Các phương pháp giải bài tập giải tích 12 nâng cao: Phần 2

Nối tiếp nội dung phần 1 tài liệu Giải bài tập giải tích 12 nâng cao, phần 2 giới thiệu tới người đọc kiến thức cần nhớ và phương pháp giải các bài tập hàm số lũy thừa, hàm số mũ và hàm số logarit; nguyên hàm - Tích phân và ứng dụng; số phức. Mời các bạn cùng tham khảo.

Trang 1

a) Dinh nghia: Vdi a ?t 0, n e Z, luy t h t e bac n cua so a la so a", xac dinh bcfi

a° = 1; a" = — (a: cof so, n: so mu)

2 C a n b $ c n v a l u y thiifa v«Ji so m u hi^u ti

a) Dinh nghia: W6\ nguyen duotng, can bSc n cua so thUc a, k i h i ^ u

la so thirc b sao cho b" = a

• N/EI = b <:5> b" = a (n nguyen difong)

Ghi chu:

• Neu n le t h i m o i so' thuc a chi c6 mot can bac n

• Neu n chSn t h i m o i so' thUc duone a c6 h a i can bac n la >/a va - N/^

b) Tinh chat: Vdri a > 0, b > 0; m, n nguyen dUcfng va hai so' p, q tuy y

* 'Vab = '^.'</h

i ' - i W ( a > 0 )

* Neu t = ± t h i =: "/a" > P = -3- Q )

n m Dac biet Va ^ Va

3 L u y thijfa vori so m u htfu ti

Cho a la so thifc duong va r la so hCfu t i Gia suT r = ^ (trong do m la so nguyen, n la so nguyen ducfng) luy thifa cua a vdi so mu r la so a', xac

^ a " = ^ dinh bori:

Ghi chu: T i n h chat ciia luy thiTa vdi so mu hUu t i (cua so thi^c duong) c6

day dii t i n h chat nhuf luy thifa vdi so mu nguyen

J B A I T A P

L (Bdi 1, trang 75 SGK) Giai

a) Sai b) Dung c) Sai

2 (Bai 2, trang 75 SGK) Gidi

Trang 2

9 (Bdi 9, trang 78 SGK) Giai

Tif tinh chat cua iQy thifa vdl so mu nguyen dUcfng, ta c6

(Ta.Tb)" = (7a)".(7b)" = ab (vdi a > 0, b > 0, n nguyen duong)

Theo dinh nghia cSn h&c n cua mot 86' ta suy ra: Tab = Ta.Tb

10 (Bdi 10, trang 78 SGK) Giai a) Cdch 1:

74 + 273 - 74 - 273 = 7(73+1)' -7(73-1)'

= 73+ i | - 7 3 - i | = 73+ 1-73+ 1 = 2

Cdch 2:

Dat X = 74 + 273 - 74-273 > 0

Trang 3

Ta chufng m i n h d i n g thiJc (1), that vay

Cho a la so' thifc diTOng va a la so' v6 t i

Xet day so' hufu t i r i , r2, T„, ma limrn = a

K h i do day so thuc a"' , a'', a'° c6 gidi han xac dinh (khong phu

thuoc vao day (rn) da chon) Ta goi gidi han do l a luy thiia ciia so' mu ot K i hieU a"

a " = lima^"

Nhd:

- Luy thCra vdi so' mu 0 va so' mu nguyen a m k h i co so' a 0

- Luy thCfa vdi so' mu khong nguyen t h i co so' a > 0

2 C o n g t h i t c l a i k e p

C = A ( l + r)^' trong do A la so' tien gijfi

Trang 4

= â^(a ')''^ ' = ậa-^^' = â^-^^-' = á = a

17 (Bdi 17 trang 81 SGK) Gidi

Sau 5 n a m ngifcfi fíy t h u duac ca von l l n l a i la:

d) (x" + y " ) ' - 4" xy = , / x ^ + 2x''y" + y^" - 4x''ý'

Vay phufOng t r i n h da cho c6 hai nghiem x = 1 va x = 16

a) x ^ < 3 o 0 < x ^ < V3 0 - V 3 < x < V3

b) x'^ > 7 « X > W c) x'° > 2 o |x| >

d) x^ < 5 « X < ^

<=> X < - hoac X >

§ 3 L O G A R I T

I \ 6 l D U I \ c A i ^ I « i 6 ' 1- D i n h n g h i a : Cho a > 0 va a 1, b > 0, So thUc a de a" = b duac goi la

l o g a r i t ccf só a cua b K i hieu logab, nghia l i : logab = a c=> a" = b

Ta c6: logal = 0; logaa = 1

logaá' = b, Vb e K; á°^-''=: b, Vb e R, b > 0

Trang 5

2 T i n h c h a t : C h o so' dUOng a 1 va cac so dUcrng b, c

* Neu a > 1 t h i logab > log^c <s> b > c

* Neu 0 < a < 1 t h i logab > logaC <=> b < c

He qua: Cho so' difong a # 1 cdc s o ' dUcfng b, c

• Neu a > 1 t h i logab > 0 o b > 1

• Neu 0 < a < 1 t h i logab > 0 <:=> b < 1

• logab = logaC C=> b = C

3 C a c q u i tAc t i n h l o g a r i t

Cho so dircrng a 1 va cAc so di/ong b, c, t a c6:

