!"#$%&'('%)*%+",+% /0%&'1%234%+#$4%5%+6789%):4&%+";4"%412%%% "<=>?@A9%BCDE%FEE%FBF%% )G@H%IJ%@6KL%M%6NO%P?@6%@6Q@%RS%TU?%6NO%V6W%%% 06?%=?X=9%2Y=6>?@IPZ[@! "! !6<\%H?]?%T^%+",+%-S_O%&?Y%5%+6789%)`@H%+6a@6%4YL% 2b@9%+<\@c%)*%d/%eCfgB% 4Ha8%=6?%9%FEfBMfFBeg% +6h?%H?Y@%>aL%ia?9%ejB%V6k=l%I6b@H%Im%=6h?%H?Y@%H?Y<%T^% n?o@%6p%TG@H%IJ%I6<\%6NO%5%"<=>?@A9%BCDE%FEE%FBF%5%06?%=?X=9%qqqZLY=6>?@IPZ[@%% 0rS%e%sFlB%T?mLtZ!#$%!$&'!()! y = mx −1 x + m (1) *! "* +$,%!( !(/!0123!.$143!5&!56!78!.$9!$&'!()!:";!5<1! m = −1 *! =* >?1!@!A&!'B.!71C'!0D.!EF!.$GBH!:";*!I12J!.GK23!HLM!:";!.N1!@!HO.!H-H!7PQ3R!.1S'!HT3!.N1! UVW*!IX'!'!7C!Y1S3!.ZH$!.M'!R1-H![UW!0\3R!"]V!5<1![!A&!R1M%!71C'!HLM!$M1!7PQ3R!.1S'!HT3*! 0rS%F%sBlg%T?mLtZ%% M; >1,1!J$P^3R!._X3$! sin x + cos2x + sin( 7π 4 − 3x ) = 1 2 *! 0; `-H!793$!J$a3!.$/H!5&!J$a3!,%!HLM!()!J$bH!c!.$%,!'d3! z + 2.z = 3− 2i *! 0rS%M%sBlg%T?mLtZ%>1,1!0D.!J$P^3R!._X3$! log 2 2 x − 1 x ⎛ ⎝ ⎜ ⎜ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ ⎟ ⎟ + log 2 x.log 2 1 x ≥ 0 *! 0rS%u%selB%T?mLtZ%>1,1!0D.!J$P^3R!._X3$! x 2 +16 − 3 x 2 −3x + 4 ≥ x +1 −3 *!! 0rS%g%selB%T?mLtZ%IZ3$!Y1S3!.ZH$!$X3$!J$e3R!R1<1!$N3!0f1!78!.$9!$&'!()! y = x 3 − 3x 2 + 4 5&!._gH! $%&3$*!!!! 0rS%E%selB%T?mLtZ%#$%!$X3$!Ah3R!._g!UW#*UiWi#i!Hj!7-K!UW#!A&!.M'!R1-H!7kG!HN3$! 2a 3 V! A'G ⊥ (ABC ) !5<1!>!._?3R!.l'!.M'!R1-H!UW#*!#N3$!043!UUi!.N%!5<1!'m.!J$e3R!:UW#;!RjH! 60 0 *!IZ3$!.$C!.ZH$!E$)1!Ah3R!._g!UW#*UiWi#i!5&!E$%,3R!H-H$!R1nM!$M1!7PQ3R!.$e3R!UUi!5&!W#*! 0rS%D%selB%T?mLtZ!I_%3R!E$o3R!R1M3!5<1!$S!._gH!.%N!7B!pqKc!H$%!$M1!71C'!U:rst"s=;V!W:"stus];! 5&!'m.!J$e3R! (P ) : 2x − y + 2z −6 = 0 *!#$b3R!'13$!_\3R!'m.!HaG!7PQ3R!EZ3$!UW!HO.!'m.! J$e3R!:v;!.$w%!R1M%!.GK23!A&!'B.!7PQ3R!._x3*!`-H!793$!.l'!5&!0-3!EZ3$!'m.!HaG!7j*! 0rS%j%selB%T?mLtZ!I_%3R!'m.!J$e3R!5<1!._gH!.%N!7B!pqK!H$%!$X3$!0X3$!$&3$!UW#y!Hj! AC = 2AB *!v$P^3R!._X3$!7PQ3R!H$z%!Wy!A&! x − 4 = 0 *!>?1!{!A&!71C'!.$GBH!7%N3!U#!.$%,! 'd3! AC = 4AE V!@!A&!._G3R!71C'!HN3$!W#*!IX'!.%N!7B!UVWV#Vy!012.! E ( 5 2 ;7),S BEDC = 36 5&! 71C'!@!3\'!._43!7PQ3R!.$e3R! 2x + y −18 = 0,x B < 2 *! 0rS%C%sBlg%T?mLtZ%#$%!7M!R1-H!7kG! 2n (n ≥ 2) HN3$*!IX'!3!012.!()!.M'!R1-H!5Go3R!.N%!.$&3$!.|! =3!7}3$!HLM!7M!R1-H!0\3R!"~]*!! 0rS%eB%selB%T?mLtZ!#$%!qVKVc!A&!H-H!()!.$/H!E$o3R!l'!.$%,!'d3! x + y + z +1 = 4xyz *!IX'!R1-! ._9!A<3!3$D.!HLM!01CG!.$bH! P = x 2 + y 2 + z 2 − 2(xy + yz + zx ) *! vvv"w+vvv% !"