Chuyên đề bồi dưỡng học sinh giỏi giá trị lớn nhất, giá trị nhỏ nhất phan huy khải (phần 7)
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Chuy6n dg BDHSG To^n gia tr| lan nha't
g\& tr| nhd nhSt - Phan Huy Kh5i
,atBB| = x ; B , C 3 =
S + S,
De lhay A 3 B B , M , CCjMA,,
1+
•iB = A 3 B 1 ; A M
S2 = ZSj + — ,-
(3)
cung chinh la
AA3B1C2.
MB1C3
la ^
cac tam giac deu vdi canh lan liTdt la x, y, z,
cua chung thi:
(5)
>—
S
(6)
3
O
S1+S2+S3 ^S;
(do a =
(difa vao nhan xet hicn nhien sau: Ne'u a + b + c = 1, thi a^ +b^
\di )
S
^
)
(z^ x ^ + y ^ + z ^
V ^ y
(x + y + z)^
x^+v^+z^
^
S.
(1)
s/
+S^\MB,C2 +SAMA,B,
ro
up
SAA,B|C| ^ S A ^ A J C J
om
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ok
bo
ce
fa
w.
ww
la dien tich tam giac ay.
Tim gia trj Idn nha't ciaa
=irs-(s,+s2+s3)
(2)
r
Hien nhien ta c6: x^ + y^ + z^ > - ( x + y + z ) ^ nen tir (1) suy ra:
>is
(3)
. T C f ( 2 ) ( 3 ) s u y r a : ^ < | . (4)
Da'u b^ng trong (4) xay ra o c6 da'u bang trong (3)
•
o x = y = z o M l a tam cua A deu ABC.
Hudng ddn gidi
Goi S la dien tich tam giac dcu ABC canh a, thi S =
( S M A , A A , +SMC2CC3 +SMB2BB1
\
a
Da'u b^ng trong (3) xay ra x = y = z = - .
Bai 2. Cho tarn giac dcu ABC canh a. M la diem tily y n^m ben trong AABC.
Xet tam giac c6 3 canh la MA, MB, MC. Goi ^
2
s'l +S2 + S3
CO dau b^ng trong (6) o O la trong tarn AABC.
B2C2, A3C3
^ fy^
+ y + z)
(7)
is
O la trong tarn AABC.
S
/g
O la trong tarn AABC.
--S.
3
Qua M ve ba doan thang A|B,,
CA (xcm hinh ve)
X
Tac6:<^=
S
S
S
1
Da'u bang trong (6) xay ra <x> ^S = S^ - ^S = 3- o
Tilf (6) (7) suy ra: P > 2S + 3
7S
S
•
.
.
Tirddtacd: S,+S2+S3=
>3'
Dau bang xay ra <=> a = b = c = ^ )
Vict lai (5) du'di dang: P = 2S +
^S2 ^S3
Ta
+ ^ = l,ncnlac6:
7S
Vay min P = — o
„
nen neu goi S,, Sj, S3 tiTdng ufng la dien tich
P = S , + S 2 + S 3 = 2 ( S , + S 2 + S 3 ) + l(sf+S^+S^)
P= — o
= C2B,
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01
/
(4)
Tir (2) (3) (4) suy ra:
O
= A 3 C 2 ; CM
Do cac tam giac M B . A , . M A . C . va
S3 = 2S3 +
O
;
Vay lam giac co ba canh M A , M B , MC
s
i
. ,
tfdoco:
(2)
s
Lap luan tiTdng tiT, ta c6:
AA3MC2
cac hinh thang can.
c2
\,
Sp
•S + S, =s 1 + — - + -^ = S + 2 S , + ^ = > S , - 2 S , + ^
S
s s^
Vi: ^ + ^
=z
y;C3C
(1)
DoS =
a'V3
4
minS = ^ ^
12
o
M la tam cua AABC.
tifdng iJng song song vdi AB, BC.
357
j^..
^ ^ . . w ^
i w u M t j i d I I I luii MiiciL v d y i d u | iiMu i i r i a i - r n a r i nuy
^nai
Bai 3. Cho lam giac ABC c6 AB = c,• AC" = b va A=n . Xet lap hdp cac du,
"^«ni.
thang A qua A va khong U-ung vc'ti hai canh AB, AC. Goi P la tich khoang each if
B va C tdiyi^A. Tim
nhat ciia
\, gia tri IdnHii(fng
danP.gidi
DiTcJng thang A qua A chia thiinh hai loai:
Nhom I:
Dirftng A cii doan BC. Dal BAH = 3 .
Kc B H l A, C K I A .
Khido:
'' '
P = BH.CK = bcsinpsin(a - p)
';
Tir(l)tac6:
P <-bc{l-co.sa) = b c s i n ^ - .
2
,
2
Dau bilng irong (2) xay ra
bcsin^ ^ , neu A la goc iD (90" < a < 180")
^bc, ne'u A la goc vuong ( a = 90")
NhiT vay ne'u A la goc nhon, ihi maxP = bccos^ y khi A la phan giac ngoai
cua goc A va ne'u A la goc tu, thi maxP = bcsin^ y, khi A la phan giac trong
cua g6c A va ne'u A 1^ goc vuong thi maxP = - b e khi A la phan giac trong
I
-•••^'A-
(2)
'
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ro
up
cua gc)c
A la dai li^ctng P dcU gia tri \&n nhat va gia tri do bSng: P| = be sin^ y.
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w.
fa
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Nlwm 2:
Difilng A qua A va citt doan B C , & day C la diem dol xuTng cua C qua A. Kc
BH va CK cimg vuong goc A.