• lOga(bc) = logab + logaC

• loga - = logab - lOgaC

He qua: V d i a, b la h a i so difong khac 1, ta c6:

• logab = — hay logab.logba = 1

• log c = — logaC (a 0; c > 0)

^" a

4 L o g a r i t t h a p p h a n

Logarit co s o 10 cua inQt s o dirong x difcfc goi la iogarit thap phan cua x

k i hieu Igx (hay l o g x )

* Logarit thap phan c6 day du cac t i n h chat cua logarit vdi ccf so lorn hem 1

^ B A I T A P •

Chon d)

24 (Bai 24 trang 89 + 90 SGK) Gidi

a) Sai b) Diing c) Sai d) Sai

a) loga (xy) = logaX + logay

log9

• logo, = 1 ^ « -0,42

log0,75

<=> 0,5 + X = 6 <=> X = 5,5

Trang 6

32. (Bdi 32 trang 93 SGK) Gidi

a) log8l2 - logglS + logg20 = logs~~ = logglG = log^,, 2' = ^

15 o

b) ^ log736 - log7l4 - 31og7 ^

= log? (6^)2 - log7l4 - l o g 7 ( 2 1 ^ ) ' = log72.3 - log72.7 - log73.7

= log72 + log73 - (log72 + log,?) - (Iog73 + log77)

a) log2 + log3 = log2.3 = log6

do log6 > log5

Vay log2 + log3 > log5

b) l o g l 2 - log5 = l o g — = log2,4

do log2.4 < log? Vay l o g l 2 - log5 < log?

c) 31og2 + log3 = log2'^ + logs = log2^3 = log24 21og5 = logS^ = log25

do log24 < l o g 2 5

v a y 31og2 + logs < 21og5 d) 1 + 21og3 = loglO + logS^ = l o g i c s ^ = log90

Do log90 > log27 Vay 1 + 21og3 > log27

1

a) logax = log«a%' x/c = logaa-'.b'

= logaá + log„b^ + logaC^ = 3 + 21og„b + - l o g c = 3 + 2.3 + - ( - 2 ) = 8

b) logaX = l0g« ^ logaâ + log, - log^

logsx = logaa" + log3b^ => logsx = logsậb'' => x = a*.b^

b) logox = 21og5a - 31og5b

=> logsx = logoá^ - loggb^ logsX = l o g 5 ^ => X = ^

a) log^gSO = log 5.10 = 2[log3 5 + log3lO]

Biet log3l5 = log33.5 = loggS + log35 = 1 + logaS Suy ra log35 = a - 1

Vay log ,| 50 = 2[a - 1 + p] = 2a + 2p - 2

b) Iog4l250 = log,, 5^2 = laogzS" + log22| = ^ [41og25 + 1| = 2a + |

a) l o g - + - l o g 4 + 41og72 = log2^^ + Iog(2')2 + log 2^ = log2^l2.2^ = l o g l

l b ) l o g ^ + ^logS6 + | l o g ^ = log(2^3-') + log(6')5 + log

Trang 7

c) log72 - 2\og~ + logN/108

= log 2-' - log(0,5^3) + log(0,5^32) ^ i o g l ^ | 5 _ 3 _ ^ i ^ g ^ = log —

So cdc chuf so cua M 3 1 k h i viet trong he thap phan b^ng so cac chuf so cua 2"

nen so cac chuf so cua M 3 1 la [31.1og2] + 1 = [9,31 + 1 = 10

Tuang tiS, so cac chC so cua M127 = 2^^'' - 1 k h i viet trong he thap phan la:

[127.1og2] + 1 = 38 + 1 = 39

So cdc chuf so cua Mi398269 k h i viet trong he thap phan la

[1398269.1og2] + 1 = 420921

41 (Bai 41 trang 93 SGK) Gidi

So t i e n ca von iSn l a i sau n qui \k

Vay sau 4 n a m , 6 t h a n g (4 nam, 2 qui) ngifdi gufi se c6 i t nha't 20 trieu tiT so

von 15 trieu dong ban dau)

(vi sau qui hai, ngifdi gijfi m d i nhan dUc/c lai)

3 L o g a r i t tu" n h i e n

L o g a r i t cof so e cua mot so diTcfng a difoc goi la logarit t i i nhien (hay logarit

ne-pe) cua so a K i hieu Ina

* Logarit tif n h i e n c6 day du t i n h chat cua logarit co so lorn hon 1

J B A I T A P

Sai tuf ln(2e) = Ine + Ine

Vi ln(2e) = ln2 + Ine ^ Ine + Ine

Trang 8

4 6 (Bai 45 trang 97 SGK) Gidi

Ti le t a n g trufdng m 6 i gicr ciia lohi v i khuS'n

Tii cong thufc S = Ạé''

300 = lOỌế'

^ ^ ^ l n _ 3 0 0 - l n l 0 0 ^ ]n3 ^

5 5

T i 1§ t a n g t r u d n g ciia loai v i khuan n^y la 21,97% m 5 i gid

Sau 10 gid, tii 100 con v i k h u a n se c6:

TiJf 100 con, de c6 200 con t h i thori gian c^n t h i e t l a

200 = lOỌê-^'^^'

, I n 2 0 0 - l n l 0 0 l n 2 o - , r : o •>,n u -

=>t= = « 3,15 gift = 3 gid 9 p h u t

0,2197 0,2197

T i le phan hiiy h a n g n a m ciia Pu^'^

Trong do S va A t i n h b k n g gam, t t i n h b&ng n a m

Vai 100 gam Pu^''^, t h d i gian cAn t h i e t de phan h u y con 1 gam la

1 = 10 e"'^''^^^'^^*''

] n l - j n l O ^ 82235 (nam)

-0,000028

Vay sau khoang 82235 n a m t h i 10 gam chat Pu^'*'' p h a n hiiy con 1 gam

§5 HAM SO MU VA HAM SO LOGARIT

I I \ O I D U I \ C A N M l I C ?