#$%&'('%)*%+",+% /0%&'1%234%+#$4%5%+6789%):4&%+";4"%412%%% "<=>?@A9%BCDE%FEE%FBF%% )G@H%IJ%@6KL%M%6NO%P?@6%@6Q@%RS%TU?%6NO%V6W%%% 06?%=?X=9%2Y=6>?@IPZ[@! =! % ,"x4%+y0"%z{4"%n.|4%)$,%$4% 0rS%e%sFlB%T?mLtZ!#$%!$&'!()! y = mx −1 x + m (1) *! "* +$,%!( !(/!0123!.$143!5&!56!78!.$9!$&'!()!:";!5<1! m =1 *! =* >?1!@!A&!'B.!71C'!0D.!EF!.$GBH!:";*!I12J!.GK23!HLM!:";!.N1!@!HO.!H-H!7PQ3R!.1S'!HT3!.N1! UVW*!IX'!'!7C!Y1S3!.ZH$!.M'!R1-H![UW!0\3R!"]V!5<1![!A&!R1M%!71C'!HLM!$M1!7PQ3R!.1S'!HT3*! "* +$,%!( !5&!56!78!.$9•! €;!ITJ!q-H!793$•! R \{1}. ! €;!•/!0123!.$143•!! ‚!>1<1!$N3!.N1!5o!H/H•!IM!Hj! lim x→−∞ y = −1 !5&! lim x→+∞ y = −1. !! !!!>1<1!$N3!5o!H/H•! lim x→1 + y = −∞ !5&! lim x→1 − y = +∞. ! •GK!_M!78!.$9!:H;!Hj!.1S'!HT3!3RM3R!A&!7PQ3R!.$e3R! y = −1, !.1S'!HT3!7b3R!A&!7PQ3R!.$e3R! x =1. !! ‚!#$1kG!0123!.$143•!IM!Hj! y ' = 2 (x −1) 2 > 0, 5<1!'?1! x ≠1. ! •GK!_M!$&'!()!783R!0123!._43!'ƒ1!E$%,3R! −∞; 1 ( ) !5&! 1; +∞ ( ) . ! ‚!W,3R!0123!.$143•! ! ! €;!„8!.$9•!! „8!.$9!HO.!Ox$.N1! −1; 0 ( ) , !HO.!Oy!.N1! (0;1). ! …$T3!R1M%!71C'! I (1; −1) !HLM!$M1!.1S'!HT3! A&'!.l'!7)1!qb3R*! =* IM!Hj•! I (−m;m) V!R?1! M (a; ma −1 a + m ) ∈ (1) V!.M!Hj!J$P^3R!._X3$!.12J!.GK23!HLM!:";!.N1!71C'!@!A&•! y = m 2 +1 (a + m) 2 (x − a)+ ma −1 a + m *! €;!v$P^3R!._X3$!.1S'!HT3!7b3R! x + m = 0 V!J$P^3R!._X3$!.1S'!HT3!3RM3R!A&! y = m *! >?1!U!A&!R1M%!71C'!HLM!.12J!.GK23!5<1!.1S'!HT3!7b3RV!W!A&!R1M%!71C'!HLM!.12J!.GK23!5<1!.1S'! HT3!3RM3R*! €;!I%N!7B!U!A&!3R$1S'!HLM!$S! y = m 2 +1 (a + m) 2 (x −a)+ ma −1 a + m x + m = 0 ⎧ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ ⎪ ⎪ ⎪ ⎪ ⇔ x = −m y = −m 2 + am − 2 a + m ⎧ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ ⎪ ⎪ ⎪ ⎪ *! x$ 'y $ y $ ∞− $ ∞+ $ 1 $ 1− $ ∞− $ €! €! ∞+ $ 1− $ x$ O$ y$ I$ 1− $ 1 $ 1 $ 1− $ !"#$%&'('%)*%+",+% /0%&'1%234%+#$4%5%+6789%):4&%+";4"%412%%% "<=>?@A9%BCDE%FEE%FBF%% )G@H%IJ%@6KL%M%6NO%P?@6%@6Q@%RS%TU?%6NO%V6W%%% 06?%=?X=9%2Y=6>?@IPZ[@! r! •GK!_M! A(−m; −m 2 + am − 2 a + m ),IA = 2(m 2 +1) a + m *! €;!I%N!7B!W!A&!3R$1S'!HLM!$S y = m 2 +1 (a + m) 2 (x −a)+ ma −1 a + m y = m ⎧ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ ⎪ ⎪ ⎪ ⎪ ⇔ x = m + 2a y = m ⎧ ⎨ ⎪ ⎪ ⎩ ⎪ ⎪ *! •GK!_M! B(m + 2a;m),IB = 2 m + a *!!! IM'!R1-H![UW!5Go3R!.N1![!343! S IAB = 1 2 IA.IB = 2 m 2 +1 = 10 ⇔ m 2 = 4 ⇔ m = ±2 *! !X=%>SQ@9%>1-!._9!Ha3!.X'!HLM!'