + 7 < 4 x + 7
42 x + 5 < 28 x + 49
⇔
a. 
8 x + 3 < 6 x + 10
8x + 3 < 2 x + 5
 2
22

x<

14 x < 44

7
⇔
⇔
Thus,solution of system of
2
x
<
7
7

x <

2
22
inequation:x <
7
1


15 x − 2 > 2 x + 3
45 x − 6 > 6 x + 1
39 x > 7
⇔
⇔
b. 
4 x − 16 < 3 x − 14
x < 2
 2( x − 4) < 3x − 14

2
7

x >
⇔
39
 x < 2

Thus,solution of system of inequation
Simplifying algebraic fractions

7
39

IV. CONSOLIDATE:
-State method to find conditions for an inequation,solve the simple inequations and
systems of inequations with one unknown

V.HOMEWORKS:
Do the exercise in workbook page 109,110

..............................................................
Preparation date:
Teaching date:

12


Period 36:SIGNS OF LINEAR BINOMIALS
I. AIMS OF THE LESSON:
- Concept of linear binomials,theorem about signs of linear binomials
-Considering signs of products and quotients of linear binomials
-Using the definition to remove absolute value bars of linear binomials

-Applications to solving product inequations, inequations containing unknowns
in denominator and inequations containing unknowns inside absolute value bars
.

II. TEACHING AIDS
-Teacher:lesson plan,text book
-Student: text book,notebook

III. TEACHING PROCEDURE
1.Orginization:
2. Checking the previous lesson:
Solve the inequations
a) 5x – 2 > 0

b) - 4x + 3 > 0

3.NEW LESSON
Activity 1: Theorem on signs of linear binomials
TEACHER’S ACTIVITIES
Call a student to read definition
Compare signs of linear binomials
with sign of coefficient a

Call a student to read theorem
b
a.f(x) = a(ax +b) = a2(x + )
a

Drawing graphical illustration

STUDENTS’ACTIVITIES
I.Theorem on signs of linear binomials
1. Linear binomials: f(x) = ax + b
a, b are two given numbers, a ≠ 0
2. Signs of linear binomials
Theorem:text book page 89



b

Proof: f(x) = a  x + ÷ ⇔ a.f(x) = a2(x +
a
x

f(x)



−∞

−

b
)
a

b
a

Opposite signs to a
a.f(x) < 0

+∞

0 the same sign as a
a.f

x) > 0
We say number x = −

A student do H2.Comment and
repeat about theorem

b

such that f(x) = 0 is the
a

solution to binomial
Graphical illustration in page 90.
3. Applications:
H2:f(x) = 3x +2,
2
3

f(x) > 0 with ∀x > − , f(x)<0 with ∀x < −
f(x) = - 2x + 5,
5
2

f(x) > 0 with ∀x < , f(x)<0 with ∀x >

13

5
2

2
3


Ex1 :text book page 90

Activity2: Considering signs of products and quotients of linear binomials
TEACHER’S ACTIVITIES

STUDENTS’ACTIVITIES
II. Considering signs of products and quotients of
linear binomials
Example 2: Consider the signs of binomial
Teacher guides method co
consider f(x)
+

O

-2

f(x) =

( 4 x − 1) ( x + 2 )
−3 x + 5

1/4 +

O

-

5/3

x

-

- Method to solve inequations:

O
1/4
+
+ inequations
+To take
about form
f(x) >H0 ( <, -2≤, ≥ )- O
5/3
+ Considering sign f(x)
+Implying the solution.

1 5


1 5


F(x) < with ∀x ∈  −2; ÷∪  ; +∞ ÷
4 3



Thusx f(x) > 0 with ∀x ∈ ( −∞; −2 ) ∪  ; ÷
4 3

Activity 3: Applications to solving inequations
TEACHER’S ACTIVITIES
STUDENTS’ACTIVITIES
III. Applications to solving inequations
1. Product inequations and inequations containing

unknowns in denominator.
Example3: text book page 92
2. Inequations containing unknowns inside absolute
value bars
Example4: text book page 93.
Exercise1

.

a. f(x) > 0 ⇔ x < - 3 or x >

b. f(x) = (- 3x – 3)(x + 2)(x + 3)
Sign of f(x)