1 K h a i n i $ m v e h a m so m u v a h a m so logarit

* Cho a > 0 va a 1

- H a m só y = a" duoc goi la h a m só mu co só a

- H a m so y = logaX dJoc goi la h a m so logarit co so a

* H a m só y = á' va y = logaX l i e n tuc t a i m o i diem ma no duac xac d i n h

• Vxo e M, l i m a" = á^

* H a m s6 y = a" C O (3ao h a m la ý = á'.lna

* H a m so y = e" c6 dao h a m la ý = e"

* H a m so y = a"'"* c6 dao h a m la y ' = ú(x).a""'\lna

* H a m sd' y = logaU(x) (vdi u(x) > 0) c6 dao h a m ý =

* H a m só y = Inu(x) (vdi u(x) > 0) c6 dao h a m y ' =

Dac bi$t: y = I n i x | (vdfi m o i x 0) t h i y ' = —

ú(x) uix).lna ú(x) u(x)

y = I n I u(x) I (vdfi m o i u(x) 0) t h i y ' = — —

- Nhm phia t r e n true hoanh

- N h a n true hoanh la t i e m can ngang

- N k m ben p h a i true tung

- N h a n true tung la t i ^ m can diJng

Dang dS thi

1

0

Trang 9

^ B A I T A P

47 (Bdi 47 ti-ang 11 SGK) Gidi

a) Khi nhiet do ciia nUdc la 100"C thi luc do P = 760

49 (Bai 49 trang 112 SGK) Gidi

a) y = (x - De'" ^ y' = e'^ + (x - l).2e''' = e'^ (1 + 2x - 2) - (2x - l)e'^

y' = - (e" + e-")' = ^ (e" - e"^)

a) Ham so y = - dong bien tren R (vi ccJ s6' - > D

H^m s6' y = .>/2 + V3 nghich bien tr§n M (vi co so V2 + Vs < 1)

51 (Bdi 51 trang 112 SGK) Gidi

a) - Ham so y = (x/2 )" xac dinh tren R

- Ham so dong bien tren R (ca so J2 > 1)

Do thi hkm so:

- Di qua cdc diem (0; 1); (1; ^); (2; 2)

- N^m phia tren true hoanh

- Nhan true hoanh lam tiem can ngang

b) Ham so' y = [3) xac dinh tren

- Nam phia tren true hoanh

- Nhan true hoanh la tiem can ngang

'2-^1

I ' 3j V 9j

y 2-

3 Nhac manh phat ra tCf loa 6,8.10** 88 dB

4 Tieng may bay phan liic 2,3.10'^ 124 dB

, , ln(l + 3x) 31n(l + 3x) ln(l + 3x) ^ , ^ a) hm = lim = 3iim = 3.1 = 3

, ln(l + x^) , b) Vi lim x-.a 5 x^ = 1 , ln(l + x') , xln(l + x') ^ nen hm = hm 5 = 0.1 = 0

Trang 10

a) H ^ m so y = logaX nghich bien t r e n (0; +x)

- Nkm bgn phdi true tung

- N h a n true tung la difdng t i | m ean diJng

- N l i m ben phSi true tung

- N h $ n true tung la t i 0 m can dufng

§6 HAM SOLUYTHLfA

1 K h a i n i ^ m h a m so l u y thxia

- H a m so luy thCfa la h a m so' dang y = x" (trong do a la h a n g so')

- H a m s6' y = x°, v6i a khong nguyen xdc d i n h t r e n (0; +oo)

2 D a o h a m c u a h a m so l u y thvta

- H a m so y = x" (vdi a e K) c6 dao h^m y' = a x " " '

- H a m so' y = u°(x) (v6i a G K) C6 dao ham y' = a u ° ' kx).u'(x)

3 V a i n e t ve su" b i e n t h i e n do thi h a m so l u y thtifa

* H ^ m so luy thiTa y = x" (vdi a 0) x^c dinh t r e n (0; +x)

* H a m so dong bien t r e n (0; +t3o) n§'u a > 0 v^ nghieh bien t r e n (0; +x)

neu a < 0

• Do t h i h a m so qua ( 1 ; 1) vdfi moi a

Mot s6 dang dd thi

O

1

M B 4 I T A P

m. (Bdi 57 trang 117 SGK) Gidi

Gia sijf (Ci) va (Cj) Ian lucft la do t h i cac h ^ m so y = x" va y = x" (a, p c6 the

- 2 h o a c - | ) Tren do t h i ta thay vdi x e (1; + x )

Do t h i (C2) d phia tren do t h i (Ci)

Nghla la k h i x > 1 ta eo bat d^ng thijfc

x" > x° o p > a

Do do j3 = - - va a = - 2

2 Vay (C2) la do t h i h a m so y = x 2

Trang 11

( 2 ^ ) ' x ' - 2 \ ( x ' ) ' 2 M n 2 x ' - 2 \ 2 x 2''(xln2-.2)