!A&! m = −2;m = 2 *!!! 0rS%F%selB%T?mLtZ% M; >1,1!J$P^3R!._X3$! sin x + cos2x + sin( 7π 4 − 3x ) = 1 2 *! 0; `-H!793$!J$a3!.$/H!5&!J$a3!,%!HLM!()!J$bH!c!.$%,!'d3! z + 2.z = 3− 2i *! M; v$P^3R!._X3$!.P^3R!7P^3R!5<1•! sin x −sin(3x + π 4 ) + cos2x = 1 2 ⇔ sin x −sin(3x + π 4 ) + cos2x −cos π 4 = 0 ⇔ 2cos(2x + π 8 ).sin(−x − π 8 )− 2sin(x + π 8 ).sin(x − π 8 ) = 0 ⇔ sin(x + π 8 ) cos(2x + π 8 ) + sin(x − π 8 ) ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ = 0 ⇔ sin(x + π 8 ) = 0 sin(x − π 8 ) = −cos(2x + π 8 ) = sin(2x − 3π 8 ) ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⇔ x = − π 8 + kπ x − π 8 = 2x − 3π 8 + k2π x − π 8 = 11π 8 − 2x + k2π ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⇔ x = − π 8 + kπ x = π 4 − k2π x = π 2 + k 2π 3 ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ *! !X=%>SQ@9!v$P^3R!._X3$!Hj!3R$1S'! x = − π 8 + kπ; x = π 4 − k2π ; x = π 2 + k 2π 3 ,k ∈ ! *!!!!!!! 0; `-H!793$!J$a3!.$/H!5&!J$a3!,%!HLM!()!J$bH!c!.$%,!'d3! z + 2.z = 3− 2i *! „m.! z = x + y.i (x, y ∈ !) ⇒ z = x − y.i !.$w%!R1,!.$12.!.M!Hj•! x + y.i + 2(x − yi ) = 3− 2i ⇔ (3x −3) + (2− y).i = 0 ⇔ 3x −3 = 0 2− y = 0 ⎧ ⎨ ⎪ ⎪ ⎩ ⎪ ⎪ ⇔ x =1 y = 2 ⎧ ⎨ ⎪ ⎪ ⎩ ⎪ ⎪ *! †X!5TK!c!Hj!J$a3!.$/H!0\3R!"V!J$a3!,%!0\3R!=*!! !"#$%&'('%)*%+",+% /0%&'1%234%+#$4%5%+6789%):4&%+";4"%412%%% "<=>?@A9%BCDE%FEE%FBF%% )G@H%IJ%@6KL%M%6NO%P?@6%@6Q@%RS%TU?%6NO%V6W%%% 06?%=?X=9%2Y=6>?@IPZ[@! ‡! !! ! 0rS%M%sBlg%T?mLtZ!>1,1!0D.!J$P^3R!._X3$! log 2 2 x − 1 x ⎛ ⎝ ⎜ ⎜ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ ⎟ ⎟ + log 2 x.log 2 1 x ≥ 0 *! „1kG!E1S3•! x > 0 x − 1 x > 0 ⎧ ⎨ ⎪ ⎪ ⎪ ⎩ ⎪ ⎪ ⎪ ⇔ x >1 *! WD.!J$P^3R!._X3$!.P^3R!7P^3R!5<1•! log 2 2 x 2 −1 x − log 2 2 x ≥ 0 *! ! ⇔ (log 2 x 2 −1 x − log 2 x )(log 2 x 2 −1 x + log 2 x ) ≥ 0 ⇔ log 2 x 2 −1 x 2 .log 2 (x 2 −1) ≥ 0 *! †<1!'?1! x >1⇒ 0 < x 2 −1 x 2 <1 ⇒ log 2 x 2 −1 x 2 < 0 *! y%!7j!0D.!J$P^3R!._X3$!.P^3R!7P^3R!5<1•! ⇔ log 2 (x 2 −1) ≤ 0 ⇔ x 2 −1≤1 ⇔ − 2 ≤ x ≤ 2 *! „)1!H$12G!5<1!71kG!E1S3!.M!Hj•! 1 < x ≤ 2 *!! !X=%>SQ@9!†TK!.TJ!3R$1S'!HLM!0D.!J$P^3R!._X3$!A&! S = 1; 2 ( ⎤ ⎦ ⎥ *!! 0rS%u%selB%T?mLtZ%>1,1!0D.!J$P^3R!._X3$! x 2 +16 − 3 x 2 −3x + 4 ≥ x +1 −3 *!! Phân%tích%lời%giải:% IM!.X'!7PˆH!$M1!3R$1S'!HLM!J$P^3R!._X3$!A&!q‰]sq‰r!Y%!5TK!.M!(6!.$/H!$1S3!J$zJ!A143!$ˆJ! R$zJ!5<1!:Mq€0;*!! „C!.X'!3$l3!.Š!R$zJ!A143!$ˆJ!:Mq€0;!5<1!'ƒ1!Hh3!.$bH!.M!A&'!3$P!(MG•! †<1!Hh3!.