Call students do exercise 1
+

+

Other student comments and
repairs
-3

1
1
, f(x) < 0 ⇔ -3 < x <
2
2

-

-1

0
-

-2

x

⇒ f(x) >0 ⇔ x ∈ ( −∞; −3) ∪ ( −2; −1)

F(x) < 0 ⇔ x ∈ ( −3; −2 ) ∪ ( −1; +∞ )
−8 + 4 x − 9 x − 3

−5 x − 11

c. f(x) = 3 x + 1 2 − x = 3 x + 1 2 − x
(
)(
) (
)(
)
Sign of f(x)
0

-11/5

+

-1/3

2

+

14

x


 11 1 
⇒ f(x) > 0 ⇔ x ∈  − ; − ÷∪ ( 2; +∞ )
 5 3
11   1 

f(x) < 0 ⇔ x ∈  −∞; − ÷∪  − ; 2 ÷
5  3 


d. f(x) = 4x2 – 1 = (2x – 1)(2x + 1)
1  1


⇒ f ( x) > 0 ⇔ x ∈  −∞; − ÷∪  − ; +∞ ÷
2  2


 1 1
f ( x) > 0 ⇔ x ∈  − ; ÷

 2 2

Exercise 2.

Call students do exercise 2

2

Other student comments and
+ 0
rapairs
1/2

+

-

1

+

1/2

0

-1/3

5

−x + 3

a. Bpt ⇔ x − 1 − 2 x − 1 ≤ 0 ⇔ x − 1 2 x − 1 ≤ 0 :
(
)(
)
3

Considering
x
sign of left side
Therefore,the
solution to the inequation is:
1
1 
+
x
 ;1÷∪ [ 3; +∞ )
2 
1
1
x ≠ ±1
2
b. x + 1 <
( 1 − x ) condition:

2
x ( x − 3)
x − 1) − x − 1
(
⇔

<
0
inequation ⇔
2
2
( x + 1) ( x − 1)
( x + 1) ( x − 1)

<0

Considering
sign of left side:
Therefore,the solution to the inequation is:
∀x ∈ ( −∞; −1) ∪ ( 0;1) ∪ ( 1;3)

.

IV. CONSOLIDATE:
V.HOMEWORKS:
- Remember theorem about signs of linear binomials
-Considering signs of products and quotients of linear binomials
-Using the definition to remove absolute value bars of linear binomials
-Applications to solving product inequations, inequations containing unknowns in
denominator and inequations containing unknowns inside absolute value bars
Học thuộc định lý về dấu của nhị thức bậc nhất, cách xét dấu của biểu thức, cách giải bất
phương trình.
-Do the exercise in text book.

..............................................................
Preparation date:

Teaching date:
Period 37:LINEAR INEQUATION WITH TWO UNKNOWNS(P1)

I. AIMS OF THE LESSON:

15


-To understand concept linear inequations with two unknowns,system of linear
inequations with two unknowns
-To understand concept solution of linear inequations with two unknowns,system of
linear inequations with two unknowns
-Representing solution sets of linear inequations with two unknowns,system of linear
inequations with two unknowns
II. TEACHING AIDS
-Teacher:lesson plan,text book
-Student: text book,notebook
Students are learned funtion y=ax+b and linear inequations with a unknown

III. TEACHING PROCEDURE
1.ORGINIZATION
2. CHECKING THE PREVIOUS LESSON
1. State theorem on signs of linear binomials
2. Considering signs of products and quotients of linear binomials
3Draw graph funtion 2x + y = 3 hay (y = 3 – 2x)

3.NEW LESSON
Activity1: Linear inequations with two unknowns; Linear inequations with two
unknowns
TEACHER’S ACTIVITIES