Kc C'K" 1 A =^ CK = C'K' =:> P = BH.CK = BH.CK'
Ap dung li luan phan 1 vao lam giac
ABC", la lhay tich BH.CK'(cung la tich
BH.CK) li'm nha'l khi A la phan giac ,
trong ciia goc BAC' (luTc A la phiin giac
ngoai ciia goc A) va gia trj Idn nhat do
= bccos — .
2
2
Theo nguycn li phan ra, thi:
maxP = max{Pi; P:} = max|bcsin^y;
358
ubccos —
2 «
s/
Vay trong cac di/ctng thdng thuoc nhom 1, thi du'clng phan giac trong
hang p. = be sin
hoSc la phan giac ngoai cua goc A.
„
Nhqn xet: Trong bSi tren da stf dung nguyen li phan ra cua bai loan tim gia tri
Idn nhat va nho nhat, di nhien ke't help vdi cac lap luan ve hinh hoc phang c6
iJng dung lifdng giac!
Bai 4. (Bai toanToricelli)
Cho tarn giac ABC c6 max {A, B, C} < 120" (tiJc la mpi goc cija tarn giac
deube hdn 120").
Hay tim trong tam giac ABC mot diem M sao cho MA + MB + MC dat gia tri
be nhat.
HUdng ddn gidi
Cdch 1: (Lc(i giai cua Toricelli)
, ,
Durng ba tam giac dcu ABC,, BCA,,
ACBi ra phia ngoai tam giac ABC. De
tha'y ba dUcJng iron ngoai ttep cua ba
tam giac deu ay dong quy tai diem T,
va T la 3 diem nhin 3 canh cua tam
giac BAC dUdi ba goc b^ng nhau va
hlng 120". Do CJTA = ABCJ = 60",
nen Afc = 120"=^ C,, T, C th^ng
hang. Li luan tiTdng tir c6 B, T, B,
thing hang va A, T, A| cung thang
hang. La'y diem I tren TB|, sao cho
TA = TI i=> AAB,I = AACT (c.g.c)
359
Ta
<=> cos(a - 2P) = 1 <=> P = Y o A lii phan giac Irong cua goc A.
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2 . cos(a-2P)-cosa
bccos^ —, ne'u A la goc nhon (o < a < 90")
Civ
=> B|I = C T => T B + T A + T C = B T + T I + I B , = BB|
(1)
V a y M ( i la d i e m nhin ba canh tam giac durdi ba goc bling nhau = 120", tuTc Mo
L a y d i e m M baft k i thuoc m i e n tam gidc A B C . Ta se chiJng m i n h r k n g :
M A + M B + M C > T A + TB + TC
, „:
!NMM M IV r i W I I Kh.il"! Viet_
la d i e m T o r i c e l l i phai t i m .
(2)
C^ch3:
That vay diCng lam giac A M E sao cho E cf nufa m a t phang hci A M khong chuTa B.
^
^
:r
m-
y
Ro rang A A M C = A A R B , => M C = E B , => M A + M B + M C > T A + T B + TC
R6 rang ta c6 b d de h i c n nhien sau day ( v i the chung t o i bo qua chtfng minh).
:=> (2) diing. D a u bang xay ra
N c u QNP la tam giac deu, I h i v d i m o i vj t r i cua d i e m M n a m trong m i e n tam
M = T
giac QNP, ta l u o n c6 long khoang each tif M t d i ba canh cua tam giac Q N P
"Toricelli".
la hang so' (c6 the tinh di/dc ngay hang so nay bSng chinh c h i c u cao cua tam
J^'i . ,
2: L a y M la d i e m tijy y trong m i e n lam giac A B C .
^ giac d c u QNP).
R A p dung bo de ay ta g i a i bai toin T o r i c e l l i nhifsau:
Thirc h i c n phep quay R" ( B , 6 0 " )
Khido:
R-(B,60")
C-yA,
B c u a tam giac dvtdi ba goc bang nhau, tuTc la: A T B = B T C = A T C = 120".
*
• Q u a A, B, C dirng ba
M ^ M '
Hdi/dng
T h e n tinh cha'l cua phep quay suy ra. M B M ' la tam giac d c u => M B = M M ' .
Ngoairado:
*
^tai
MC = M'A,.
(1)
up
jjnj^
s/
I
V a y M A + M B + MC = M A + M M ' + M'A,.
'' 7jy (*)
vuong
TB,
TC.
goc
Ba
Hdirdng nay c^t nhau
diTcfng gap khiic, ta c6:
(2)
Q, N , P. De thay
QNP la tam gi^c deu.
L a y M la d i e m tiiy y
trong A A B C . G o i h , ,
h2, hi la khoang each
N h i r vay tir (1) (2) ta da chi^ng m i n h difcfc rhng v d i m o i vj t r i d i e m M thuoc
tijr M t d i ba canh NP,
m i e n A A B C , ta luon c6: M A + M B + M C < A A , .
ok
(3)
A, M , A2 thang hang.
bo
Da'u bang xay ra o
cua
phep quay ta c6
goc
tao b d i M,)C va
bkng 6 0 "
=j^MoMoC = 6 0 "
B M o C = 120".
M„A|
fa
chat
w.
tinh
A M „ B = I20".
ww
Theo
ce
Gia sur Mo la vj t r i cua M ma A, M„, A, thang hang
Do BMoMo = 60"
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/g
M A + M M ' + M'A,
ro
vfiy
•TA,
R-(B,60")
MC^M'A,,
Do A A B C CO m a x ( A , B , C) < 120", nen ton tai duy nhat d i e m T nhln ba canh
•
>.'1'VV,,,.,
Ta
Cdch
• '
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V a y d i e m M can t i m chinh la d i e m T n 6 i tren. D i e m 66 thi^dng g o i 1^ diem
PQ, Q N cua AQNP.
m.