^' = , = = 5

y ' ( l ) = 2(ln2 - 2) « 2,61

a) Goi ( C i ) va (C2) M n lugt la do t h i cdc h ^ m so y = ax^ va y =

M(xo; yo) la d i e m bat k i D i e m doi xufng ciia M qua true t u n g la M'(-xo; yo),

ta c6:

/ J N

M e (Ci) o yo = a"" o yo = M ' e ( C )

Dieu nay chufng to ( C i ) va (C^) doi xiifng nhau qua true tung

b) Goi (C,'j) va ( C 4 ) Ian luot la do t h i cae h a m so y = log^x va y = logj x

Goi M(xo; yo) la diem bat ki.* Diem doi xutng cua M qua true hoanh la M"(xo; -yo)

Ta CO M e (C3) <=> yo = l o g a X o c=> y„ = - log, x„ o M " e (C4)

a

D i e u n ^ y chufng to (C;J), ( C 4 ) do'i xufng nhau qua true hoanh

^ a) H a m so y = logonx xdc d i n h t r e n (0; +oc) y'

• H a m so' nghich bien t r e n (0; + « )

Do t h i h a m so:

- D i qua di§m ( 1 ; 0); (0,5; 1)

- N a m ben p h a i true tung

- N h a n true tung la t i e m can dufng

a) Can cuf vao do t h i , ta c6: logo,5X > 0

.3-<=> 0 < X < 1

•(ling v6i phan do t h i d phia tren true hoanh)

I b) - 3 < logo.sx < - l c : > 2 < x < 8

(lifng vdi nhufng diem t r e n do t h i c6 tung do thuoe niifa khoang (-3; -1))

2 (Bdi 62, trang 118 SGK) Gidi

- H a m so y = (yfSY xae d i n h t r e n R

~ H a m so y = (^f3)'' dong bien t r e n K

Do t h i h a m so:

- D i qua diem (0; 1); ( 1 ; Vs )

- N ^ m phia t r e n true hoanh

- N h a n true hoanh la t i e m can ngang

a) ( ^ / 3 ) ' ' < l c ^ x < 0 (ufng v6i edc diem tren do t h i c6 tung

do nho hcfn hay b k n g 1)

b) ( S ) " > 3 o X > 2

(ufng v d i cAc diim t r g n do t h i c6 tung dp 16n hofn 3)

Trang 12

§7 PHLfdNG TRINH M O V A LOGARIT

1 + PhUcfng trinh mu cat ban: a" = m

• Neu m < 0 t h i phiforng t r i n h a" = m v6 nghi#m

• Neu m > 0 t h i phu'cfng t r i n h a" = m <=> x = l o g e m

+ Phitcfng trinh logarit cd ban: log^x = m

• logaX = m <=> a"' = X ( d i l u k i e n xac d i n h x > 0)

* Co the sijf dung c^ch dSt bien sS' phu

* Phiicfng phap logarit hoa

* Suf dung t i n h d6ng bien hay nghich bien ciia h ^ m so'

2 b) 2"'-^''^' = 4 « 2"'-'"*' = 2' » x ' - 3x + 2 = 2

Vay phuong t r i n h c6 nghi$m x = 0

64. (Bdi 64, trang 124 SGK) Giai

a) log2[x(x - 1)] = 1 (1)

fx < 0 Vdi d i l u k i f n x(x - 1) > 0

[x > 1

(1) <=> x(x - 1) = 2' » x^ - x - 2 = 0 <=> X = - 1 hoac x = 2 (thoa dieu ki?n)

V$y phuong t r i n h c6 h a i nghi$ra x = - 1 x = 2

o x = - 1 h o a c X = 2 (cau a) <» x = 2 Vay phUdng t r i n h c6 n g h i e m x = 2

65. (Bdi 65, trang 124 SGK) Giai

a) Theo gia t h i e t k h i d = 0 thi F 53 (kHz)

k h i d - 12 t h i F 160 (kHz)

K a " = 5 3 (1)

K a ' ' = 160 (2) Tif (1) suy r a k = 53

12

o a" = 3,019 <^ 121oga = log3,019 o 121oga * 0,480

o loga * 0,04 <:=> a « 10°'^^ * 1,096

b) G i a i phiiOng t r i n h k a ^ = F

<=> logKa"' = logF a- logK + loga** = logF loga'' = logF - logK

<=> dloga = logF - log53 o d = - — - ( l o g F - log53)

Trang 13

6T (Bdi 67, trang 124 SGK) Gidi

a) logzx + log4X = logi N/3 ( D i ^ u k i e n x > 0)

Vay n g h i e m ciia phUcfng t r i n h l a x = 0

a) log^x^ - 201og7x + 1 = 0 Dieu k i e n x > 0

1

o 91og-x - 201ogx2 + 1 = 0 91og^x - lOlogx + 1 = 0

Dat t = logx, ta c6 phi/ofng t r i n h

2t 4 (2 + t) Dat t = log2X, t a c6 phifong t r i n h :

1 + t 3 ( 3 + t )

o 6t(3 + t ) = 4(1 + t)(2 + t ) 18t + et"* = 4(t^ + 3t + 2)

o 6t^ + 18t - 4t^ - 12t - 8 = 0 <o 2t^ + 6t - 8 = 0 c:>t^ + 3 t - 4 = 0 c = > t = l hoac t = - 4

