Chủ đề 1 : Khảo sát hàm số Câu 1: 4 2 2y x x= - + !"#$%&!'()&*+, 4 2 2 0x x m- + = Câu 2: 4 2 2 3y x x+ -= - - .'()&*+'#/01234&5 Câu 3: 4 2 1 4 y x x= - 2 .'()&*+'#/012#&34& 3 4 - Câu 4: 4 2 2 1y x x= - - 2 !"#$%&!'()&*+ 4 2 2 0 x x m- - = Câu 5:,/678 9 78 : .'()&*+'#/; !&2;4&7< Câu 6: 4 2 1 5 3 2 2 y x x= - + .'()&*+'#/='#/&&>;,/6?98:@ Câu 7:/68 9 78 :5=2" .'()&*+'#/>='#/#A&&2>;,/6 1 6 - 8:@ Câu 8: , 2 1 x y x - = - B&*4&>C&*(D&E&;, y x m= - + "#AF01'G! Câu 9: 1 1 x y x + = - H+1(D&E&;,/68:F01'G! Câu 10: 2 1 1 x y x + = - 2 .'()&*+'#/>0IJK Câu 11: 2 2 1 x y x - = + .'()&*+'#/>='#/2!&2L6?K Câu 12: 2 1 2 x y x + = - - .'()&*+'#/01*2#&3 3y = - Câu 13:,/6 1 1 x x - + 2 .'()&*+'#/0&1>*M#& Câu 14: 3 5 2 2 x y x + = + 2 - .'()&*+'#/01234& Câu 15: 2 1 1 x y x + = - 2 .'()&*+'#/>0&1>*MN8 Câu 16:,/6O86 2 3 1 x x + - - .'()&*+'#/='#/#A&&2>;, 1 2011 5 y x= - + Chủ đề 2 : TÌM GIÁ TRỊ LỚN NHẤT, GIÁ TRỊ NHỎ NHẤT CỦA HÀM SỐ Bài 1. H+PHQR=PHRR#, 2 4 3y x x= + + 3 4 4 3y x x= - 4 2 2 2y x x= + - ; 2 2y x x= + - % 2 1 2 2 x y x x - = - + O 2 2 4 5 1 x x y x + + = + & 2 1 ( 0)y x x x = + > 2 1 1 x x y x - + = + 4 2 1 ( 0) x x y x x + + = > Bài 2. H+PHQR=PHRR#, 3 2 2 3 12 1y x x x= + - + *S7JKT 3 3y x x= - *S7J5T 4 2 2 3y x x= - + *S75JT ; 4 2 2 5y x x= - + *S7JT % 3 1 3 x y x - = - *S@JT O 1 1 x y x - = + *S@J9T & 2 4 7 7 2 x x y x + + = + *S@JT 2 1 1 x x y x - + = + *S@JT 2 100y x= - *S7<JUT L 2 4y x x= + + - Bài 3. H+PHQR=PHRR#, 2sin 1 sin 2 x y x - = + 2 sin 2sin 3y x x= + + 2 2sin cos 1y x x= - + ; cos2 2sin 1y x x= - - % 2 4 1 2 x x y + = OO86 2 2x x+ - & x y xe - = * [ ] 0;2 2 8ln y x x= - *SJ%T 2 2 1 2 x x y - - = *S@JT V lny x x= - L 2 2 x y x e= - * [ ] 1;1- " lny x x= - x x e y e e = + * [ln2 ; ln4] /6878* ; 2 2 p p- é ù ê ú ê ú ë û /687*S@JWT Chủ đề 3 : PHƯƠNG TRÌNH, BẤT PHƯƠNG TRÌNH MŨ VÀ LÔGARÍT Bài 1. P'()&*+#, 2 6 8 3 1 x x+ + = J ( ) 2 2 1 3 27 x x - + = J 5 2.9 3 3 x x - = J 9 1 9 2.3 9 0 x x+ - + = K 2 4 5 25 x x- + = J < 2 2 4 3 1 2 2 x x - - æö ÷ ç = ÷ ç ÷ ç è ø J X 7 1 2 1 1 . 