M
Ro rang ta c6:
M A + M B + M C > h| + h2 + h j
H
•
•
T h e o bo de neu tren (ap dung vao tam giac deu QNP), ta c6:
h , + h2 + h j = T A + T B + T C
Tir do suy ra: M A + M B + M C > T A + T B + T C
(*)
•
Da'u b^ng xay ra trong (*)<=> M s T . Do chinh la dpcm.
m:(ich4:
Sis dung b6 &G h i e n nhien sau day (ChuTng m i n h ra't ddn gian va x i n danh cho
cac ban doc)
B6 de: Gia si^ A M B = B M C = C M A = 120" va M A = M B = M C t h i :
M A + M B + M C = 0
- :
Do gia thie't m a x ( A . B , C) < 120", nen t6n tai duy nhaft d i l m T trong tam giac
sao cho A T B = B T C = C T A = 120" .
160
361
Cty TIMHH MTV DVVH Khang Vi^t
ChuySn de BDHSG Toan g i i t r i Ion nhat va g i i Iri nh6 nha't - Phan Huy KhAi
'
.
,
TA TB TC
An dung bo de suy ra:
1
1
=0
' • ^
Gia sur M e BC.
Goi H, I , J ti/(tng i?ng la hinh chic'u cua M xuong AB, BC, CA.
(*)
.
T A TB
TBTCTC
TA
Do do dai cua ba vccUl
, — , — dcu bang 1
TA TB TC
Lay M 1^ diem bat ky trong mien tarn giac ABC, ta c6:
Khi do ta eo: M I = x; MJ = y; M H = k.
Gia siJ K e AC sao cho AB = CK (do a > b > c). *
G o i L = M K n BC
Su" dung cac bat dang thtfc hien nhien sau:
1. a|+a2+... + a„
.
<
.
+ «2 + ...+
Delhay: VF(2) =
(2)'
MC.TC
TC
>
MA.TA
TA
+
MB.TB
TB
MC.TCi,
+•
TC
(MT +TA)TA
(Mf +TB)TB
TA
TB
= MT
TA
TB TC
TA
TB TC
r^ - o a b c a a 2a
Do vay S = — + — + - = — + — = — .
(2)
X
TC
MeBC
+TA+TB+TC.
. T i i f ( * ) v a ( 3 ) s u y r a : VF(2) = TA + TB + TC.
z
X
(3)
'i
(4)
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trj Idn nha't, nho nha't trong cac biii toan khac noi chung (trong hinh hoc
bo
phang noi rieng) cflng c6 rat nhicu each gitii khac nhau. Bai toan tren la mot
ce
vi du dien hinh minh chufng cho tinh da dang ciia cac phiTdng phap tim gia tri
fa
w.
BC, CA, AB ti/dng uTng. Gia suT M la mot diem di dpng tren diTctng tron ngoai
tie'p tam giac ABC. Goi x, y, z Ian liTdt la khoang each tif M den cac canh
max M l M,)Io
MeBC
Ta c6: M,,!,, = BI„tan MoBI,, = - t a n ^ Tiif do suy ra: min S = —
MeBC
,
2a
A
=:4C0t—.
a
tan A
2
2
Cung gio'ng nhu" trong cac bai loan dai so', giai tich, v6'\c bai toan tim giii
B a i 5. Cho tam giac ABC vdi a > b > e, d day qua a, b, c ki hicu do dai cac canh
(3)
2
Lap luan ttfdng tu", co:
i
B
C
' min S = 4cot— va min S - 4 c o t — .
MeAC
2
MeAB
(4)
2
Theo nguyen l i phan rii, ta co:
min S = min min S, min S, min S
MeBC
M6AC
MeAB
(5)
Tif (3) (4) (5) suy ra:
BC.CA, AB.
A
B
C
min S = min 4col—,4cot—,4cot—
2
2
2
Tim gia tri be nha't cua dai ii/dng S = — + — + - .
X
y z
Do a > b > c =^ 180" > A > B > C > 0
HUdngddngidi
2a
2a
MeBC
om
Nfuln xet: Da'u bang xay ra <=> M s T. Do chinh la dpcm.
ww
X
d day Mo la Irung diem ciia BC
ro
Ttr (1) (2) (3) (4) suy ra: M A + MB + MC > TA + TB + TC.
Idn nha't va nho nha't ciia mot dai lufdng cho tri/dc.
X
^2a^
Nhuf vay min S = min
{MT + TC)TC
^
y
Ta
TB
+
p a cung CO ACLM - AABM => - = —
z
X
s/
MB.TB
up
TA
•+
(1)
Tir(l)(2)=^^ +^ =- ^ l i l ^ =i .
y z
X
X
2. a.p <|a|. p ; ta c6;
MA.TA
De thay A B L M - AACM ^ - = — .
y
X
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K^ro y^Ar^ MA.TA MB.TB MC.TC
M A + MB + MC +
+
TC
TA
TB
„
B
9 0 0 >A^ >B- ^ C
>|>0^cot-
(6)
(7)
Goi (~(f) la dircJng tron ngoai tie'p AABC.
Khid6(^)= BCUCAUAB.
62
(1)
. Vav lir (6) (7) di den: min S = 4cot— 44. M la trung diem cua BC
Ctiuygn de BDHSG Tcrin gia
tf| I6n nha't
g\i
trj nh6 nliil
Cty TNHH MTV DVVH Khang Vi?t
Phan Huy KhJi
Bai 6. Cho hlnh tron ban kinh r. Xet ta't ca cac tvt giac ABCD ngoai tiep dtfcj^
tron. Tim gia trj nho nhat cua dai luTOng P = AB + CD.
,
Hitdng ddn gidi
Goi M la tam dtfSng tron npi tiep ti? giac => M
giao diem cua cac diTdng phan gi^c trong cua cac
goc A, B, C, D cua ti? giac.
Ve di/dng tron ngoai tiep tam giac ABM va gpi
ABN la tam giac can npi tiep c6 dinh la N sao
cho ANB = AMB.