$bH!7aG!.143•! x 2 +16 = ax + b ⇒ x = 0 ⇒ 0 2 +16 = a.0+ b x = 3 ⇒ 3 2 +16 = 3a + b ⎧ ⎨ ⎪ ⎪ ⎪ ⎩ ⎪ ⎪ ⎪ ⇔ a = 1 3 b = 4 ⎧ ⎨ ⎪ ⎪ ⎪ ⎩ ⎪ ⎪ ⎪ *! †TK!.M!(6!R$zJ! x 2 +16 !5<1! 1 3 x + 4 = x +12 3 s!.P^3R!./!H$%!$M1!Hh3!.$bH!Hx3!AN1!.M!.X'!7PˆH!! x 2 − 3x + 4 → 2; x +1 → x + 3 3 *! Lời%giải:%%%% †12.!AN1!0D.!J$P^3R!._X3$!YN3R!3$P!(MG•! x 2 +16 − x +12 3 −3( x 2 −3x + 4 − 2) ≥ x +1− x + 3 3 ⇔ 3 x 2 +16 −(x +12)−9( x 2 −3x + 4 − 2) ≥ 3 x +1−(x + 3) ⇔ 9(x 2 +16)−(x +12) 2 3 x 2 +16 + x +12 − 9(x 2 −3x ) x 2 −3x + 4 + 2 ≥ 9(x +1)−(x + 3) 2 3 x +1 + x + 3 *! ⇔ (x 2 −3x ) 8 3 x 2 +16 + x +12 − 9 x 2 −3x + 4 + 2 + 1 3 x +1 + x + 3 ⎡ ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ≥ 0 (*) *!! ! !"#$%&'('%)*%+",+% /0%&'1%234%+#$4%5%+6789%):4&%+";4"%412%%% "<=>?@A9%BCDE%FEE%FBF%% )G@H%IJ%@6KL%M%6NO%P?@6%@6Q@%RS%TU?%6NO%V6W%%% 06?%=?X=9%2Y=6>?@IPZ[@! u! +Z!$1SG!U!A&!01CG!.$bH!._%3R!3R%mH!5Go3R!.M!Hj•! ! 3 x 2 +16 + x +12 ≥ x + 24;3 x +1 + x + 3≥ x + 3,∀x ≥−1 *! y%!7j•! A ≤ 8 x + 24 + 1 x + 3 − 9 x 2 −3x + 4 + 2 = 9x + 48 x 2 + 27x + 72 − 9 x 2 −3x + 4 + 2 = (9x + 48)( x 2 −3x + 4 + 2)−9(x 2 + 27x + 72) (x 2 + 27x + 72)( x 2 −3x + 4 + 2) *! y%!'‹G!()!HLM!U!YP^3R!343!.M!H$}!Ha3!7-3$!R1-!YDG!._43!.Š!HLM!U!0\3R!H-H$!3$j'!.$&3$! $\3R!7e3R!.$bH!3$P!(MG•! (9x + 48)( x 2 −3x + 4 + 2)−9(x 2 + 27x + 72) = 9 (x + 16 3 ) x 2 −3x + 4 − x 2 −25x − 184 3 ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ = 3 − 1 2 {(x + 16 3 )− x 2 −3x + 4} 2 − 381x + 812 9 ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ < 0,∀x ≥−1 *! y%!7j! (*) ⇔ x 2 −3x ≤ 0 ⇔ 0 ≤ x ≤ 3 *!! !X=%>SQ@9!†TK!.TJ!3R$1S'!HLM!0D.!J$P^3R!._X3$!A&! S = 0;3 ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ *! 46Q@%}~=9!IGK!3$143!5<1!AQ1!R1,1!3&K!01CG!.$bH!(13$!_M!(MG!E$1!A143!$ˆJ!E$j!7-3$!R1-!5&!7x1! $Œ1!.Z3$!E143!._X!5&!E$z%!Az%!3$P!._43V!Y%!5TK!.M!Ha3!$N3!H$2!71C'!3&K*!#$•!Ž!A&!.M!Hj!.$C!512.! 0D.!J$P^3R!._X3$!YP<1!YN3R•! x 2 +16 + 3 ≥ x +1 + 3 x 2 −3x + 4 *! y%!7j!H,!$M1!52!HLM!0D.!J$P^3R!._X3$!E$o3R!l'V!A•H!3&K!.$/H!$1S3!J$zJ!3l3R!AG•!.$|M!=!HLM! $M1!52!7PM!5k!0D.!J$P^3R!._X3$!H$bM!Z.!Hh3!.$bH!$^3V!(MG!7j!.$/H!$1S3!.P!.Pf3R!A143!