2 2 2 x x+ - æö æö ÷ ÷ ç ç = ÷ ÷ ç ç ÷ ÷ ç ç è ø è ø J U 2 3 2 1 2 3 x x x- + - = Y 2 3 5 9 3 x x + = J @ 1 9 2.3 9 0 x x+ - + = J 2 1 1 3.5 2.5 0,2 x x- - - = J 2 5 2 3 3 2 x x+ + = + 5 25 5 24 5 x x - = J9 3 4.3 3 0 x x- - + = J K 1 3 .2 72 x x+ = J < 2 3 1 2 4 x x- - = X 1 2 4 3 7.3 5 3 5 x x x x+ + + + - = - U 3 2 3 7 9.5 25 9.7 x x x x + = + Y 2 2 5 3 2(5 3 ) x x x x = + + J @ 2 1 2 2 1 1 2 2 3 5 2 3 5 x x x x x x- + + + + + = + + 2 2 2 2 1 2 1 2 2 3 3 x x x x- + - + = + J 1 1 5 6. 5 – 3. 5 52 x x x+ - + = 5 2 4 5.4 4 0 x x - + = 9 1 4 2 8 0 x x+ + - = K 1 1 4 6.2 8 0 x x+ + - + = J < 4 8 2 5 3 4.3 27 0 x x+ + - + = X 16 17.4 16 0 x x - + = J U 1 49 7 8 0 x x+ + - = Bài 2. P'()&*+#, "& 5 8:K76@ 1 2 2 log (3 5 ) 3 0x x- + = 5 5 5 log (2 1) log (2 ) 0x x+ + - = 9 2 2 log ( 5) log ( 2) 1x x+ - + = K 1 2 2 2 log ( 5) log (2 1) log 6x x- + + = <"&K:"&8:@?6"&8?@?"&8? X"& U 8:"& <9 86 1 2 U 5 5 5 log log ( 6) log ( 2)x x x= + - + Y 5 25 0,2 log log log 3x x+ = @ 2 log (2 5 4) 2 x x x- + = 2 2 log ( 3) 1 log ( 1)x x+ = + - 1 2 3 log (log ) 0x = 5 [ ] 1 2 2 1 log log (log 9 0 x- = 9 2 log (9 2 ) 3 x x- = - K 1 2 3 3 3 log ( 5) log 2 log (3 20)x x- - - - < 5 log ( 20).log 5 1 x x + = X 3 2 1 log (2 2 3 1) 3 x x x x + + - + = U 4 2 2 4 log (log ) log (log ) 2x x+ = Y 4 2 4 log ( 3) log ( 1) 2 log 8x x+ - - = - @ 1 3 3 3 log log log 6x x x+ + = 2 2 lg 3lg lg( ) 4x x x- = - 2 4 log log 3 2x x= + 5 2 3 3 2(log ) 5log 9 3 0x x- + = 9 1 1 3 3 log 3 log 2 0x x- + = K 1 2 2 2 log ( 5) log (2 1) log 6x x- + + = < 2 log log 8 2 x x - = - X 7 6 4 log 2 log 0 x x- + = U 1 2 1 4 lg 2 lgx x + = - + Bài 3. PZ'()&*+#, 2 3 4 1 2 2 2 2 5 5 x x x x x+ + + + + - - > - 1 2 3 3 3 11 x x x - - + - < 5 2 2 3 2 3 2 9 6 0 x x x x- + - + - < 9 2 3 7 3 1 6 2 .3 x x x+ + - < K 2 2 2 1 2 2 4 .2 3.2 .2 8 12 x x x x x x x + + + > + + < 2 1 2 6. 3 . 3 2.3 . 3 9 x x x x x x x + + + < + + X 1 2 1 2 9 9 9 4 4 4 x x x x x x+ + + + + + < + + U 1 3 4 2 7.3 5 3 5 x x x x+ + + + + £ + Y 2 1 2 2 5 2 5 x x x x+ + + + < + @ 1 2 2 .3 36 x x- + > 8 ?5 8: :5[@ U 8 ≤997 8 5K 8 [<K 8 7K 99 8 : 8: 7U@\@ K K 8 7K 8: \9 < ( ) 2 4 15 13 3 4 1 2 2 x x x - + - < X ( ) 2 3 2 1 1 2 2 21. 