Gpi h la khoang each tif N xuong AB, con h, la
khoang each tir M xuong AB. Khi do ta c6:
h > hi va hi = r.
T
Ta CO' AABD = 2htan ANB = 2htan AMB
(2)
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Dau h\ng trong (2) xay ra o M each deu C va D.
AMB htan CMD
Tir (1) (2) c6: P = AB + CD > 2r tan
(3)
2
2
Dau bkng trong (3) xay ra o dong thcfi c6 dau bkng trong (1) (2).
Taco: AMB-f CMD = 1 8 0 " - A l l + 180" - C + D = 180^'
tan AMB = cot CMD
tan
1 - . Vi theo bat dang thiJc Cosi, suy ra:
CMD
1 H-tan
^ AMB
CMD
tan
1-tan
tan CMD
364
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Ti/Png ixi ta c6: CD > 2r t a n ^ ^ .
Ta
(1)
s/
AB>2rtan AMB
Dau bang trong (1) xay ra o h = hi
o M each deu A va B.
AMB
Dau bang trong (4) xay ra <=> - ^ — = 4 3 .
Tir (3) (4) di den: P > 4r
(5)
Dau bang trong (5) xay ra <=> dong thcti c6 dau bang trong (3), (4)
M each deu A, B; M each deu C, D va AMB = CMD = 90"
o ABCD la hinh vuong ngoai tiep dufdng tion ban kinh r da cho.
Tom lai min P = 4r. Gia tri nho nhat dat du-dc khi v^ ehi khi ABCD la hinh
vuong ngoai tiep diTdng tron ban kinh r cho trifdc.
Bai 7. Xet tat ea cae ti? giac ABCD ehi c6 duy nhat mot canh Idn hdn 1. Tim gia
tri Idn nhat cua S, d day S la dien tich tu* giac ABCD.
Hiidng ddn giai
^
Gia siJ AD > 1. Khi do ta c6: AB < 1, BC < 1, CD < 1.
Dat AC = X va gpi M la trung diem cua AC
Ta eo: A B ' + B C ' = 2BM^ + AC^
>2.
(4)
.2
=^2BM^+ — < 2 = > B M < - V 4 - x ^
2 ~
2
(1)
Do AC < AB + BC < 2 0
2
Dau b^ng trong (1) xay ra o AB = BC = 1. Ke BH 1 AC. Ta c6:
(2)
-1A C . B H < -1x . B M .
< X <
SARr=
2""
"~2
Dau b^ng trong (2) xay ra o H s M o AB = BC
Tir (1) (2) c6: SABC < - x V 4 - x^ .
(3)
4
Dau bang trong (3) xay ra o dong thcfi c6 dau bang trong (I) (2)
o A B = BC=I.
(4)
Tac6:SACD= -CA.CDsinACD<-x.
Dau b^ng trong (4) xay ra o CD = 1 va AC 1 CD.
Tir (3) (4) ta c6: S = SASBCD = SABC + SACD < ixVTI x^+ix,
2
2x + x V 4 - x ^
(5)
hay S <
Dau b^ng trong (5) xay ra o dong thdi c6 dau b^ng trong (3), (4)
o AB = BC = CD = 1 va ACD = 90".
365
Cty TNIiH MTV DV'VH Khang Vigt
ChuySn gg BDHSG Toan g\i trj Idn nhS't va g\i tr| nh6 nha't - Phan Huy KhSi
_
,
Ta co:
2x + x V 4 - x ^
4
=
H i c n nhicn ta c6: A B < CE = aV2 .
x/, , ,
r
f\
+ 1 + V4 - x"^ j .
4
A p dung ba't dang thiJc Bunhiascopski, ta c6:
( l + l + V 4 ^ ) < 3 ( 2 + 4 - x ^ ) hay 2 + V 4 ^ < 7 3 . 7 6 ^
+ x\/4-x^
fZ
73
T i r ( 6 ) (7) s u y r a :
4
4
Da'u bang trong (8) xay ra <=> c6 dau bang trong (7)
Da'u bang trong (2) xay ra o
CD=:1
N/3
"(2)'
A B la diTcJng chco cua hinh vuong.
M
Huding ddn gidi
ro
/g
cac
bo
hinh vuong sao cho A va B nam tren cac canh cua hinh vuong ay. G o i / la tong
ce
cac khoang each tiT A den cac dinh cua hinh vuong. T i m gii tri nho nha't ciia /.
fa
Gia sijf C D E F la m o t hinh vuong tuy y canh a va gia suT A e C D .
Dat A B M = (v, N B C = 3.
D o do canh \dn nha't trong cac doan A C , A D , A E ,
,;l
Do M B N = 45" =i> a + 3 = 45"
tan((v + 3) = 1
:A *
tan (x + t a n 3
,
x+ y
=
1=^
1 - x y^ = 1
l - t a n a + tan3
>x
Ta
+ y = 1 - xy.
c6:
_ ^
X
y
~
2
2
D a t t = xy.
K h i do
TacotCfd):
(1)
••.yy.A,.
S = SBMN = SABCB -
ww
w.
Dat A M = X, C N = y, nhiT vay x, y e [0; 1 ].
om
ok
B a i 8. T r e n mSt ph^ng cho hai d i e m A , B v6i A B = d. X e t tap hdp tS't ca
AE>a, AF>a.
'
gia tri Idn nhat va nho nha't ciia S.
A B C D la nii-a luc giac deu canh 1.
Taco: AC + A D = a
72) = d ( l + 72).
A B va C D sao cho M B N -=45*'. Gia sijf S la di?n tich tarn giac B M N . T i m
(10)
HUdng ddn gidi
+
i 9. Cho hinh vuong A B C D canh bang 1. D i e m M va N Ian liTdt di dpng tren
.c
o
+ AB = A B ( I
72
Gici trj nho nha't dat diTdc khi A B la diTdng cheo cua hinh vuong.