STUDENTS’ACTIVITIES
I. Linear inequations with two unknowns
1. Definition in text book 95
II. Linear inequations with two unknowns
1. Definition solution sets of linear inequations with
two unknowns
2. Rules for geometric representation of solution
sets ax + by < c ( ≤, >, ≥ )
+ Draw line ax + by = c ( ∆ )
+ Take a point M0(x0; y0) ∉ (∆)
+ Compute ax0 + by0 and compare with c.
+ Conclusion:
Teacher call a student draw line
Ex1Produce a geometric representation solution sets
( ∆ ): 2x + 3y = 3 ?
of linear inequations with two unknowns 2x + 3y ≤
3.
- Draw line ( ∆ ): 2x + 3y = 3
- O(0;0) , O ∉ ( ∆ ) và 2* 0 + 0 ≤ 3 so the half plane
Call a student repeat rules for
on one side of the boundary line ∆ containing the O
geometric representation of solution is the solution domain of the given inequation:
sets ax + by < c
2x + y ≤ 3 ?
Teacher guide student steps
Ex2Produce a geometric representation solution sets
+ Draw line (d): -3x + 2y = 0
of linear inequations with two unknowns - 3x + 2y >
+ Take a point M (x0; y0); M ∉2 (d) 0
sao cho

3/2
-3x0+ 2y0 > 0
+ The half plane on one side of the
1
x
boundary line containing the O is O

-2

16


the solution domain of the given
inequation: .

Activity 2: System of linear inequations with two unknowns
TEACHER’S ACTIVITIES
STUDENTS’ACTIVITIES
III. System of linear inequations with two unknowns
1. Definition in text book 96
2. Produce a geometric representation solution sets
of system of linear inequations with two unknowns:
3 x + y ≤ 6
x + y ≤ 4


x ≥ 0
 y ≥ 0

Call students representing solution

sets of linear inequations with two
unknowns

IV. CONSOLIDATE:
- Produce a geometric representation solution sets of system of linear inequations
with two unknowns

V.HOMEWORKS:
- Memorize concept linear inequations with two unknowns,system of linear
inequations with two unknowns
.
- Produce a geometric representation solution sets of system of linear inequations
with two unknowns
- Produce a geometric representation solution sets of system of linear inequations
2 x − y ≤ 3
2 x + 5 y ≤ 12 x + 8

with two unknowns: 

..............................................................
Preparation date:
Teaching date:

Period 38:LINEAR INEQUATION WITH TWO UNKNOWNS(p2)
I. AIMS OF THE LESSON:
-To understand concept linear inequations with two unknowns,system of linear
inequations with two unknowns
-To understand concept solution of linear inequations with two unknowns,system of
linear inequations with two unknowns
-Representing solution sets of linear inequations with two unknowns,system of linear

inequations with two unknowns
II. TEACHING AIDS
-Teacher:lesson plan,text book
17


-Student: text book,notebook
Students are learned funtion y=ax+b and linear inequations with a unknown
III. TEACHING PROCEDURE
1.ORGINIZATION
2. CHECKING THE PREVIOUS
1.State concept linear inequations with two unknowns,system of linear inequations
with two unknowns
2. Representing solution sets of linear inequations with two unknowns:
2 x − y ≤ 3

2 x + 5 y ≤ 12 x + 8

3. State steps solve exercises by set up system of linear equations

3.NEW LESSON
TEACHER’S ACTIVITIES

STUDENTS’ACTIVITIES

IV. Applications to economic problems
Let x, y be the number of tons of type I and II
products producted a day ( x ≥ 0, y ≥ 0 )
Conditions of x,y?
L = 2x + 1,6y

The profit per day
3x + y
The working period of machine x + y
M1?
3 x + y ≤ 6
The working period of machine M1?  x + y ≤ 4
Teacher guide student do problem 1

We have system of linear inequations with two
3 x + y ≤ 6

Problem become solve system of

linear
inequations
with
two unknowns :  x + y ≤ 4
unknowns .Find solution such that
x ≥ 0
 y ≥ 0
2x + 1, 6y max
+ Representing solution sets of
system linear inequations with two
unknowns
+Find (x0, y0) such that L = 2x0 +
1,6y0 max.