2 0 x x + + - + > U 2 3.7 37.140 26.20 x x x - £ Y 1 2 2 log ( 4 6) 2x x- + < - @ 1 2 2 log ( 1) log (2 )x x+ £ - 1 5 4 6 log 0 x x + ³ 1 2 log (2 1) 1x + > - Chủ đề 4 : Tích phân - Ứng dụng của tích phân Bài 1:H]]'G#G/, ( ) 6 3 0 cos 2sin 1 xdx x p + ò ( ) 1 3ln 2 e dx x x + ò 2 3 6cos 1sinx xdx p p + ò ; 19 2 3 0 8 xdx x + ò Bài 2: H]]'G#G/, ( ) 1 2 0 2 4 5 x dx x x - - + ò ( ) 2 2 6 3cot 1 sin dx gx x p p + ò 4 2 2 0 cos tgx e dx x p ò ; 4 2 1 1 x dx e x + ò Bài 3:H]]'G#G/, 3 3 0 cos tgxdx x p ò 6 4 4 0 sin2 cos sin xdx x x p - ò 2 2 3 6 sin cosx xdx p p ò ; ( ) 4 2 0 cos2 sin cos xdx x x p + ò Bài 4:H]]'G#G/, 3 3 4 0 sin cos xdx x p ò 6 0 sin2 2sin 1 xdx x p + ò 3 2 3 0 1x x dx+ ò ; 4 3 6 dx tgx tg x p p + ò Bài 5:H]]'G#G/, ( ) 0 2 1 sinx xdx p + ò % ( ) 1 2 2 0 1 x x e dx+ ò ( ) 2 0 2 cosx x xdx p + ò O 1 0 3 2 x x dx e - ò 4 2 0 cosx xdx p ò & 1 0 ( 3)2 x x dx- ò ; 4 2 0 cos xdx x p ò ( ) 1 2 0 x x e dx+ ò Bài 6:H]]'G#G/, ( ) 3 2 1 3 1 lnx xdx+ ò 2 1 ln e xdx ò ( ) 1 0 ln 1x x dx+ ò ; ( ) 1 2 0 ln 1x x dx+ ò Bài 1:H];!]+'E&&>0^(D&& ( ) ( ) 2 : 3C y x x= - *MN8 Bài 2:H];!]+'E&&>0^(D&& ( ) 4 2 :C y x x= - *MN8 Bài 3:H];!]+'E&&>0^(D&& ( ) 3 : 3 1C y x x= - + (D&E& : 3d y = Bài 4:(D&& ( ) 3 2 : 3 4C y x x x= - + .'()&*+'#/ d ( ) C 0&C3NH_2];! ]+'E&&>0^ ( ) C d Bài 5:'*" ( ) 2 : 6 5P y x x= - + .'()&*+'#/ ( ) P 0&1 ( ) P >*MN8 H];!]+'E&&>0^ ( ) P '#/2^G# Bài 6:H];!]+'E&&>0^'*" ( ) 2 : 4P y x= (D&E& : 2 4d y x= - Bài 7:(D&& ( ) 4 2 :C y x x= - PC`"+'E&&>0^ ( ) C *MN8H]1]+*a 8/(b*Lc#/`8#&c#*MN8 Chủ đề 4 : Số Phức Bài 1:H+8=/ ( ) ( ) ( ) ( ) 2 3 1 2 3 2 2 4 3x y x y i x y x y i+ + + - + = - + + - - ( ) ( ) 2 1 1 2 2 3 2x y i x y i+ + - = - + - 5 ( ) ( ) ( ) ( ) 4 3 2 2 1 3x y i y x i+ + - = + - - 9 ( ) ( ) 2 2 2 2x y y x i x y x y i+ - - = + + + Bài 2H+A#='B"b'1;d*e'E&C3'B# :5 5? 5?9: 9??5 Bài 3 :H] :K:?5: 5:5? 