$ˆJ!R$zJ! 5<1!:Mq€0;!3$P!._43*! Cách%2:% „1kG!E1S3•! x ≥−1 *!WD.!J$P^3R!._X3$!.P^3R!7P^3R!5<1•! x 2 +16 + 3≥ x +1 + 3 x 2 −3x + 4 ⇔ x 2 + 25+ 6 x 2 +16 ≥ x +1+ 9(x 2 −3x + 4) + 6 (x +1)(x 2 −3x + 4) ⇔ 8x 2 − 26x +12+ 6 (x +1)(x 2 −3x + 4) −6 x 2 +16 ≤ 0 ⇔ 4x 2 −13x + 6 + 3 (x +1)(x 2 −3x + 4) −3 x 2 +16 ≤ 0 ⇔ 4(x 2 −3x ) + (x +12−3 x 2 +16) +3 x +1( x 2 −3x + 4 − 2) + 6 x +1 −(2x + 6) ≤ 0 *!! ! ! ⇔ (x 2 −3x ) 4+ 3 x +1 x 2 −3x + 4 + 2 − 4 2x + 6 + 6 x +1 − 8 x +12+ 3 x 2 +16 ⎡ ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ≤ 0 ⇔ 0 ≤ x ≤ 3 *! Wf1!5X! A > 4− 4 4 − 8 12+ 3.4 > 0,∀x ≥−1 !._%3R!7j! KHOÁ%GIẢI%ĐỀ%THPT%QUỐC%GIA%MÔN%TOÁN%–%Thầy:%ĐẶNG%THÀNH%NAM%%% Hotline:%0976%266%202%% Đăng%ký%nhóm%3%học%sinh%nhận%ưu%đãi%học%phí%%% Chi%tiết:%Mathlinks.vn! "! A = 4 + 3 x +1 x 2 − 3x + 4 + 2 − 4 2x + 6 + 6 x +1 − 8 x +12 + 3 x 2 +16 !#! !Kết%luận:!$%&!'%(!)*+, !/01!23'!(+45)*!'67)+!89! S = 0;3 ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ #!! Câu%5%(1,0%điểm).%:;)+!<,-)!';/+!+7)+!(+=)*!*,>,!+?)!2@,!AB!'+C!+9.!DE! y = x 3 − 3x 2 + 4 F9!'6G/! +H9)+#!!!! I+45)*!'67)+!+H9)+!AJ!*,1H!A,K.L! x 3 −3x 2 + 4 = 0 ⇔ (x +1)(x − 2) 2 = 0 ⇔ x = −1 x = 2 ⎡ ⎣ ⎢ ⎢ #! $7!F%&! S = x 3 −3x 2 + 4 dx −1 2 ∫ = (x 3 −3x 2 + 4)dx −1 2 ∫ = ( x 4 4 − x 3 + 4x ) 2 −1 = 27 4 #! Câu%5%(1,0%điểm).%M+H!+7)+!8N)*!'6G!OPM#OQPQMQ!/R!AS&!OPM!89!'1.!*,S/!ATU!/?)+! 2a 3 V!+7)+! /+,WU!FUX)*!*R/!/01!OQ!'6Y)!.Z'!(+=)*![OPM\!'6])*!F>,!'6^)*!'_.!'1.!*,S/!OPM#!M?)+!2Y)! OOQ!'?H!F>,!.Z'!(+=)*![OPM\!*R/! 60 0 #!:;)+!'+K!';/+!`+E,!8N)*!'6G!OPM#OQPQMQ!F9!`+Ha)*!/S/+! *,b1!+1,!A4c)*!'+=)*!OOQ!F9!PM#! ! d^,!d!89!'6^)*!'_.!'1.!*,S/!OPMV!F9!e!89!'6U)*! A,K.!PM!'1!/RL! A'G ⊥ (ABC ) #! fH!dO!89!+7)+!/+,WU!FUX)*!*R/!/01!OOQ!'6Y)![OPM\! )Y)!*R/!*,b1!OOQ!F9!.Z'!(+=)*![OPM\!2g)*!*R/! A' AG ! = 60 0 #! h\! AM = AB.sin60 0 = 2a 3. 3 2 = 3a, ! ! AG = 2 3 AM = 2a,A'G = AG tan60 0 = 2a 3 #!! S ABC = 1 2 AM .BC = 1 2 .3a.2a 3 = 3a 2 3 ⇒V ABC .A' B 'C ' = A'G.S ABC = 3a 2 3.2a 3 = 18a 3 #! h\!:1!/RL! BC ⊥ AM ,BC ⊥ A'G ⇒ BC ⊥ (A' AM ) #! ij!ek!FUX)*!*R/!F>,!OOQ!'?,!k!'+7!'1!/RL! MH ⊥ AA' MH ⊥ BC ⎧ ⎨ ⎪ ⎪ ⎩ ⎪ ⎪ ⇒ d (AA ';BC ) = MH #! :1.!*,S/!FUX)*!Oek!/RL! MH = AM .sin 60 0 = 3a. 3 2 = 3a 3 2 #! Kết%luận:!$%&! d (AA';BC ) = 3a 3 2 #!!!!!!! Câu%7%(1,0%điểm).!:6H)*!`+X)*!*,1)!F>,!+-!'6G/!'H?!AJ!lm&n!/+H!+1,!A,K.!O[opqrps\V!P[rpqtpu\! F9!.Z'!(+=)*! (P ) : 2x − y + 2z −6 = 0 #!M+v)*!.,)+!6g)*!.Z'!/wU!A4c)*!`;)+!OP!/x'!.Z'! (+=)*![I\!'+yH!*,1H!'U&W)!89!.J'!A4c)*!'6z)#!{S/!AC)+!'_.!F9!2S)!`;)+!.Z'!/wU!AR#! h\!:H?!AJ!'6U)*!A,K.!|!/01!OP!89!|[spqopr\V! R = IA = 6 #! I+45)*!'67)+!.Z'!