IV. CONSOLIDATE:
- State method to solve system of linear inequations with two unknowns

V.HOMEWORKS:
- Do the exercise1, 2, 3 page 99

..............................................................
Preparation date:
Teaching date:

Period 39:

Practise

I. AIMS OF THE LESSON:

18


- Consolidate concept linear inequations with two unknowns,system of linear
inequations with two unknowns
-To understand concept solution of linear inequations with two unknowns,system of
linear inequations with two unknowns
-Representing solution sets of linear inequations with two unknowns,system of linear
inequations with two unknowns

II. TEACHING AIDS
- Teacher:
textbook,workbook
- Student learned graph funtion y = ax +b; linear inequations with two
unknowns,system of linear inequations with two unknowns

III. TEACHING PROCEDURE

1.ORGINIZATION
2. CHECKING THE PREVIOUS LESSON
-State method to representing solution sets of linear inequations with two
unknowns,system of linear inequations with two unknowns

3.NEW LESSON
Activity 1: Exercise1
TEACHER’S ACTIVITIES

STUDENTS’ACTIVITIES
Exercise1 :textbook page 99
a.
Inequation ⇔ x + 2 y − 4 < 0
+ Draw line x + 2y - 4 = 0. (d)
+ O(0; 0) ∉ (d)
+ We have 0 + 2.0 - 4 < 0
Satisfy

Call 2 students go blackboard do
exercise 1a, b.
- Suggest draw line d
- Suggest to students represent
solution sets of linear inequations y

2
4
O

x

so the half plane on one side of the boundary line
d containing the O is the solution domain of the
given inequation
y
b. Inequation ⇔ −2 x + 4 y − 8 < 0
+ Draw line - 2x + 4y - 8 = 0. (d)
+ Given
O(0; 0) ∉ (d)
2
+ We have - 2.0 + 4.0 - 8 < 0
Satisfy
O
x
So the half plane on one side of the boundary line
d containing the O is the solution domain of the
given inequation

-4

y
3
y-x=3
Activity 2: Exercise2
2
TEACHER’S ACTIVITIES

-5

x-2y=o

O

-3

-2

STUDENTS’ACTIVITIES

Exercise2 :textbook page 99
D

x+3y=-2

E

19


a.
x − 2 y < 0

 x + 3 y > −2
y − x < 3


Call 2 students go blackboard do
exercise 2a, b.
- Suggest draw line d
- Suggest to students represent
solution sets of linear inequations

Area of solution is the area without line

IV. CONSOLIDATE:
- State method to representing solution sets of linear inequations with two
unknowns,system of linear inequations with two unknowns

V.HOMEWORKS:
- Memorize method to representing solution sets of linear inequations with two
unknowns,system of linear inequations with two unknowns

..............................................................
Preparation date:
Teaching date:
Period 40 SIGNS OF QUADRATIC TRINOMIALS(P1)

I. AIMS OF THE LESSON:
-To understand theorem on signs of quadratic trinomials
-By applying theorem on signs of quadratic trinomials so solve quaratic inequations
with one unknown
II. TEACHING AIDS
- Teacher:textbook,lesson plan
- Students:textbook,notebook

III. TEACHING PROCEDURE
1.ORGINIZATION
2. CHECKING THE PREVIOUS LESSON
1. State method to solve equation: ax2 + bx + c = 0
20

2
2. Given quadratic trinomials f ( x ) = x − 5 x + 4 . Compute f ( 1) ; f ( 2 ) ; f ( 3)
- Comment sign?
- Find solution of quadratic trinomials ?

3.NEW LESSON
Activity 1: Quadratic trinomials
TEACHER’S ACTIVITIES

STUDENTS’ACTIVITIES

* Examine the graphs in fig 32 page
101
Answer the following questions
1 Find the intersection of (P) and the
axis Ox. It follows the solution
equation f (x) = 0, imply sign ∆ .
2.Show the intervals on which the
graph above the horizontal axis.
3. Point out that the range on the
graph at the bottom of a horizontal
axis.)
* Based on the results on the fill in the
following table:
TH1 : ∆ > 0 ( H 32 a )
−∞
x1
x2
+∞

x
f(x)
? 0 ? 0 ?

I. Theorem on signs of quadratic trinomials

Comment sign of f(x) and sign of a?
TH2 : ∆ = 0 ( H 32 b )

x

x

−∞

b
−
2a

1. Quadratic trinomials

Definition: f ( x) = ax + bx + c ( a ≠ 0 )
H1: ∆ > 0 ( H 32 a )
−∞
x1
x2
+∞
x
f(x)
+ 0 - 0 +

2

TH2 : ∆ = 0 ( H 32 b )
−∞

f(x)

−

b
2a

+∞

+ 0 +

+∞

f(x)
? 0 ?
Comment sign of f(x) and sign of a?
TH3: ∆ < 0 ( H 32c )
−∞
+∞
x
f(x)
?
* Comment sign of f(x) and sign of a?