A B C D la nufa luc giac deu canh 1.
T 6 m l a i max S = —
+ ^
T o m h i i : m i n / = ( l + 72)d.
A C I C D
o
>4^
AF
Nhir vay ta d i den: / > ( l + 7 2 ) d .
s/
AC =
7^-
72
up
-o
72
AB
Ta
AB = BC =
Ket hcfp l a i ta c6: S =
/ = AC + A D + A E +
.
x ' = 6 - x^ <=> x = N/3 .
/3
I
AB
do
TiJf (1) ta c6: A F > A B , nen ket hdp l a i suy ra
(9)
Tir do ket hcJp v d i (5) co: S < — . 7 3 = 4
4
4i
(7)
=lox-V3.
Lai iheo ba't dang thiJc Cosi, ta c6;
cVri-x^ <
= 3.
Da'u bang trong (9) xay ra o
,
>a>
AE
(8)
T
o74-x^
72
iL
ie
uO
nT
hi
Da
iH
oc
01
/
T v / - / ; \ / ' 7 \x
AB
TiS do suy ra: A C + A D = a > -
(6)
2
ta c6:
[x + y
SACM - SBCN -
(l-x)(l-y) ^
S
=
1-t
A x M
SMDN
1-xy
2
(2)
1-t
[xy = t
A F la A E hoSc A F (thi du la AF)
nen theo dinh l i V i e t x, y la hai nghi^m thoa man dieu k i e n 0 < x, < X2 < 1
Ro rang trong cac d i e m tren bien cua hinh vuong
cua phiTdng trinh:
thi d i e m each xa A nha't phai la mot trong cac
dinh.
x2_(i _t)x + t = 0
(3)
De (3) CO nghiem nhiT vay, ta can c6:
Tir do suy ra:
AB
(1)
367
ChuySn dg BDHSG Jo&n gia tr| Idn nha^t vi g\i tr| nh6 nhS't - Phan
fA>0
-6t + l>()
(X,-1)(X2-1)>0
<=>
X| +X2
<2
(*)
>0
X| +X2
>0.
l-t<2
a - p = 45"
t
<=>0
t>0
. ,
a - p = -45"
(3)
,
„
<=>
a = 45";P
= 0
a = 0;p = 45"
42
:
2
"M = D ; N = C
M = A; N = D
:S M
Nhdn xet: Mot Ian ni^a ta thafy du'dc tinh da dang cua cic phtfcJng phap d6ng de
giai bai toan lim gia tri Idn nha't va nho nha't trong cac linh vifc khac nhau:
Tir(2) (3) suyra: V 2 - 1 < S < (4)
i '
2
X = 0; y = 1
M = A;N = D
S = - o t = 0<=>
2
M = D;NsC
x = l;y = 0
dai so, giai tich, so" hoc, hinh hoc, liTdng giac...
ai 10. Cho nufa di^cfng Iron bdn kinh R, di/dng kinh AOB. C la mot diem tily y
up
s/
Ta
tren nufa di/dng tron khong trung vdi A va B. Trong hai hinh quat BOC va
ro
o M , N tiTdng tfng la chan diTdng phan giac cua ABD, DEC .
om
/g
1
M = A,N = D
Vsiy maxS = - <=>
.c
minS= V ^ - I o M = M„;NsNo,
bo
ok
cl day BM,, va BN,, Ian liTdt la phan giac cua cdc goc ABD va DBC .
ce
Nhgn xet: Ta c6 each giai khac bai toan tren nhiTsau: (diTa vao li/cfng giac)
1
2
w.
42
ww
\
2 cosa cosp 2
fa
1
Ta c6: S = SBMN = - B M . BNsinMBN
2
cos(a + 3 ) - c o s ( a - P )
4i
>/^ + 2cos(a-(3)
(5)
(do a + p = 45")
V i a, p e [0; 45"], nen - 4 5 " < a - p < 45"
V2
=> — < cos(a - p) < I
368
(7>
Ta thu lai ket qua tren!
l-t>0
1 1
«cos(a-p) =
iL
ie
uO
nT
hi
Da
iH
oc
01
/
S=V2-1
t^ - 6 l + l > 0
(*)o
'
Lai c6: S = - o cos(a - p ) = l < = > a = P = 22"30'
2
'
Ap dung dinh l i Viet vdi (3) ta c6: x, + X2 = I - t; x,X2 = t nen
t-(l-t) + l>0
242 2
2 + V2
hay V 2 - 1 < S < ^ .
',
X|X2>0
>0
X, +X2
Tir (5) (6) suy ra:
X | X 2 - ( X | + X 2 ) + 1>0
<2
X| +X2
X|X2
i.ii
ihry
^' ^
AOC ve hai di/dng tron npi tiep. Gpi M v^ N 1^ hai tiep diem cua hai diTdng
tron ay vdi diTcJng kinh AB cua nuTa diTcJng tron da cho. DSt 1 = M N . Tim gia
tri nho nha't cua 1.
• >
HUdngddngiai
Gpi Oi, O21^ tarn ciaa hai di/cJng tron.
Dat CON = 2a (nhir vay 0 < a < 90")
D 3 t M 0 , = R; NO2 = R2.
De thay O ^
= ^CON - a ,
O.OM = - C O M = 90" -
a.