Trang 14

Phiitfng trinh c6 hai nghiem x = — va x =

-7 0 (Bdi -70, trang 125 SGK) Gidi

x = - l o g 3 2 - l (thda dieu ki#n)

Phuong trinh c6 hai nghiem x = 2 va x = -(log32 + 1)

Ta chufng m i n h phifcfng t r i n h k h o n g c6 n g h i e m n^o k h a c

- Ham so 7 = 2" dong bien tren K

- Ham so' y = 3 - X nghich bien tren R

Do do 2" < 3 - X tren khoang (-oo; 1)

Khong CO gia t r i nao ciia x thoa phiTcfng trinh da cho

Vay phuorng trinh da cho c6 nghiem duy nhat la x = 1

b) log2X = 3 - X. Dilu ki^n xdc dinh x > 0

D l t h a y x = 2 l a n g h i e m duy n h a t c u a phiTcfng t r i n h (vi log22 = 3 - 2<=>l = l)

Ta chufng m i n h p h U c f n g t r i n h k h o n g c6 n g h i e m n a o k h a c

- Ham so' y = log2X dong bien tren (0; +<»)

- Ham so y = 3 - X nghich bien tren (0; +oo)

• Vdri X > 2, log2X > log22 log2X > 1

V 6 i x > 2 , 3 - x < 3 - 2 o 3 - x < l

Do do log2X > 3 - x

Phifang trinh da cho khong c6 nghiem tren (2; +x)

• Vdi 0 < X < 2, log2X < log22 o log2X < 1

V d i 0 < x < 2 , 3 - x > 3 - 2 < = > 3 - x > l

Phi/cfng t r i n h da c h o k h o n g c6 n g h i e m t r e n (0; 2)

Vay p h u a n g t r i n h da c h o c6 n g h i f m duy n h a t l a x = 2

Trang 15

§8 HE PH J O N G TRiNH MU VA LOGARIT

BUNG C A I \

Gidi he phucmg trinh mu vk logarit ta dung cdc phufcfng phdp the, cong dai

so, dat phu

Xet phiJong trinh (2) log4X + log4y = 1 + log49

<=> log4X.y = log44.9 oxy = 36

4- 2 X + 4 - 2 y ^ Q 5 ( 4 )

Tii phifcfng trinh (3): y = 1 - x

The vao phifang trinh (4): 4-^" + 4-2'^-"' = 0,5 o 4-2" + 4'\4'^'' = 0,5

• Tir do: X = - (thoa dilu kien)

Vay tap nghiem cua he phiTcfng trinh Ik S =

b) log2(9 - 2") = 10'°^='-'" ( 2 )

H Vdi dieu kien x < 3

^ ( 2 ) o logaO - 2") = 3 - X o log2(9 - 2") = log22'-''

H Phuong trinh c6 nghiem x = 0

m 7iogx _ giogx+i ^ 3 5iogx-i _ i3_7iogx-i ( 3 )

B Vdi dieu kien x > 0

B (3) « + 13.7'°^-^ = 3.5'"^"^ + 5'°^*'

B ^ ylogx ^ 2^ ^logx _ 3 gi^g, ^ g

• 7 5

Trang 16

<=> logx = 2 <=> logx = loglO^

<=> X = 100 (thoa dieu kien)

Vay phuong trinh c6 nghiem x = 100

d) 6^ + 6''^^ = 2" + 2"^! + 2"*'

o 2''.3'' + 6.2^3'' = 2^+2.2" + 4.2"

o 7.2''.3'' = 7 2 " « 3" = 1 (do 7.2" > 0)

o X = 0

Nghiem phiTOng trinh lA x = 0

o X = log382 - log381 o X = log382 - 4 (th6a dieu ki0n)

V|[y tap n g h i l m cua phiictng t r i n h la S = {log328; log382 - 4)

t > 0

o t = 0 hoac t = 25

t = 0 hoac t = 25

• t = 0; log2(-x) = 0 o - x = l o x = - l (thda d i l u kien)

• t = 25; log2(-x) = 25 o - X = 2^^ o X = -2^^ (th6a dieu ki$n), V$y tap nghi$m cua phuctog t r i n h la S = (-2^^ -1}

Vdri d i l u kien x > 0

(4) ^ 3"*«'".32 +3'°^*".3'2 = o 3'°*«".(32 +3'^) = 4 ^

^/3

V3 l3 <=> log4X = l o g a - ^ 3 v 3

o X = 4 (thda digu ki§n)

V|ly tap nghiSm cua phifcfng t r i n h la S = (4

Trang 17

X = - 1 la mot nghiem cua phuong t r i n h

Ta chijfng m i n h phufong t r i n h (1) k h o n g c6 nghiem nao khac, t h a t vay

Trang 18

D i l u n^y chiJng to tren khodng ( - 1 ; + « ) phuong t r i n h (1) khong c6 nghiem

Vay X = - 1 1^ n g h i e m duy nha't cua phUong t r i n h da cho

f T T Y f TtV

b) s i n - + c o s - = 1 ( 2 )

\ y 5J

D l t h a y x = 2 l a m o t n g h i e m cua phifcfng t r i n h v i s i n ' - + cos' - = 1