5::K? 9 7 2 2 5 4 3 3 6 i i æ ö æ ö ÷ ÷ ç ç - + - + ÷ ÷ ç ç ÷ ÷ ç ç è ø è ø K5:?K: <9?K?5? X?7K? U@=K?5=K:=K:@=K:?7 Y9?7??5:? @:9:5????K ( ) ( ) 4 3 2i i- + ( ) ( ) 1 3i i+ - - 5 ( ) ( ) 2 3 0,2 0,5 3i i i- - 9 ( ) 2 1 i- K 1 1 i i + - < 2 1 i i- X 3 1 2i+ U ( ) ( ) ( ) 2 1 4 3 3 2 i i i i + + + - + Y ( ) ( ) 3 4 1 2 4 3 1 2 i i i i - + + - - Bài 4 :H] ( ) ( ) ( ) 2 4 3 5 7 4 3i i i+ - + - ( ) ( ) ( ) 2 1 2 2 3 3 2i i i- - - + 5 ( ) 10 1 i+ 9 ( ) 2 3 4i- K ( ) 3 2 3i+ < ( ) ( ) [ ] 2 4 5 4 3i i+ - + X ( ) 2 2 3i- U ( ) 2008 1 i+ Y ( ) 2010 1 i- @ 3 1 3 2 2 i æ ö ÷ ç ÷ - + ç ÷ ç ÷ ç è ø 3 1 3 2 2 i æ ö ÷ ç ÷ + ç ÷ ç ÷ ç è ø Bài 5 :P'*$''B+8 58::5?6K:9 ( ) ( ) ( ) 5 7 3 2 5 1 3i x i i- + = - + 5 ( ) ( ) 5 2 3 4 1 3ix i i- = + - 9 ( ) ( ) ( ) 3 4 1 2 4i x i i+ = + + K 2 3 5 4ix x i+ = + < ( ) ( ) 2 3 1 2 1 3i x ix i i- + = + + Bài 6 :P'#*$''B 8 :8:X6@ 8 :58:96@ 558 ?8:X6@ 98 5 ?U6@ K8 5 :U6@ <8 9 :58 ?K6@ Bài 7 :H+'$2&!" 1 2 ;1 2i i+ - 3 2 ; 3 2i i+ - 5 3 2 ; 3 2i i- + - - Bài 1 :H] ( ) 5 2 3 7 6i i+ - - + ( ) 1 2 3 3 2 i i æ ö ÷ ç - + ÷ ç ÷ ç è ø 5 ( ) 2 1 2i+ 9 2 15 3 2 i i - + K ( ) ( ) 2 2 1 3 1 3P i i= + + - Bài 2 :H]'B#=+A#=8'fB='f2 ( ) ( ) ( ) 0 2 3 7 8z i i i= - - - + + , 5 5z = ( ) ( ) ( ) 0 2 3 5 2 : 377z i i i kq z= - + + = 5 6 481 : 3 2 13 i z kq z i - = = + 9 ( ) ( ) 2 2 7 3 2 : 2813z i i kq z= - - - = K ( ) 3 4 3 1 : 29z i i kq z= - + - = Bài 3 :'Bg6975H] 2 z , 7 24i- 1 4 3 : 25 25 kq i z + 5 z 9 2 3 : 33 144z z z kq i+ + - - Bài 4 :P'#*$''B 2 6 29 0x x- + = 2 1 0x x+ + = 5 2 2 5 0z z- + = 9 2 4 7 0z z- + = K 2 6 25 0z z- + = < 2 2 3 5 0x x- + = X 2 3 10 9 0x x- - - = U 2 72 1297 0x x- - - = Y 2 600 2008 2009 0x x+ + = Chủ đề 5 : Phương Pháp Tọa Độ Trong Không Gian Bài 1:H*&LA&&>!C3N8/g;2'", 5 x y z− + − = = 'h2'",87/:5g:6@ H+'; H+C3&1I;'h 5H+&1'h*MC3 9.'; #A&&2'h0I K.''iB;#A&&2'h Bài 2 : H*&LA&&>!C3N8/g' ( ) α ,8:/:g?6@ ( ) , x y z d − = = − 1I?9JKJ?< .''hc#I&&' ( ) α .''ic#I#A&&2; 5.'; c#I&&; 9.'; c#I#A&&2' ( ) α KH+&1;' ( ) α <H]L&_I' ( ) α XH]1]LB;!j k=j= ="&1()&B&' ( ) α >*MC3N8=N/=Ng= ak"&;'N8/ U.'ef#lc#91j= ==k Bài 3 :H*&LA&&>!C3N8/g ( ) ( ) ( ) J@J = JJ = @JJ@A B C− PCP"*C&G&j .'NP .'ef#lc#91N=j= = 5.''