TH3: ∆ < 0 ( H 32c )

−∞
+∞
x
f(x)
+

.

Activity2: Signs of quadratic trinomials
TEACHER’S ACTIVITIES
* Looking at Figure 33 part a <0,
proceed similarly.
+ Call Hs read theorem, summarized

STUDENTS’ACTIVITIES
2. Signs of quadratic trinomials
Theorem
If ∆ < 0 then a.f(x) > 0 with ∀x ∈ R

21


To observe the 33 Hs Teacher
explains the case for students.

If ∆ = 0 then af(x) > 0 with ∀x ≠ −

b
2a

If ∆ > 0
Then af(x) > 0 with ∀x ∈ ( −∞; x1 ) ∪ ( x2 ; +∞ )

af(x) < 0 with ∀x ∈ ( x1 ; x2 )
Note : We replace ∆ ' with ∆ .
Geometric representation:
3. Applications:
Example1:
Consider the sign of trinomial
Example2:
Consider the sign of trinomial

.

IV. CONSOLIDATE:
- State theorem on signs of quadratic trinomials

V.HOMEWORKS:
- Memorize theorem on signs of quadratic trinomials
- Do the exercise 1, 2 page 105.

..............................................................
Preparation date:
Teaching date:
Period 41 SIGNS OF QUADRATIC TRINOMIALS(P2)
I. AIMS OF THE LESSON:
-To understand theorem on signs of quadratic trinomials
-By applying theorem on signs of quadratic trinomials so solve quaratic inequations
with one unknown

II.TEACHING AIDS
- Teacher:textbook,lesson plan
- Students:textbook,notebook

III. TEACHING PROCEDURE
1.ORGINIZATION
2. CHECKING THE PREVIOUS LESSON
15 minutes test:
1. Consider the sign of the following expressions:
a. f(x) = x2 - 6x - 7

(2 đ)

11x − 5 x + 6
−x
(8 đ)
x2 + 5x + 6
Answer : a. f(x) = (x +1)(x - 7) ⇒ f ( x) > 0 with ∀x ∈ ( −∞; −1) ∪ ( 7; +∞ )
2

b. f(x) =

f(x) < 0 with ∀x ∈ ( −1;7 )

b. f ( x) =

(1)
(1)

11x 2 − 5 x + 6 − x 3 − 5 x 2 − 6 x

− x 3 + 6 x 2 − 11x + 6 − ( x − 1) ( x − 2 ) ( x − 3)
=
=
(3)
( x + 2 ) ( x + 3)
x2 + 5x + 6
x2 + 5x + 6

Consider the sign
+
-3

-

+
-2

+
-1

0

22

1

-

2

3

-

x


(3)
Thus: f(x) > 0 with ∀x ∈ ( −∞; −3) ∪ ( −2;0 ) ∪ ( 2;3)

(1)

f(x) < 0 với ∀x ∈ ( −3; −2 ) ∪ ( 1; 2 ) ∪ ( 3; +∞ )

3.NEW LESSON
TEACHER’S ACTIVITIES

(1)

STUDENTS’ACTIVITIES
II. Quadrartic inequations with one unknown
1. Quadrartic inequations
Quadrartic inequationsis an inequation of
the form ax2 + bx + c < 0 ( ≤, >, ≥ ) a ≠ 0
2. Solving quadrartic inequations

Call student up the blackboard
solve the inequation in ex3.
student comment homework.
Recalling the exact solutions.

When has quadratic equation has
two opposite solution ?

+ Consider the sign of trinomial f(x) = ax2 + bx
+ Based on the sign of trinomial ,solution of
inequations is interval such that sign of f(x) same
with sign inequation
Example 3: Solve the following inequation
a. f(x) = 3x2 + 2x + 5 > 0
b. f(x) = - 2x2 + 3x + 5 > 0
c. f(x) = - 3x2 + 7x - 4 < 0
2
d. f ( x ) = 9 x − 24 x + 16 ≥ 0
Example 4:
Equation has two oppositely solutions
⇔ af (0) < 0

⇔ 2m 2 − 3m − 5 < 0 ⇔ −1 < m <

5
2

Thus , Equation has two oppositely solutionsKết
luận if and only if −1 < m <

5
2

IV. CONSOLIDATE:

- State theorem on signs of quadratic trinomials
- State method to solve the following inequation.