Ta chtJng m i n h phUcfng t r i n h ( 2 ) khong c6 n g h i e m n^o kh^c, t h a t vky do

3 + log.,y = log2 5 ( l + 31ogsx) ( 2 )

V d i dieu ki§n x > 0 , y > 0

log2 2^ + log2 y = log2 5 + 31og2 x [logj 8y = log^ 5 + Slogg x b) ( I I )

logjxy = log6l0 logaSy = log2 5.x^

xy = 10 8y = 5x^

§9 BAT P H l / d N G TRINH M O V A L O G A R I T

I l>fOI DUIVG CAM NH6

G i i i b a t phuong t r i n h mO b a t phiTong t r i n h logarit, c i n nhdf cdc h ^ r a so

y = a" y = log^x dong b i e n k h i a > 1 n g h i c h b i e n k h i 0 < a < 1

4

Trang 19

a) loggOx - 1) < 1 (1) Dieu kien xac dinh 3x - 1 > 0

logsOx - 1) < 1

<=> logsCSx - 1) < logsS 3x - 1 < 5 (do tinh dong bien ciia h^m so logarit

CO so 5)

(1) » '3x - 1 > 0 3x - 1 < 5 x<2 ^ 3 o - < X < 2

Vay tap nghiern ciia bat phiTOng trinh 1^: S =

b) log, (5x -1) > 0 (2) Dieu ki|n xdc dinh 5x - 1 > 0

Vay t$p nghi^m cua phifcrng trinh la S =

c) logo,5(x^ - 5x + 6) > -1 (3) Dieu ki|n xdc dinh x^ - 5x + 6 > 0

logo,5(x' - 5x + 6) > -1

o logo.5(x^ - 5x + 6) > logo.50,5'' o x^ - 5x + 6 < 2 (do tinh nghich bieii

ciia ham so logarit co so 0,5)

x' - 5x + 6 > 0 Jx^ - 5x + 6 > 0 x' - 5x + 6 < 2 jx' - 5x + 4 < 0

Tap nghi^m bat phUdng trinh da cho la S = 3' 2)

I (Bdi 82, trang 130 SGK) Gidi

a) logogX + logggX - 2 < 0 Dilu kien xac dinh x > 0

Dat t = logo.sx

Ta c6: t'' + t - 2 < 0 » -2 < t < 1 <=> (1)

l t < l

• t > -2 o logo.sx > -2 o logo,5X > logo.sO.S"^ <=> logo.sx > logo,54

<=> x < 4 (do tinh nghich bien ciia ham so logarit co so 0,5)

• t < 1 <=> logo.sx < logo,50,5 <=> X > 0,5 (do tinh nghich bien ciia h^m so logarit CO so' 0,5) fx<4 1

83 (Bai 83, trang 130 SGK) Gidi

-Vs < X < -2

1 <x < \[5

Dieu kien xdc dinh x^ - 6x + 5 > 0 2 - x > 0

Trang 20

a) A = 92'<'K3'l + 41og,,2 _ g2Iog34 Q41og8,2 _ g41og34 g j 2 1 o g „ 2

I Vay l o g 2 3 > log34

(Bdi 88, trang 130 SGK) Gidi

I c la canh huyen, a va b 1^ h a i canh goc vuong cua tarn giac vuong nen a >

b > 0, c > 0

Ta CO a^ + b^ = c^ (Pitago)

=^ a^ = c^ - b^ = (c - b)(c + b)

I log„a^ = loga(c - b)(c + b) (do b + c > 0, c - b > 0)

=> 21ogaa = log,(c - b) + loga(c + b) => log„(c - b) + log3(c + b) = 2

+ — - — = 2 => loge+ba + logc-bB = 21ogc-ba.log,.^ba

loge-ba l o g „ b a

Trang 21

so g6c ciia tiep tuyen tai A

y'(0) = i = tga (a = OBA)

Trong tam gidc OBA, tga

-Suy ra OB = 20A =

OA 1

OB " 2

ln2 Di§n tich AOAB la:

3^.t^ - 4.3h + 3 ^ = 0

A' = {2.3^f - 3 * 3 ^ = 4 3 ^ ° - 3^' = 3^°i4 - 3 ) = 3^°

V A ^ = 3 ^ _ 2 3 ° + 3 ' _ 3 3 ' _ o 2 , _ 2 3 ' - 3 ' _ 3 ' _ _ 3

3 « ^ " 3 ^ = "^ ~ ^ ^ ¥ ^ t = 3 - ^ <=> 3^" = 3 - '3 2 x D - 2 o 2x = - 2 c=> x = - 1

t = 3 - ' o 3 ' " = 3 - ^ o 2x = - 3 <=> X = - -

Vay tap nghiem cua phuong trinh la S =•{-—; if-

Trang 22

94 (Bai 94, trang 132 SGK) Giai

t = 3 < » 3 " = 3 c ^ x = l (thoa dieu kif n)

Nghif m phucfng trinh da cho la x = 1

Vay nghiem phiicfng trinh da cho la x = 3

4" - 3" = 1 (1)

Dl thay x = 1 la nghiem phifomg trinh (vi 4^- 3' = 1)