#A&&2>NP'8m>l Bài 4 : H*&LA&&>!C3N8/g ( ) ( ) ( ) ( ) 9J5J = 5J@J@ = @J5J@ = @J@J5A B C D .''hc#51 ==kB&nj k"B;! .' ( ) ∆ c#j*C&GP& k 5.'ef#l2Gj'8m>' k Bài 5:H*&LA&&>!C3N8/g ( ) ( ) ( ) J@J@ = @J5J@ = @J@J<A B C .''hc#j= = H++#&C3N"'hl#/*C31N o 8B&>Nc#'h 5.';c#1j= 9H++#1";l#/*C31 o 8B&>Nc#; KPCP"*C&G&j .'ef#l22(D&L]NP Bài 6 :H*&LA&&>!C3N8/g ( ) ( ) ( ) JJ = @JJ = J@J9A B C− B&&j #A&.'j PCI"1 MB MC= − uuur uuuur .''c#I#A&&2> Bài 7:l, 2 2 2 2 2 4 3 0x y z x y z+ + - + + - = 1 2 : 1 x t y t z t ì = ï ï ï ï D = - í ï ï = ï ï î J 2 1 : 1 1 1 x y z- D = = - - * =∆ ∆ p# H+&1 ∆ >'C3 5.''h'8m&&>* Bài 8:H*&LA&&jJ?J= J5J=9J5J=k9J?J *j= ==k"91&'E& PCI=R=h"f"(b"+# "'C3N8/=N/g=N8g.''c#51I=R=h 5PCj o "+##A&&2j"'N8/`q/'lc#91j o = ==k 9.'' ( ) α '8ml01j o r, 5 9 @x y z+ + + = Bài 9 :H*&LA&&1j= ==k2C38^!B, ( ) ( ) J9J = 9 = J9J5 = A OB i j k C OD i j k− = + − = + − uuur r r ur uuur r r r *j #A&&2jJj#A&&2jkJjk#A&&2j H]1]B;!j k .'(D&#A&&2#& ∆ j k 5.'ef#lc#91j= ==k.'' ( ) α '8ml&&>'j k Bài 10:H*&LA&&>!C3N8/g ( ) ( ) J@J@ = JJ = J J 5 5 5 A B C ÷ .'' ( ) α #A&&2>N0B&51N= =E&& .';"+##A&&2j "' ( ) α Bài 11:H*&LA&&h,8?5/:9g7K6@l, 5 9 K < @x y z x y z+ + + + − + = sGt]L]u H]L&_t'h 5H_2#/**4&'hF%&#/"3#D&vL]!#"sG`L]* (D&*a Bài 12 :sp*]()& , 9 5 x t d y t z t = + = + = + >'#, h ,8:/:g:6@ h ,98:U/:g:X6@B&n#A&&2' 5h 5 ,8?/:g:K6@ 9h 9 ,8?/:9g?@6@ Bài 13 :s=1'h,98:/:<g?@6@=i,8?/?g:96@&&H]L&&w'/ Bài 14 :H++##A&&2IJ?5J"'h,8:5/?g:6@l#/*CI o 8B&>I c#'h Bài 15 :H++##A&&2IJ?J" 1 2 : 1 2 x t d y t z t ì ï = + ï ï ï ï = - - í ï ï ï = ï ï î l#/*CI o 8B&>Ic#; Bài 1 :sGL]2', 2 2 2 8 2 1 0x y z x y+ + - - + = t9JJ@u69 2 2 2 2 1 2 0 3 9 x y z x y+ + - + + = 5 2 2 2 4 2 6 5 0x y z x y z+ + + - + + = ( ) 2;1; 3I - - u65 Bài 2 :l2', 2 2 2 2 4 4 0x y z x y z+ + - - - = sGL]tJJu65 s&1l>*MC3N8=N/=Ng Bài 3 :Q$''*&*(D&b'#, 2GtJ?J?9L]u6< 2Gt@?J5c#1IJJ?9 52Gt5J?5Jc#1IKJ?J 92Gj9J?9Jc#&CN K2(D&L]j >j?JJ= @JJ5 <2(D&L]j >j<J9J?5= JUJ X2Gt?JJ'8m>'h,8:/?g:K6@ U2GIJ?J5'8m>'h,8?/?g:56@ Y2GtJ?J5'8m>'N8/ @2GtJ?J5'8m>'N8g 2GtJ?J5'8m>'N/g Bài 4 :Q$'''h*&*(D&b'#, c#51 ( ) ( ) 2;0; 1 , (1; 2;3), 0;1;2A B C- - c#51 ( ) ( ) ( ) 1;2;3 , 2; 4;3 , 4;5;6A B C- - 5c#51 ( ) ( ) ( ) 1;0;0 , 0; 2;0 , 0;0;3M N P- 9c#1IJ5J?