2
A
Ta CO / = M N = O M + ON :p R,cot(90" - a) + R2Cota
= Rjtana + R2Cota
Trong tarn gidc vuong 0 | M 0 , ta c6 R) = OO|Sin(90" - a) = (R - Ri)cosa
=> R i ( l + cosa) = Rcosa =:> R| = Rcosa
(2)
1 + cosa
Hoan to^n ti/dng tiT, ta c6 R2 =
(6)
(1)
T i r ( l ) ( 2 ) (3) suy r a / =
Rsina
(3)
1 + sina
Rcosa sina
Rsina cosa
1 + cosa cosa
1 + sina sina
369
l + cosa
•H
Rcosa
s i n a + coscx + l
= K1 + sina
(1 + s i n a ) ( l + c o s a )
a
2 cos - sin 2
= R
2cos2^
. a
sin
2
Cty TNHH MTV DVVH Khang Vigt
Dau bSng trong (2) xay ra <=> A = B = C
a
+ COS ~
2
+COS
2R
-.
s i n a + cosa + 1
L a i Iheo ba't dang thuTc Cosi, thi
R
A
B
B
C , C,
A ^ . J
2A , 2B. 2C
1 = t a n — t a n — + tan—tan — + tan —tan — > 3 : V t a n — t a n —tan —
a ( . a
cos
sin + cos
2 I
2
2
a
2
2 2
(3)
<^a = 4 5 "
2
la ba goc cua m o t tarn giac A B C .
ro
s/
up
E. Diem quarngtso bai todn tim gid tri Idn nhd't, nho nhd't trong
luang gidc
Chu y rhng P > 0 vdi m o i tarn giac A B C . T i f do ta c6 maxP = VmaxP^
, ^2
Taco
2A
B
C I ,
. 2A
P = cos —cos—cos—= — 1 - sin —
2
2
2
2)
/g
B a i 1. Clio tarn giac A B C . T i m gia trj nho nhat cua d a i liTdng sau:
om
\ 2 A V . AA
1-sin —
2
ok
.c
P = tan^ A + tan^ ^ + tan^ ^ - tan^ ^ t a n ^ ^ t a n ^ ^
2
2
2
2
2
2
bo
HuHng ddn giai
ce
tan^ — + tan^ — > 2 t a n — t a n —
2
2
2
2
. 2B
2C ,
B
C
tan^ — + t a n ' ' — > 2 t a n — t a n —
2
2
2
Docos^—^<1,
2
sin — + cos
2
nentir(2)c6
Da'u bang trong (3) xsiy ra o cos
w.
ww
tan ^
2
.2A
C
A
+tan^ — > 2 t a n I —tan
—.
—
2
2
2
-2 A .
2B
B
B
tan ^ + tan^ ^ + tan'' ^ > tan—tan — + tan — t a n — + tan — t a n — .
2 , 2
2
2 2
2 2
2 2
Trong m o i tarn giac A B C , thi V P ( 1 ) = 1 , nhiT vay ta c6 :
1.
(2)
Ta
B-C
2
, + sin
• —
Al
2 j
1-sin^-^
2
2
CO —
1^
1
Tijr do theo ba't dang thuTc Cosi, c6
^
1-sin^A
2)
A;
1-sin^A^
2)
,
1
1
. A)
+ sin —
2)
+ sin —
2
Da'u b^ng trong ( 4 ) x a y ra o
cos-
1
B +C
B-C^
• + cos -
B-C^l
2
P^<-|l-sin^2 1 2
fa
Thco bat dang ihu-c Cosi, ta c6
Hitfing ddn giai
Ta
V a y min/ = 2 R ( N/2 - 1), k h i C la di em chinh giffa cua A B .
Tir do suy ra:
B
C
cos —cos— d day A , B, C
2
2
B a i 2 . T i m gia trj Kin nha'l cua bicu thuTc P = '^"'^
<?>C\A triing d i e m ciia A B
2
2
26
Nhi/vay minP = — o A B C la tarn giac deu.
/ = 2 R ( N ^ - 1 )
2
2
o A = B = C.
N/2+1
2
V
Da'u b^ng trong ( 4 ) xay ra o dong thcJi c6 da'u bang trong (2), (3)
• = 2R(N/2-1)
tan^ ~ + tan^ — + tan^ — >
2
26
Da'u bang trong (3) x a y ra o A = B = C. T i f (2) (3) c6 P > — .
=> 0 < sina + cosa < yfl ,\\e tiT (3) c6 :
*
2
2A
2B
2C
1
tan'' — tan —tan — < —
2
2
2 27
Do sina + cosa = yfl cos(a — 4 5 " ) , ma 0 < a < 90".
2R
2 2
iL
ie
uO
nT
hi
Da
iH
oc
01
/
Rsina
= 1
,
1
. A^
+ sin —
o B =C
A\ ,
• A
2-2sin1 + sin —
2 A
2
. • A'
2-2sn I
2>
• A—
1 + sin
I
(.
• A ^ (,
• A^
+ I 1 + sin 2 ; + I 1 + sin 2)
n3
4
,116
'27
A
A
A 1
2 - 2 s i n — = 1 + s i n — <=> s i u y = - .
(4)
Cty TNHH MIV DVVH Khang Vijt
Chuyen dg BDHSG Join gii trj I6n nha't vh gia tri nhd nhat - Phan Huy KhAi
A
B
C
Bai 4. T i m gia tri nho nhat cua dai lifdng S = tan'' — + tan^ — + tan'' —, d day A,
27
B, C la ba goc ciaa mot tarn giac.
P^=-^<:>B = C v a s i n ^ = i
27
2 3
V a y tiTCl) ta c6 m a x P =
16
27
I
<=> A B C la tarn giac can dinh A v d i A = 2arcsin ^ .
\,
'ii
B a i 3. X e t cac tarn giac A B C thoa man he thtfc tanA + tanC = 2tanB.
gidi
I
cosB = 2cosAcosC (do sinB = sin(A + C)^Q)
=>
=>
cosB = cos(A + C) + cos(A - C)
2C0SB = C O S ( A - C).