Ta chufng minh phiiomg trinh (1) khong c6 nghiem nac khac, that vay

= 1 (1) C 5 1 -

nghich bien tren

Tren khoang (-co; 1), phuofng trinh (1) khong c6 nghiem

Vay phiicfng trinh da cho c6 nghiem x = 1

98 (Bai 96, trang 132 SGK) Giai

log^ix - y) = 5 - log2(x + y)

^) (I) \x - log4

= - 1

(1) (2) logy - log3

Dieu ki#n xac dinh x > y > 0

(1) o log2(x - y) = log22^ - log2(x + y)

o log2(x - y) + log2(x + y)log22®

o log2(x - y)(x + y) = 2*^ o x^ - y2 = 32

(2) <=> logx - log4 = -logy + log3

<r> logx + logy = log4 + log3

o logxy - logl2 C5> xy = 12

Trang 23

t < l

5 < t < 6

5 < 6^ < 6 logg5 < X < 1 ' V a y t a p n g h i e m c i i a bS't phi/ong t r i n h S = ( - « ; 0 ] u [loggS; 1) c) l o g , ( x ' - 6 x + 18) + 21og5(x - 4) < 0 (*)

5

h a y - l o g 5 ( x ^ - 6 x + 18) + log5(x - 4 ) " < 0

( x - 4 ) ^

l o g ' x^ - 6 x + 18 < 0

(3)

G i a i b a t p h i i o m g t r i n h :

(1) X " - 6 x + 18 > 0 d u n g v d i m o i x e 5

( 2 ) x - 4 > 0 <= > 4

Trang 24

Vay t$p nghi|m bat phifong t r i n h la S = (4; + « )

BAI TAP TRAC NGHIEM KHACH QUAN^

Dieu ki§n xac d i n h x - 4 > 0

logo,4(x - 4) + logo,40,4 > 0 c:> logo,40,4(x - 4) > 0

Vay x - 4 > 0 x > 4 X - 4 < 2,5

0,4(x - 4) < 1 ' Vay tap n g h i e m bat phifang t r i n h S = (4; 6,5]

101. (Bdi 101, trang 132 SGK) Gidi

'10

103. (Bdi 103, trang 133 SGK)

Chon (C)

Gidi

(0,5)log225 + Iog2(l,6) = log2252 +log2(l,6)

= log25.1,6 = log28 = log22^' = 3

= (4 + log2l5)(log2l5 - 2) - log2l5.(2 + log2l5)

= 21og2l5 + log2'l5 - 8 - 21og2l5 + log^l5 = - 8

Tap n g h i e m bat phUcfng t r i n h [ 1 ; +oo)

Tii d o t h i t a c6 neu x > 0 t h i a" > c'' > b* => a > c > b

Chon (C)

Tii do t h i t a CO a > 1, b > 1, c < 1

K h i X > 1 t h i logaX > logbX > 0 => log^a < logxb => a < b

Vay b > a > c

Trang 25

1 D i n h nghia: Cho h a m so' f xac d i n h t r e n tap K

Ham so F duqfc goi nguyen ham cua f tren K neu F'(x) = Rx) v6i moi x e K

2 T i n h chat: Gia siJf ham so F la mpt nguyen ham ciia ham so f tren K, k h i do:

a) Vdi moi hkng so C, ham so y = F(x) + C cung la mot nguyen ham f, tren K

b) NgiT^c l a i v d i mfii nguyen h^m G ciia f t r e n K t h i ton t a i mpt h k n g so C sao cho G(x) = F(x) + C vdi moi x e K

K i hieu Jf(x)dx de chi mpt nguyen ham bS't k i cua f

•' k b) fcoskxdx = + C

•' k c) fc'"'dx= + C

•' k d) fa-dx - - ^ + C ( 0 < a ? l )

Trang 26

Thii' bling each lay dao h a m cua (A), (B), (C), ro r a n g

( - X C O S X + sinx + C)' = -(cosx - xsinx) + eosx = xsinx

Diing v i - Vx 1^ mot nguyen h ^ m ciia f

§2 M O T S O P H a d N G PHAP T I M N G U Y E N HAM

1 Phtfcfng p h a p doi b i e n so Cho h ^ m so u = u(x) c6 dao ham lien tuc tren K va ham so' f(u) lien tuc sao cho |f[u(x)] xdc d i n h t r e n K

K h i do neu F la mpt nguyen h a m ciia f, nghia la |f(u)du = F(u) + C t h i Jf[u(x)].u'(x)dx = F[u(x)] + C

G h i g o n jf(u).du = F(u) + C

-2 PhiTorng p h a p lay n g u y e n ham tuTng p h a n

Neu u, V Ik hai h ^ m so c6 dao h ^ m lien tuc t r e n K t h i Judv = uv - jvdu

Trang 27

2 Dat u = X => du = dx

Do d6 J2xsinxdx = uv - vdu = -2xcosx + J2cosxdx = -2xcosx + 2sinx + C i

Vay jf(x)dx = x^sin2x - [-2xcosx + 2sin2x + C i l

= x^sm2x + 2xcos2x - 2sin2x + C c) ftx) = x.e"