V.HOMEWORKS:
. Memorize theorem on signs of quadratic trinomials
- Memorize method to solve quadratic inequations
- Do the exercises page 105 and end of chapter IV exercises

..............................................................

Preparation date:
Teaching date:
Period 42

PRACTISE

I. AIMS OF THE LESSON:

23


- Consolidate quadratic inequations,applying rule consider signs of trinomials
to solve inequations
II. TEACHING AIDS
- Teacher:textbook,lesson plan
- Students:textbook,notebook

III. TEACHING PROCEDURE
1.ORGINIZATION
2. CHECKING THE PREVIOUS LESSON

- State theorem on signs of quadratic trinomials
3.NEW LESSON
Activity 1: Exercise 1
TEACHER’S ACTIVITIES

STUDENTS’ACTIVITIES

Exercise 1
a.assess the signs of the following
quadratic trinomials : f(x) = -x2 + 4x
+ 5.
.

Exercise 1:
a.Solution
we have ; quadratic trinomial has form
a – b + c = -1 – 4 + 5 = 0
quadratic trinomial has 2 solution: x1 = - 1 ; x2 = 5.
⇒ f(x) < 0 with x < - 1 or x > 5
f(x) > 0 with x ∈ ( −1;5 )
b. quadratic trinomial has
∆’ = 62 –(- 4).12 = 36 – 36 = 0
quadratic trinomial has a solution : x1 = x2 =

b. assess the signs of the following
quadratic trinomials:
f(x) = - 4x2 + 12x – 9

−

b 12 3
=
= .
2a 8 2

since a = - 4 < 0 ⇒ f(x) < 0 ; ∀x ≠ −

b 3
=
2a 2

Activity 2: Exercise 2
TEACHER’S ACTIVITIES

STUDENTS’ACTIVITIES
Exercise 2: a.
∆ = (-5)2 – 4.2.2 = 25 – 1 6 = 9 .
quadratic trinomial has 2 solution:

Exercise 2; find x
a) f(x) = 2x2 – 5x +2 <0

x1 =

5+3 8
5−3 1
= = 2; x2 =
= .
2.2 4
4

2

x

-∞

VT

½
+

0

2
-

0

+∞
+

1
Thus f(x) < 0 with x ∈ ( ;2)
2

b. :
∆’ = 202 – 16.25 = 400 – 400 = 0 .
24

−

b
40
40
5
=−
=−
=− .
2a
2.16
32
4

since a = 16 > 0

2

b) f(x) = 16x + 40x + 25 > 0

hence left side possitive with x ≠ −

5
4

 5
 4

therefore x ∈ R\ − 
c.

∆’ = (-2)2 – 3.4 = 4 – 12 = -8 < 0
Since a = 3 > 0 , left side possitive
to R.
hence x ∈ R

with x belong

c) f(x) = 3x2 – 4x + 4 ≥ 0

IV. CONSOLIDATE:
- State theorem on signs of quadratic trinomials
- State method to solve the quadratic inequation

V.HOMEWORKS:
- do the end of chapter IV exercises

..............................................................
Period 43

End of chapter iv exercises (P1)

I. AIMS OF THE LESSON:
- To understand and apply properties of inequalities , Cauchy’s inequalities and
inequalities with absolute value bars
- To undestand condition of inequation ; theorem about signs of linear binomials
theorem on signs of quadratic trinomials
- To undestand method to solve inequations and system of inequations

II. TEACHING AID
- Teacher:textbook,lesson plan

- Students:textbook,notebook
III. TEACHING PROCEDURE
1.ORGINIZATION
2. CHECKING THE PREVIOUS LESSON
1. State concept and properties of inequalities , Cauchy’s inequalities
2. State theorem about signs of linear binomials theorem on signs of quadratic
trinomials
3. State method to solve the quadratic inequation

3.NEW LESSON
TEACHER’S ACTIVITIES
1.Call student answer question from
1 to 5 page 106
Teacher guide to students

STUDENTS’ACTIVITIES

Exercise 1:a. x > 0; b. y ≥ 0; c. |a| ≥ 0 with ∀x ∈ R;

25