/wU!A4c)*!`;)+!OP!89! (S ) : (x − 2) 2 + ( y + 3) 2 + (z −1) 2 = 6 #! KHOÁ%GIẢI%ĐỀ%THPT%QUỐC%GIA%MÔN%TOÁN%–%Thầy:%ĐẶNG%THÀNH%NAM%%% Hotline:%0976%266%202%% Đăng%ký%nhóm%3%học%sinh%nhận%ưu%đãi%học%phí%%% Chi%tiết:%Mathlinks.vn! }! :1!/RL! d (I ;(P )) = 2.2−(−3) + 2.1− 6 2 2 + (−1) 2 + 2 2 = 1< 6 )Y)![~\!/x'![I\!'+yH!*,1H!'U&W)!89!.J'!A4c)*!'6z)! /R!'_.!kV!2S)!`;)+!6#! h\!k!89!+7)+!/+,WU!FUX)*!*R/!/01!|!'6Y)![I\#!•4c)*!'+=)*!|k!FUX)*!*R/!F>,![I\!)+%)!F'('!/01! [I\!89![spqrps\!89.!F€/!'5!/+•!(+45)*V!<H!AR! IH : x = 2+ 2t y = −3−t z = 1+ 2t ⎧ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ ⎪ ⎪ ⎪ ⎪ #! :+1&!mV&Vn!'‚!(+45)*!'67)+!/01!|k!F9H!('!/01![I\!'1!A4ƒ/L! 2(2+ 2t ) − (−3−t ) + 2(1+ 2t ) − 6 = 0 ⇔ 9t + 3 = 0 ⇔ t = − 1 3 ⇒ H ( 4 3 ;− 8 3 ; 1 3 ) #! PS)!`;)+!/01!A4c)*!'6z)!*,1H!'U&W)! r = R 2 − d 2 (I ;(P )) = 6−1 = 5 #!! Câu%8%(1,0%điểm).!:6H)*!.Z'!(+=)*!F>,!'6G/!'H?!AJ!lm&!/+H!+7)+!27)+!+9)+!OPMf!/R! AC = 2AB #!I+45)*!'67)+!A4c)*!/+€H!Pf!89! x − 4 = 0 #!d^,!„!89!A,K.!'+UJ/!AH?)!OM!'+Ha! .…)! AC = 4AE V!e!89!'6U)*!A,K.!/?)+!PM#!:7.!'H?!AJ!OVPVMVf!2,W'! E ( 5 2 ;7),S BEDC = 36 F9! A,K.!e!)g.!'6Y)!A4c)*!'+=)*! 2x + y −18 = 0 !F9!A,K.!P!/R!'U)*!AJ!)+†!+5)!s#! ! d^,!|!89!'_.!+7)+!27)+!+9)+!OPMf#! Phát%hiện%tính%chất%hình%học:% :1!/RL!„e!FUX)*!*R/!F>,!Pf#! Cách%1:!{€'!+1,!'1.!*,S/!OP„!F9!OMP!/RL! A ! (chung) AC AB = AB AE = 2 ⎧ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ ⎪ ⎪ ⎪ ⎪ #!fH!AR!OP„!AB)*!<?)*!F>,!'1.!*,S/!OMP#! ~U&!61! BC = 2BE = 2BM ⇒ ΔBEM /_)!'?,!P#! eZ'!`+S/L! IE = IM = AB 2 )Y)!|!)g.!'6Y)!A4c)*!'6U)*!'6‡/! /01!„eV!+1&!P|!89!'6U)*!'6‡/!/01!„e!<H!AR!„e!FUX)*!*R/! F>,!Pf#! Cách%2:!M+v)*!.,)+!2g)*!F€/!'5!! !!!! h\!•4c)*!'+=)*!„e!A,!ˆU1!„!F9!FUX)*!*R/!F>,!Pf!/R!(+45)*!'67)+!89! y −7 = 0 #! :H?!AJ!A,K.!e!'+Ha!.…)!+-! 2x + y −18 = 0 y −7 = 0 ⎧ ⎨ ⎪ ⎪ ⎩ ⎪ ⎪ ⇔ x = 11 2 y = 7 ⎧ ⎨ ⎪ ⎪ ⎪ ⎩ ⎪ ⎪ ⎪ ⇒ M ( 11 2 ;7) #!!!! :1!/RL! AB ! "!! = (3; 2b −14 3 ),AE ! "!! = ( 3 2 ; 7− b 3 ) #! :1!/RL! S BEDC = S BED + S BCD = 3S BED = 36 ⇒ S BED = 12 ⇒ BD = 2S BED d (E; BD ) = 16 #! KHOÁ%GIẢI%ĐỀ%THPT%QUỐC%GIA%MÔN%TOÁN%–%Thầy:%ĐẶNG%THÀNH%NAM%%% Hotline:%0976%266%202%% Đăng%ký%nhóm%3%học%sinh%nhận%ưu%đãi%học%phí%%% Chi%tiết:%Mathlinks.vn! ‰! h\!d^,!P[Šp2\!'+UJ/!A4c)*!'