,
(1)
s/
Tur (3) suy ra S > 0, nen m a x S = V m a x S ^ .
.
.
CO
2
+ z ' > (x^ + y^ +
2
ce
(X
.»1,
(2)
7})'.
2
+ y + z)
1
=^
= -.
(3)
(4)
ViU.
= B = C.
/g
o A
9
T i r ( l ) ( 4 ) CO
S =i o A =B =C
9
1
V a y minS = ^
fa
;
D a u bang trong (4) xay ra o dong thcfi c6 dau b i n g trong (2) (3)
la tarn giac deu.
w.
B a i 5. X c t cac tarn giac A B C v d i A la goc Idn nha't.
ww
T i m gia t r i Idn nhat cua dai lifdng S = sin2B + sin2C + sin A
Hiidng
•
r
ddn gidi
2
]_
Ta CO S = 2sin(B + C)cos(B - C) + — —
sin A
= 2sinAcos(B - C) + sin A
=^c:>2-2cosB = l +2cosBocosB = - .
8
4
c o s ( A - C ) = 2cosB
max S =
- r - .
TCf do suy ra x ' + y^ + z^ > ^ .
ok
bo
^?::ii^^i?Ki±l£^
2
x-" + y + z >
(4)
V i the theo bat dang thiJc Co si di den
CO
om
.c
(3)
2
Vx, ^ y .
+ y + z)(x' + y ' + z') > (x^ + y^ + v?)^
ro
(2)
Tir(l)(2)suyraS=2sin|^i±^.
TiT do ta
up
A +C
A - C - . B l + cos(A-C)
= 2cos
cos
= 2 sin—J
2
2
2\
8
CO ( X
Lai
L a i CO S = cosA + cosC
n2
ta
x^ +
,.
(1)
Da'u b^ng trong (1) xay ra <:5> A = B = C.
A
B
B
C
C
A
D a t X = t a n y t a n - j , y = tan — t a u y , z = t a n y t a n y thi x + y + z = 1
..^^
=>
• ( 2 - 2 c o s B ) + (l + 2cosB)
ddn gidi
Apdungba'tdangthuTcBunhiacopskichohaiday
sin(A + C)
2sinB
—
=
cos A cos C
cosB
Taco S ^ = 4 s i n 2 - i i ^ ^ =
2
2
'
Ta
. A ,.
\ r.
tanA + tanC = 2tanB
'
(chiiy x > 0 , y > ( ) , 7 . > 0 )
T i m gia t r i Idn nhat cua dai liTdng S = cosA + cosC.
T„
'
Ta CO
' *'
,
TA
^B
,B
iC
TC.
^A
De lhay S > tan — t a n — + t a n — t a n ' — + tan- — t a n " — .
2
2
2
2
2
2
4N/3
HUdng ddn
'*
Hiidng
iL
ie
uO
nT
hi
Da
iH
oc
01
/
Tfif (3) (4) di den
<=>
cosB = —
4
<=>
3
B = arccos—
4
• ,
, j
f
Do sinA > 0 va cos(B - C) < 1, nen lif (1) cd:
S < 2sinA +
2
sin A
(2)
373
Chuyfin 66 BDHSG Toan gii trj Idn nha't vi giA trj nh6 nhat - Phan Huy Kh&\
Cty TIMHH MTV DWH Khang ViSt
Da'u bang trong (2) xay ra <=> cos(B - C) = 1 <=> B = C.
Dau bang trong (4) xay ra
Do A la goc l(^n nhii't trong lam giac => A > ^ => ~
Xct ham so f(x) = x ' + x - — vdi 0 < x < 1
< sin A < 1
ci>
B=C
'
>3 ; i i b i f
<^
X
(do A > 90" ^ — > 45"
X c l ham so i\x) = 2\ - wYi —
:
X
2
• • ,;
2
2x' - 2
Ta CO f'(x) = 2 — - =
— , ncn c6 bang bicn thicn sau
Ta CO f'(x) = 2x +1 + -J- > 0 Vx e (0; 1], ncn c6 bang bicn thicn sau
iL
ie
uO
nT
hi
Da
iH
oc
01
/
1
0
X
r(x)
X
, ,Vay ^ l a x i(\) = { yf3]
v 2 .
4;—
__
1
S<
3
73
/g
ro
up
S=:——-oB = C v a s i n A =
•
2
oA=B=C
Da'u bhng trong (1) xay ra o B = C.
.;,
ww
w.
^ ., j,
fa
ce
Hiidng dan gidi
bo
Tim gia trj Idn nha't ciia dai li/iing P = tanBtanC + 2colA
T i f d 6 P < c o t ' A +2cotA.
2
.
.
,
1-lan^ A
,
i-ian^
Do colA = tan A
A •
2 tan
TCr do ta CO maxP = 1 o ABC la lam giac vuong can tai A.
''hiiy: {I) chiJng minh nhU'sau
'f
Do A > 90" =o B + C < 90"
Ta CO cosBcosC = ^[cos(B + C) + cos(B -C)J
.
> -[cos(B + C)cos(B-C) + cos(B-C)]
P =>cosBcosC > - c o s ( B - C ) l l + cos(B + C)] = c o s ( B - C ) c o s ^ - ^ i ^
2
.. 2 A
=> cosBcosC > sin^ —cos(B - C )
2
=> cosBcosC > sin^—(cosBcosC + sinBsinC)
2
=>(1 - sin" — )cosBcosC > sin^ — sinBsinC.
2
2
Do B, C nhon => cosBcosC > 0.
„2A
cos
L .
A
col
2
9
.,
sin B sin C
„
2A
^ >
=>tanBtanC
cosBcosC
sm
2
Da'u bang xay ra <=> cos(B - C) <=> B = C => (1) di/cJc chiJng minh! '
Tir do ta CO
Thay (3) vao (2) ta di den P < c o l ' - + c o l 2
2
'
cos(B + C) > 0.
ok
Bai 6. X e l tarn giac ABC vdi A > 90".