Dat u = X =5> du = dx

dv = eMx => V = e"

jf (x)dx = JeMx = Jiidv = uv - Jvdu = xe" - Je'dx xe" - + C

d) fix) = x^ln(2x) Dat u - ln(2x) =o du = - d x

= ^ } ~ ^ ^ = o = ^-t^" ^ C = itan(3x + 2) + C 3-'cos'^(3x + 2) 3-'cos-u 3 3

X X

cl)f(x)=:sin^3^os-Dat u = sin • du = —cos^x

3 3

JiF(x)dx = 3j|sin^|cos|dx= 3fuMu= 3-^ + C= - W | + C

8- (Bdi 8, trang 145 SGK) Gidi

\

a) f(x) = X' 18 - - 1

Dat u = — - 1 => du =

18 Jf(x)dx = 6 j -

Trang 28

JxeM^' = xe' - e" + Cj (Bai tap 6 c))

Vay |f (x)dx = x^e" - Slx^e" - 2(xe'' - e") + C]

= x^e'' - Sx^e" + 6xe^ - 6e" + C

d) f(x) =

Dat u = V 3 x - 9 ^ du = ^

2u => dx = 2"du 3 .—du = - fue-du

D^t u = sinx => du = cosxdx

^ iF(x)dx = Jsin'' xcosxdx = Ju'' du = ^ + C = — + C d) f(x) = xcos(x')

Dat u = x^ => du = 2xdx ff(x)dx = jxcos(x^)dx = — 2x.cos(x^)dx

1 Dinh nghia: Cho ham so f lien tuc tren K va a, b la hai so' bat k i thupc K,

neu F la mpt nguyen ham cua K t h i hi|u so F(b) - F(a) diiac gpi la tich

phan cua f tiT a den b, k i hieu la f(x)dx (a, b la hai can ciia tich phan, a la can dudi, b la can tren)

• Neu a < b, ta goi J^f(x)dx la tich phSn ciia f tren doan [a; b]

• Neu F la m6t nguyen ham ciia f tren K t h i

f f(x)dx = F(x)|

a

2 Dinh It: Cho ham so y = f(x), lien tuc va kh6ng am tren doan [a; b] k h i d6

di^n tich S ciia hinh thang cong gidri han bdi do t h i h a m so y = f(x), true hoanh va hai diidng t h i n g x = a, x = b la

S = £f(x)dx

Trang 29

Tich phan nky hkng di^n tich hinh

thang ABCD gidi han hdi do thi ham

so y = 7^ + 3 , true hoknh va cac difdng y = ^ + 3

S = 1(2 + 5).6 = 21 (dvdt)

2 Vay ^ 3 dx = 21

b) xldx

Tich phan n^y hkng tong di^n tich

eac tam giae OAB va OCD gidi han

bdi do thi h^m so y = I x l , true hoanh

I Tich phan nay b^ng dien tich gidi han

I bdi do thi h^m so' y = v 9 - x ^ hay

+ y^ = 9 (niifa dudng trbn tam O, b^n kinh 3), v^ true hoknh, cae dudng thing

^ 13 (Bdi 13, trang 153 SGK) Gidi

a) [nx)dx la di^n tich hinh thang cong gidi han bdi d6 thi h^m so y = fix),

true hoanh va cae difdng t h i n g x = a, x = b nen:

j['f (x)dx > 0

b) Dat h(x) = fix) - g(x) khi do h(x) > 0 (do fix) > g(x))

TheoeSua) jfh(x)dx > 0=:> £[f(x) - g(x)]dx > 0 ^ j['f(x)dx - £g(x)dx > 0 Vay j['f(x)dx > j['g(x)dx

a) Quang dudng vat di chuyen trong khoang thd; gian tiT t = 0 den t = 3n (don vi giay) la

Trang 30

b) Goi to Ik thcfi d i ^ m vat dtrng l a i

Turc la V(to) = 0 160 - lOto = 0 => to = 16

Quang dirdng vat di chuyen trong khoang t h d i gian tii t h d i diem t = 0 den

a) Goi V{t) la van toe ciia vien dan

1)

2u

" 5 "1 r _ 1 2 _ 5 ~ 4 8

Trang 31

Dat u = x + l = > d u = dx

dv - e^dx => V = e"

| ' ( X 4 DeMx = (x ^ De" jVdx = (x + l)e'

= [(1 + De' - (0 + l)e"l - Ic' e"l = 2e - 1 o + 1 c) £e''cosxdx

Dat u = cosx du = - sinxdx

D o d o f e " s i n x d x = (e"sinx) fe"cosxdx = (e"sinx) C

Suy ra 2C = (e^cosx)' + ( e s i n x ) " = (e'cosrr - e'cosO) +

e' + 1 Vay J^e^ cosxdx - -

i3at u = 5 - 4co:ji du = 4siui(lt u(0) = 5 - 4cos0 = 1; u(;t) - 5 - 4cosTt = 5 + 4 = 9 , 5(5- 4cost)-sintdt = f u ' d u

Trang 32

P , r C O S X ,

•e) h~.—dx

j l + sinx Dat u = 1 + sinx =:> du = cosxdx u(Oj = 1 + sinO = 1

Trang 33

~ 3 ' '9 ^ 9 ~ ' ' 9 '

§5 CfNG DUNG CUA TICH PHAN DE TINH

DIEN TICH HlNH PHANG

x = h(y) (g, h l a hai h a m lien tuc tren [c; d] va hai dudng t h i n g y = c, y = d la

S = (• g(y) - h(y) dy

Tiêu đề	Các Phương Pháp Giải Bài Tập Giải Tích 12 Nâng Cao: Phần 2
Thể loại	Tài liệu

Định dạng
Số trang	66
Dung lượng	12,51 MB