&&'h87/:5g:96@ Kc#1R@JJ@&&'h8:5/79g76@ <c#1jUJYJ?@&&'N8/ Xc#1jUJYJ?@&&'N/g Uc#1IJ5J?#A&&2 > @JJ?5=J?9J Yc#1IJJ#A&&2; 2 1 1 1 2 2 x y z+ - + = = - @c#1R?JJ?5#A&&2; 1 6 2 2 1 3 x t y t z t ì ï = + ï ï ï ï = - í ï ï ï = - - ï ï î c#1IJ5J?#A&&2>*MN/ c#1IJ5J?#A&&2>*MNg 5c#1IJ5J?#A&&2>*MN8 9"'*#&*0j >jJ5JX= 9JJ5 K"'*#&*0hi>hJ?9J=iXJJ?K <c#1j@JJ= ?J@J#A&&2'i,87/:g:6@ Xc#1IJJ5=RJ?J9&&*MN/ Bài 5 :Q$''=';*&*(D&b'#, c#1jJ5J?= JJ9 c#1IJ?J5=R5J@J@ 5c#&C3N1IJ9J?< 9c#1I5JJ?&& 1 1 : 2 3 4 x y z- + D = = - Kc#1kJ@J?5&& 1 2 : 3 3 4 x t y t z t ì ï = + ï ï ï ï D = - + í ï ï ï = ï ï î <c#1I?5JJ#A&&2'h8?/:9g:6@ Xc#h@J@?5@&&*MNgx"$'' Uc#1jJ@J?#A&&2'h8?/:g:Y6@ Yc#1 KJJ?9#A&&2'C3N8/ Bài 6 :sp*]()& 1 1 5 : 2 3 1 x y z- + - D = = >#, 1 3 2 6 : 4 6 2 x y z- - - D = = J 2 4 1 3 : 6 9 3 x y z- - - D = = 5 3 3 2 6 : 4 3 5 x y z- - - D = = J 9 4 1 2 1 : 4 2 2 x y z- + + D = = Bài 7 :sp*]()& 3 2 : 2 3 6 4 x t d y t z t ì ï = - + ï ï ï ï = - + í ï ï ï = + ï ï î ' ' ' ' 5 : 1 4 20 x t d y t z t ì ï = + ï ï ï ï = - - í ï ï ï = + ï ï î R#m&F#++C3&1 Bài 8 :sp*]()& 1 1 : 2 1 1 x y z d - + = = - ' ' ' ' 3 : 2 1 x t d y t z t ì ï = - ï ï ï ï = í ï ï ï = - + ï ï î R#m&F#++C3&1 Bài 10 :H+C3&1 1 2 : 1 x t d y t z t ì = + ï ï ï ï = - + í ï ï = - ï ï î 'h,8:/:g?6@ Bài 11 :H+C3&1 : 1 2 1 x t d y t z t ì ï = ï ï ï ï = + í ï ï ï = - ï ï î 'h,8:/:g?56@ Bài 12 :H+C3&1 1 2 : 2 1 3 x y z d - + = = - 'h,8:/:g?6@ Bài 13 :H++##A&&2IJ?J"'h,87/:g:6@ Bài 14 :H++##A&&2kJJ"'h,X8:K/:g?5X6@ Bài 15 :H++##A&&2&C3N""'hc#51jJJ= ?JJ?=J?J? Bài 16 :H++##A&&2I9J?5J" 2 3 : 2 2 x t d y t z t ì = - + ï ï ï ï = - + í ï ï = - ï ï î Bài 17 :H++##A&&2jJJ" 2 1 1 : 1 2 2 x y z d + - + = = - Bài 18 :H++##A&&2j?J5J";c#1 9J@J?5=KJ?J9 Bài 19 :H*&LA&&>!C3N8/g ( ) ( ) ( ) ( ) J@J = @JJ@ = JJU = 5JJA B C D − .'='j .'' ( ) α c#51j= =B&nj k"B;! H]3;(D&B;!j kLy_k 5.'2(D&L]j 9.'Gkc#1 K.'Gk'8m>' ( ) α . = Y 2 600 2008 2009 0x x+ + = Chủ đề 5 : Phương Pháp Tọa Độ Trong Không Gian Bài 1:H*&LA&&>!C3N8/g;2'",