Phuong Trinh Mu

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NguyÔn ThÞ Ngäc Minh THPT Chuyªn Lµo Cai

GVto¸n

Ph¬ng tr×nh bÊt ph¬ng tr×nh mò A/Ph¬ng ph¸p ®a vÒ cïng c¬ sè vµ ®Æt Èn sè phô: Bµi 1: Gi¶i c¸c ph¬ng tr×nh vµ bÊt PT sau:

25 x

1/ 3/ 32 x

2

 2 x 1

 8.3x 

Bµi 2: 1/



3/ 2

x4

 

 6.2 x 

Bµi 3:



1/ 3  5



 2 x 1

1 23 x 3





1/ 5  21



x



cosx

 4 2/

2 x  x2





 7. 5  21

x

2

1

 9.2 x

2

x

 22 x  2  0



2 3

  cosx



2 3

cosx

4

2

 21 2 x  x  0 2/ 2(5 x  24)  5 x  7  5 x  7



x

4/

11.3x 1  31 5 4.9 x  11.3x 1  5

 2 x 3 2/125 x  50 x  23 x1

x



2/ 22 x 2

x

3/ 3x 1  22 x 1  12 2  0 4/ 3x  9.2 x  5.6 2 Bµi 5: 2 2 1/ 9sin x  9cos x  10 2/ 26  15 3

2 x

 0 4/ 32 x  2.3x  x  6  32 x 12  0

3/ 13x  5  2(13x  12)  13x  5 Bµi 4:

2

12 1 2x



 3 5

 34.15 x 2

x4

74 3



2 x  x2

2

 91

cosx

74 3 3x

 9 x



 2. 7  4 3



x



 2. 2  3

HD: PT (3) t¬ng ®¬ng víi PT





x



 1. 3/ 3  2 2



2 1

2x







2 1

x

  x

3



x

2  1  3 (3) §Æt





x

2  1  t >0

PT trë thµnh :t3-3t-1=0 .§Æt t=2y >0 PT trë thµnh : 8y3-6y-1=0 <=> 2(4y3-3y)=1(3’) XÐt c¸c nghiÖm cña PT trong ®o¹n [-1;1] ®Æt y=cosu do y thuéc ®o¹n [-1;1] =>  2   u  k 3 9 3 u   0;   PT trë thµnh 2(cos u-3cosu)=1<=> 2cos3u=1 <=>   u   0;    => u 

 7 5 ;u  ;u  => 9 9 9

 7 5 y  cos ; y  cos ; y  cos 9 9 9

Lµ 3 nghiÖm ph©n biÖt cña PT (3’) .V× PT bËc 3 cã tèi ®a 3 nghiÖm do ®ã ngoµi 3 nghiÖm trªn kh«ng cßn nghiÖm nµo kh¸c ë ngoµi ®o¹n ®ang xÐt. Mµ do y>0 x     => t  2co s => 2  1  2cos  x  log 2 1  2cos  9 9 9  x x x+1 4/cotg2 =tg2 +2tg2 (§HAN-99)





B/Ph¬ng ph¸p sö dông tÝnh ®¬n ®iÖu cña hµm sè : C¬ së lý thyÕt : 3 Bµi to¸n c¬ b¶n (Nªu ,c/m ,híng dÉn häc sinh sö dông) Bµi 1 : Gi¶i c¸c PT 1/ 5 x  12 x  13x 2/ 5cosx  12cosx  13cosx 3/



5 3

  x

5 3



x

 2x



  x

4/ 10 2  14  10 2  14



x

 16 x

Bµi 2: Cuéc sèng kh«ng ®ßi hái b¹n ph¶i giái nhÊt ,nhng ®ßi hái b¹n ph¶i nç lùc nhiÒu nhÊt

NguyÔn ThÞ Ngäc Minh THPT Chuyªn Lµo Cai

GVto¸n x

1/ 2 x  3x  5 x  10 x

4/ 2 x  3x  5 2

2/ 3x  4 x  12 x  13x

5/ 9 x  5 x  4 x  2

x

3/ 2 x  3 2  1 Bµi 3: 1/ 3.25 x  2  (3x  10)5 x 2  3  x  0

6/

2

x

2

3 2

4/ 2 x 1  2 x 2

  x

20



x

3 2

   5 x

x

2/ 25 x  2.(3  x)5 x  2 x  7  0

3/ 2.2 x  3.3x  6 x  1 5/ 2sin





2

 3sin x  2cos x  3cos x  2cos 2 x

2

x

  x  1

2

6/ 9 x  2.( x  5)3x  9(2 x  1)  0

Bµi 4: 1 x 2 1 2 x 2 x 1  2 x  1 1 1 x2 x2 1/ 2/  0 2  2   x 2 1 2 x x x 3/ 5  4 x  1 4/3 +2x=3x+2 HD: XÐt hµm sè f(x)= 5 x  4 x  1 HD: Hoµn toµn t¬ng tùphÇn 3 PT cã hai Cã f’(x)=5x.ln5-4 => f’(x)=0 <=> nghiÖm x=0 ;x=1 4 x  log 5 => lËp BBT xÐt tÝnh §§ 6/3x+5x=6x+2 ln 5 Cña h/s nh vËy H/S cã mét kho¶ng §B Mét kho¶ng NB nh vËy PT cã kh«ng qu¸2 nghiÖm vµ dÔ thÊy PT cã 2 nghiÖm x=0;x=1 .HoÆc AD B§T BÐcnuli 2 5/ 27 x  6 x 2  4 x  1 .9 x . HD:Hoµn toµn t¬ng tù sau khi chia c¶ hai vÕ cho 9x vµ ®¹t