+=)*!PfV!F7!e!89!'6U)*!A,K.!PM!)Y)! C (7;14− b) #! fH! AC ! "!! = 4AE ! "!! ⇒ 7− x A = 4( 5 2 − x A ) 14−b − y A = 4(7− y A ) ⎧ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ ⎪ ⎪ ⎪ ⎪ ⇔ x A = 1 y A = 14+ b 3 ⎧ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ ⎪ ⎪ ⎪ ⎪ ⇒ A(1; 14+ b 3 ) #! ~U&!61! I (4; 28−b 3 ) ⇒ IB = 8 ⇔ 28− 4b 3 = 8 ⇔ b =1 b =13 ⎡ ⎣ ⎢ ⎢ ⇒ B (4;1),D (4;17) #! :‚!AR!DU&!61!O[rpt\!F9!M[}pro\#! Kết%luận:!$%&!O[rpt\V!P[Špr\V!M[}pro\!F9!f[Špr}\#!!!! P7)+!8U%)L!‹+c!(+S'!+,-)!+7)+!+^/!„e!FUX)*!*R/!F>,!Pf!'1!/R!8c,!*,a,!AŒ(!/01!29,!'HS)V!)*H9,! 61!/R!'+K!'+‡/!+,-)!2g)*!`+Ha)*!/S/+!)+4!D1UL! ! :+‡/!+,-)!'45)*!'‡!'1!/R!/])*!`W'!ˆUa#! Câu%9%(0,5%điểm).%M+H!A1!*,S/!ATU! 2n (n ≥ 2) /?)+#!:7.!)!2,W'!DE!'1.!*,S/!FUX)*!'?H!'+9)+!'‚! s)!A•)+!/01!A1!*,S/!2g)*!r‰u#!%% h\!•1!*,S/!ATU!A…!/+H!)J,!',W(!A4c)*!'6z)#!MR!'3'!/a!)!A4c)*!`;)+#! eJ'!'1.!*,S/!FUX)*!'?H!2@,!.J'!A4c)*!`;)+!F9!.J'!A•)+!`+X)*!)g.!'6Y)!A4c)*!`;)+!AR#! h\!M+^)!61!.J'!A4c)*!`;)+!/R!)!/S/+V!/+^)!61!r!'6H)*![s)qs\!A•)+!`+X)*!)g.!'6Y)!A4c)*! `;)+!F‚1!/+^)!/R![s)qs\!/S/+#! $%&!DE!'1.!*,S/!FUX)*!'?H!'+9)+!'‚!s)!A•)+!/01!A1!*,S/!89! n.(2n − 2) = 2n 2 − 2n #! :1!/R!(+45)*!'67)+L! 2n 2 − 2n =180 ⇔ n 2 −n −90 = 0 ⇔ n =10(t / m) n = −9(l ) ⎡ ⎣ ⎢ ⎢ #! Kết%luận:!$%&! n =10 89!*,S!'6C!/w)!'7.#!!!! KHOÁ%GIẢI%ĐỀ%THPT%QUỐC%GIA%MÔN%TOÁN%–%Thầy:%ĐẶNG%THÀNH%NAM%%% Hotline:%0976%266%202%% Đăng%ký%nhóm%3%học%sinh%nhận%ưu%đãi%học%phí%%% Chi%tiết:%Mathlinks.vn! •! Bình%luận.!ŽH?,!'HS)!`+S/!8,Y)!ˆU1)!AW)!F%)!<G)*!A4c)*!`;)+!/01!A4c)*!'6z)!)*H?,!',W(!A1! *,S/!ATU!s)!/?)+!89!&YU!/wU!k7)+!/+b!)+%'#! ~E!+7)+!/+b!)+%'!'?H!2@,!/S/!A•)+!/01!A1!*,S/!ATU!s)!/?)+!2g)*! C n 2 #!! Câu%10%(1,0%điểm).!M+H!mV&Vn!89!/S/!DE!'+‡/!`+X)*!_.!'+Ha!.…)! x + y + z +1 = 4xyz #!:7.!*,S! '6C!8>)!)+3'!/01!2,KU!'+v/! P = x 2 + y 2 + z 2 − 2(xy + yz + zx ) #! :+yH!*,a!'+,W'!'1!/RL!mV&Vn!<45)*!F9!/RL! 2x + 2y + 2z + 2 = 2x.2y.2z ⇔ (2x +1)(2y +1)(2z +1) = (2x +1)(2y +1)+ (2y +1)(2z +1)+ (2z +1)(2x +1) ⇔ 1 2x +1 + 1 2y +1 + 1 2z +1 = 1 !#! ~•!<G)*!23'!A=)*!'+v/!M1U/+&!•~/+‘16n!<?)*!(+_)!'+v/!'1!/RL! ! 1= 1 2x +1 ∑ = 1 x 2 2 x + 1 x 2 ∑ ≥ ( 1 x + 1 y + 1 z ) 2 2( 1 x + 1 y + 1 z ) + 1 x 2 + 1 y 2 + 1 z 2 !#! ~U&!61L! ! 2( 1 x + 1 y + 1 z ) + 1 x 2 + 1 y 2 + 1 z 2 ≥ ( 1 x + 1 y + 1 z ) 2 ⇔ 1 x + 1 y + 1 z ≥ 1 xy + 1 yz + 1 zx ⇔ xy + yz + zx ≥ x + y + z #!! $%&!'1!/+v)*!.,)+!A4ƒ/L! xy + yz + zx ≥ x + y + z V!F9!