Tuần 15 giao an tieng anh hpt10 2016

Tuần 15 Tiết 29-30 Ngày 11 / 12 / 2015 Lớp 10T LESSION PLAN THE SYSTEM OF EQUATION 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 Subject Matter - Reference: Algebra - Analysis for High-school Textbook - Materials: Sheets of paper, Puzzles Procedure CONTENT Teacher and Students’ activities Exercise 1: Solve the system of equation �xy  x   0(1) � x  x y  x  y  xy  y  0(2) � Solution We have (2) � (2 x  xy)  ( x y  y )  ( x  y )  � (2 x  y  1)( x  y )  y  2x 1 � �� y  x2 � By substituting y  x into (1), we obtain: x3  x   � x  � y  By substituting y  x  into (1), we obtain: � 1  x � 2 � x  x 1  � � 1  x � � �1 � � ;� � � � � Thus, the solution set is  1;1 ; � � Gv: Cao Bảo Đằng Exercise 2: Solve the system of equation �xy  x  y  x  y (1) � � �x y  y x   2( x  y )(2) Solution �x �1 �y �0 Conditions: � From (1) � x  ( y  1) x  y  y  0(*)   (3 y  1)2 Therefore ,(*) has two roots x  y  1; x   y Since x �1 and y �0 then x  y  (no solution) We substitute x  y  into (2), we obtain: ( y  1)( y  2)  y  1( does not satisfy condition) � �� y2 � Thus, the solution set is  5;2  Exercise 3: Solve the system of equation �y  (5 x  4)(4  x )(1) �2 �y  x  xy  16 x  y  16  0(2) Solution From (2), we have y  (4 x  8) y  x  16 x  16  0(*)   b  4ac  � (4 x  8) � � � 4.(5 x  16 x  16)  16 x  64 x  64  20 x  64 x  64  36 x Therefore, (*) has two roots y 4x   6x 4x   6x  5x  , y   4 x 2 By substituting y=5x+4 into (1), we obtain � (5 x  4)  (5 x  4)(4  x) � (5 x  4)  (5 x  4)(4  x)  � (5 x  4)(5 x    x)  � (5 x  4).6 x  � 5 5x   x �y0 � �� �� � x0 � x 0� y 4 � By substituting y = 4-x into (1), we obtain � (4  x)  (5 x  4)(4  x) � (4  x)(4  x  x  4)  x 4� y 0 � �� x 0� y 4 � �4 � ;  0;  ;  4;0  Thus, the solution set is � ;0 � �5 � Gv: Cao Bảo Đằng Exercise 4: Solve the system of equation x  y   xy (1) � �2 �x  2( xy  8)   y (2) Solution We have from (2) : x  xy  16  y  0(*)   y  4(16  y )  y  64  y  64  Therefore, (*) has two roots y2   y2  2 y2  x  y2  x By substituting x  y  into (1) , we obtain 4( y  4)  y   4( y  4) y � y  32 y  64  y   y  16 y � y  48 y  60  ( no solution) By substituting x  y  into (1), we obtain  y2  4  y4   y2  y2  4 � y  32 y  64  y   y  16 y � y  48 y  60  � y2  � � 10 � y2  � Hence, the solution set is � 10 ��2 10 � 2; ; 2;  ; �  ; ; ;  �� � � 3 ��3 � � �� �    Exercise 5: Solve the system of equation � �x  y  xy � � x 1  y 1  Solution �x �1 � Conditions: � y� � � We have from (1):  x  2y   xy  � x  xy  y  xy � x  xy  y  0(*)   (5 y )  4(1).(4 y )  25 y  16 y  y Therefore, (*) has two roots Gv: Cao Bảo Đằng x   y (no solution) x  4y …… Gv: Cao Bảo Đằng ...  1) x  y  y  0(*)   (3 y  1)2 Therefore ,(*) has two roots x  y  1; x   y Since x �1 and y �0 then x  y  (no solution) We substitute x  y  into (2), we obtain: ( y  1)( y  2)