Èn phô 3x -2x=u PT trë thµnh 3u=2x+1 C/Ph¬ng ph¸p l«garÝt ho¸ : Bµi 1: 2

x 1

x 1

1/ 5 x.x 1 8 x  100 2/ 3x.2 x 1  72 3/ 5 x.8 x  500 NghiÖm: x=10; x= 1/10 D/Ph¬ng ph¸p ®¸nh gi¸ vµ ,mét sè PP kh«ng mÉu mùc 2 2 1/ 8sin x  8cos x  9  2cos 2 y



2/ 2  2



sin 2 x



 2 2



co s 2 x



 2 2



cos 2 x

cos 2 x



2   1   2   

3/ x4-8ex-1>x(x2ex-1-8) 4/ 4 x 2  x.2 x 1  3.2 x  x 2 .2 x  8 x  12 5/ 4 x 2  x.3 x  31 x  2 x 2 .3 x  2 x  6 6/ 51+x+51-x=3x+3-x+41-x+41+x 2

7/ 196cos

2

x

2



2

2

 196sin x  100cos x  100sin x  16cos x  16sin

8/ 1  sin x.2 2

2

2

cos 2 x

 

2

2

x

1  sin 2 2 x  cos 2 x 9/ 4 x2 3 x  2  4 x 2  6 x 5  42 x2 3 x  7 2



x

x

 1 a2   1 a2   10/     1  2a   2a 

0 0 x 11/ cos 72  cos36  3.2 12/ 6x+2x=5x+3x HD: Sö dông §L Lagr¨ng PP:B1:Gäi  lµ nghiÖm cña PT B2:BiÕn ®æi PT trë vÒ d¹ng f(a)=f(b) tõ ®ã chØ ra ®îc HS f(x) kh¶ vi vµ liªn tôc f (b)  f (a) 0 trªn [a;b] khi ®ã AD §L Lagr¨ng  c  (a; b) sao cho f '(c)  ba B3: Gi¶i PT f’(c)=0 tõ ®ã t×m ®îc 

Cuéc sèng kh«ng ®ßi hái b¹n ph¶i giái nhÊt ,nhng ®ßi hái b¹n ph¶i nç lùc nhiÒu nhÊt

NguyÔn ThÞ Ngäc Minh THPT Chuyªn Lµo Cai

GVto¸n

B4 :Thö l¹i Trë l¹i PT trªn : Gäi  lµ nghiÖm cña PT => cã 6  5  3  2 (1) xÐt HS f(t)= (t  1)  t  víi t>0 tõ (1) => f(5)=f(2) ,do ®ã theo §L Lagr¨ng tån t¹i c thuéc kho¶ng (2;5) sao cho f (5)  f (2) f '(c)   0 =>   (c  1) 1  c 1   0 <=>  =0 hoÆc  =1 52 Thö l¹i thÊy tho¶ m·n E/Ph¬ng tr×nh bÊt PT cã chøa tham sè 1/T×m a sao cho BPT sau nghiÖm ®óng víi  x  0



a.2 x 1  (2a  1) 3  5

  3 5 x

x

0

2/ T×m m sao cho BPT sau nghiÖm ®óng víi  x  0



12(2m  1) 7  3 5

  7 3 5 x

x

 (2m 2  m).2 x 3

2

2

3/Gi¶i vµ biÖn luËn PT : 5 x  2 mx  2  52 x  4 mx  m  2  x 2  2mx  m 4/ Gi¶i vµ biÖn luËn BPT : m2-25x+1-8m.5x>0 5/T×m m ®Ó PT sau cã nghiÖm:





m 2x  2  1  1  2x

6/T×m m ®Ó BPT sau cã nghiÖm: 16 -(m-1)22x+m-1=0 2 2 7/T×m m ®Ó PT sau cã nghiÖm: 9 x 2 x  3( x 1)  m 8/T×m m ®Ó BPT sau sau nghiÖm ®ung víi mäi x: 25x-(2m+5)5x+m2+5m>0 9/Cho BPT : 9x-(2m-1)3x+m2+m2-m  0 .T×m m ®Ó BPT nghiÖm ®óng víi mäi x  2 m 2m  1  m  4  0 .T×m m ®Ó PT cã hai nghiÖm x1 ;x2 10/Cho PT : x  4 2x tho¶ m·n : -1<x1 <0<x2 ax 1  ax 11/Gi¶i vµ biÖn luËn: x (Víi 0
13/T×m m ®Ó PT sau cã ®óng mét nghiÖm:





x

5 1  m

14/ Gi¶i vµ biÖn luËn BPT : m2-9x+1-8m.3x rel="nofollow">0 15/ Gi¶i vµ biÖn luËn BPT : m2-2.4x+1-m.2x+1 < 0





x

5 1  2x

(T¬ng tù bµi 4) 1

1

16T×m m sao cho PT sau cã hai nghiÖm d¬ng ph©n biÖt : 91 x 2  m.31 x2  2  0 2 17/T×m m ®Ó BPT sau ®©y v« nghiÖm : 2( m 1) x  2 x 4 m 3  (2  m 2 ) x  3  4m 2

HD:Sö dông PP hµm sè .ViÕt l¹i BPT díi d¹ng 2( m 1) x  (m 2  1) x  2 x 4 m 3  x  4m  3 XÐt hµm sè f(t)=2t+t lµ hµm sè §B trªn R …….KQ:m=1



18/Cho PT 3  2 2



tgx



 3 2 2



tgx

m

a/ Gi¶i PT víi m=6 b/ T×m m ®Ó PT ®· cho cã ®óng hai nghiÖm thuéc kho¶ng

 

 

  ;  2 2

Cuéc sèng kh«ng ®ßi hái b¹n ph¶i giái nhÊt ,nhng ®ßi hái b¹n ph¶i nç lùc nhiÒu nhÊt

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