/R! x + y + z ≥ 3 #!! $9! x 2 + y 2 + z 2 = (x + y + z) 2 − 2(xy + yz + zx ) ≤ (x + y + z) 2 − 2(x + y + z) #! fH!ARL! ! P ≤ (x + y + z) 2 − 2(x + y + z) −2(x + z + z), ! •Z'! t = x + y + z ≥ 3⇒ P ≤ f (t) = t 2 − 2t − 2t, f '(t) = t −1 t 2 − 2t − 2 < 0,∀t ≥ 3 #! $7!F%&! P ≤ f (t ) ≤ f (3) = 3 −6 #!f3U!2g)*!A?'!'?,! x = y = z = 1 #!! Bình%luận:!$>,!*,a!'+,W'!29,!'HS)!/R!'+K!F,W'!8?,!<4>,!<?)*L! ! 1 ka + m + 1 kb + m + 1 kc + m = 1 m #! :1!/R!'+K!D•!<G)*!23'!A=)*!'+v/!M1U/+&!•~/+‘16n!<?)*!(+_)!'+v/!AK!'7.!.E,!8,Y)!+-!*,b1!21! A?,!84ƒ)*!AE,!mv)*! a + b + c;ab + bc + ca;abc #!!! Bài%tập%tương%tự%% Bài%số%01.!M+H!1V2V/!89!/S/!DE!'+‡/!<45)*!'+Ha!.…)! a + b + c + 2 = abc #!M+v)*!.,)+!6g)* ! ab + bc + ca ≥ 2(a + b + c ) #! Bài%số%02.!M+H!1V2V/!89!/S/!DE!'+‡/!<45)*!'+Ha!.…)! 1 a 4 +1 + 1 b 4 +1 = c 4 c 4 +1 #!M+v)*!.,)+!6g)*! KHOÁ%GIẢI%ĐỀ%THPT%QUỐC%GIA%MÔN%TOÁN%–%Thầy:%ĐẶNG%THÀNH%NAM%%% Hotline:%0976%266%202%% Đăng%ký%nhóm%3%học%sinh%nhận%ưu%đãi%học%phí%%% Chi%tiết:%Mathlinks.vn! ru! ! abc(a + b + c ) ab + bc + ca ≥ 2 #!! Cách%2:!:+yH!*,a!'+,W'!'1!/RL x + y + z = 4xyz −1 #!! ! x 2 + y 2 + z 2 = (x + y + z) 2 − 2(xy + yz + zx ) ≤ (x + y + z) 2 − 2 3xyz(x + y + z) = (4xyz −1) 2 − 2 3xyz(4xyz −1) #! fH!AR! P ≤ (4xyz −1) 2 − 2 3xyz(4xyz −1) − 2 3xyz(4xyz −1) ≤ 3 −6 #!!! :U&!)+,Y)!'+‡/!+,-)!'+yH!/S/+!)9&!/+•!S(!<G)*!A4ƒ/!F>,!.J'!DE!29,!'HS)!&WU!F9!F,-/!`+aH!DS'! +9.!DE!D,)+!61!D’!`+R!`+N)!+5)#!‹+4!F%&!'1!/R!'+K!)+%)!m€'V!.J'!29,!/‡/!'6C!)WU!8ƒ,!FT!AS)+! *,S!D’!`+R!`+N)!FT!+9.!DE!F9!)WU!8ƒ,!FT!+9.!DE!Az,!+†,!`“!'+U%'!AS)+!*,S!/+U”)!mS/!F9!/+Z'! )+3'#! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! . 3 *! Wf1!5X! A > 4− 4 4 − 8 12+ 3.4 > 0,∀x ≥−1 !._%3R!7j! KHOÁ%GIẢI%ĐỀ%THPT%QUỐC%GIA%MÔN%TOÁN%– %Thầy: %ĐẶNG%THÀNH %NAM% %% Hotline:%0976%266%202%% Đăng%ký%nhóm%3%học%sinh%nhận%ưu%đãi%học%phí%%% Chi%tiết:%Mathlinks.vn! "! . 6 #! I+45)*!'67)+!.Z'!/wU!A4c)*!`;)+!OP!89! (S ) : (x − 2) 2 + ( y + 3) 2 + (z −1) 2 = 6 #! KHOÁ%GIẢI%ĐỀ%THPT%QUỐC%GIA%MÔN%TOÁN%– %Thầy: %ĐẶNG%THÀNH %NAM% %% Hotline:%0976%266%202%% Đăng%ký%nhóm%3%học%sinh%nhận%ưu%đãi%học%phí%%% Chi%tiết:%Mathlinks.vn! }! :1!/RL! . S BCD = 3S BED = 36 ⇒ S BED = 12 ⇒ BD = 2S BED d (E; BD ) = 16 #! KHOÁ%GIẢI%ĐỀ%THPT%QUỐC%GIA%MÔN%TOÁN%– %Thầy: %ĐẶNG%THÀNH %NAM% %% Hotline:%0976%266%202%% Đăng%ký%nhóm%3%học%sinh%nhận%ưu%đãi%học%phí%%% Chi%tiết:%Mathlinks.vn! ‰! h!d^,!P[Šp2!'+UJ/!A4c)*!'+=)*!PfV!F7!e!89!'6U)*!A,K.!PM!)Y)!