Tuần 15 giao an tieng anh hpt10 2016

Aim of The Lesson: a.. Knowledge: Students use knowledge: the system of equation and solution system of equation b.. Skills: Know how to solution, and applications solution system of e

Tuần 15 Ngày 11 / 12 / 2015 Lớp 10T Tiết 29-30

LESSION PLAN THE SYSTEM OF EQUATION

1 Aim of The Lesson:

a Knowledge:

Students use knowledge: the system of equation and solution system of equation

b Skills:

Know how to solution, and applications solution system of equation

c Attitude:

 Careful, accuracy in calculations and reasoning

 Positive sense of learning, creative thinking

 Understand and apply solution system of equation

2 Subject Matter

- Reference: Algebra - Analysis for High-school Textbook

- Materials: Sheets of paper, Puzzles

3 Procedure

Exercise 1: Solve the system of equation

2 0(1)

xy x

  

�

�      

�

Solution

We have

(2)�(2x 2 ) (xy  x y y ) ( x y) 0

2 (2x y 1)(x y) 0

�

2

2 1

y x

�

� �

�

By substituting y x 2 into (1), we obtain:

x   x � x � y

By substituting y2x1 into (1), we obtain:

2

1 5 2

1 0

2

x

x

�  

�

�

   �

�  

�

� Thus, the solution set is  1;1 ; 1 5; 5

2

Exercise 2: Solve the system of equation

2 2 (1)2

�

�

�

Solution Conditions: 1

0

x y

�

�

� �

� From (1)� x2  (y 1)x2y2 y 0(*)

2 (3y 1)

   Therefore ,(*) has two roots

2 1;

Since x�1 and y�0 then x y 0 (no solution)

We substitute x2y1 into (2), we obtain: (y1)( 2y 2) 0

1( does not satisfy condition) 2

y y

 

�

� � � Thus, the solution set is  5;2

Exercise 3: Solve the system of equation

2

(5 4)(4 )(1)

�

�

�

Solution From (2), we have

(4 8) 5 16 16 0(*)

2 4 (4 8)2 4.( 5 2 16 16)

16x 64x 64 20x 64x 64 36x

Therefore, (*) has two roots

By substituting y=5x+4 into (1), we obtain

2

2

(5 4) (5 4)(4 ) (5 4) (5 4)(4 ) 0 (5 4)(5 4 4 ) 0 (5 4).6 0

5

4 0

x

�

�

�

�



�

�



By substituting y = 4-x into (1), we obtain

2 (4 ) (5 4)(4 ) (4 )(4 5 4) 0

�

�

�

� �� �  Thus, the solution set is 4;0 ; 0; 4 ; 4;0   

5

Exercise 4: Solve the system of equation

�

�

�

Solution

We have from (2) : 2 2 4

4y 4( 16 y ) 4y 64 4y 64 0

Therefore, (*) has two roots

2

2

2

2

4 2

4 2

y

y





By substituting x y24 into (1) , we obtain

�

9y 48y 60 0

By substituting x y24 into (1), we obtain

 2 2 4 2 2 

2

2

2 10 3

y y

�

�

�

� 

� Hence, the solution set is

Exercise 5: Solve the system of equation

2

�  

�

�

�

Solution Conditions:

1 1 2

x y

�

�

�

�

�

�

We have from (1):

2

(5 ) 4(1).(4 ) 25 16 9

�

�

Therefore, (*) has two roots

(no solution) 4

 



……