Tr\idc hct ta c6 tanBtanC < cot" A
' u ;
om
<=> ABC la tarn giac dcu.
P - l o B - C v a c o t —= 1
2
« ABC la tarn giac vuong can tai A
.c
7V3
1
P<1
773
Vay maxS =
+
0
773
TCr (16 suy ra (kcl hctp vdi (2))
1
1
1
Vay max r(x) = 1(1) = 1. Tif (4) suy ra
Ta
i
f(x)
f(x)
-
s/
f'(x)
0 < cot — < 1). *
Chuy6n dg BDHSG Toan gia Irj I6n nhS't vi gia Irj nhd nha't - Phan Huy KhJi
Cly TNHH MTV DWH KhangVi§t
Bai 7. Xet tap hcJp cac tarn giac ABC khong phai la tam giac c6 goc tu.
Tir do C O bang bien thien sau:
Tim gia tri nho nha't cua da i liTrtng P ^ sin A + sinB + sinC
cos A + cosB + cosC
Hudng ddn gidi
Do vai tro binh dang cua A, B, C nen c6 the gia suf A = max{A; B; C}
71
.
3
X
g(x)
Ta c6:
Vay
cosA + cosB + cosC
^
,4 COS+2COS
2
B+ C
2
B
—
cos - ~
sin A + 2 cos — cos
2 ^
2
2__
B-C
[ ^ . A
B ^
cos
cosA + 2sm—cos
^=
2
2
2
(0
'
!
'
71
371
X6thams6'g(A)=
Ta
CO
sinA + 2cos—
2_.
.
^ . A
cosA + 2sm —
2
smA + 2cos~~
1- vdi - < A < - .
• —
A
3
2
cosA + 2sin
(sau khi rut gon) g'(A)
\A ,
sin
2
'
1
AV
cosA + 2sin 2J
376
<0.
ro
/g
om
bo
1
^
w.
o
^
ww
Dovay min f(x) = f(l) =
^
fa
1
^
ok
1
ce
1
.c
0
f(x)
up
Do - < A < - = > - < — < — =>cos — < 0
3
2
2
2
4
2
Tur do CO f'(x) < 0 V 0 < X < 1, nen CO bang bien thien sau:
f'(x)
"3
(1)
B-C
=1
cos2
A= ^
2
o ABC la tam giac vuong can dinh A.
3A
. • • i- • •
1
:
Ta
3A
-
Dau bkng trong (1) xay ra <=>
s/
7t
I
V2 _ V2 , ,
x= l
sinA + 2xcos—
2 cos
Xethamsof(x)=
2- v d i O < x < 1 z:>f'(x) = —
2
A
• A
cos A + 2x sm —
(cosA + 2 x s i n y ) ^
2
^
1
Tiif do suy ra f(x) > — + 1 .
3A
7t ^
min g(A) = g
2"
C
Datx=cos
I +
2
iL
ie
uO
nT
hi
Da
iH
oc
01
/
=> - < A < - .
sin A + 2 sin
i
i
1
g'(x)
71
sinA + sinB + sinC
n"'"
71
\/2
Ttf d6 ta C O ket qua sau: minP = ^ +
o ABC la tam giac vuong can dinh A.
Bai 8. Xet ABC la tap hdp cac tam giac nhpn.
Tim gia tri nho nha't cua bieu thuTc:
A B C
P = tanA + tanB + tanC + tan —tan—tan —.
2
2
2
Hiidng ddn gidi
R5 rang trong mot tam giac nhon, ta c6:
A
B
C
tanA + tanB + tanC > cot — + cot — + cot —.
Z
Z
Zr
(1)
.
Dau bang trong (1) xay ra <=> A = B = C.
A
B
C
1
NhiTvay P > cot — + cot — + cot — +
A
B C
2
cot — cot—cot —
2
2
2
A B C
A
B
C
cot—cot—cot — = cot — h cot — + cot —,
Trong moi tam giac ta c6
2
2
2
2
2
2
cot— + cot — + cot — > 3 VJ.
2
2
2
A
B
C
Tiy(2) (3) (4) s a u k h i d a t x = cot — + cot—+ cot—, ta c6:
2
2
2
377
ChuySn dg BDHSG 7oAn gia Irj Idn nha't
gia trj nh6 nhU't - Phan Huy Khii
!S/hdn xet: Xet cac each giai khac sau day:
1. Xet cac vectd ddn vi e , , C 2 , C 3 nhtfhinhve.
Ta
Xethams6'f(x) = x + 1
CO
(e, + 62
f'(x) = l - ± , nen c6 bang bien thien sauc,
3v^
f'(x)
f(x)
+00
1
1
1
28N/3
9
B ^
cos(62,33) > 0 .
-
»- ••; •
= 1 => 3 - 2(cosA + cosB + cosC) > 0
=>S = cosA + cosB + cosC < —.
2
3
Lai CO S = — o C | + 62 + 63 = 0
28^3
o A = B = C.
o e , + e 2 = -e3
Ta
s/
up
Bai 9. Tim gia tri Idn nhat cua dai lifdng S = cosA + cosB + cosC,
ro
d day A, B, C la ba goc ciia tarn giac ABC.
/g
Hitdng ddn gidi
om
S = 2cos^^-tlcos^—^ + l - 2 s i n 2 2
2
2
^ . C
A-B , ^ . 2C
= 2sm—cos
t-l-2sin —
.c
ok
w.
•2A-B
-sir-—-+3
Tir(l)suyra
3
Tom lai maxS = 2
'^"^ Siac deu.
Ta thu lai ket qua tren.
IT
IT
•2'2
thi
cox+cosy ^
< cos
Da'u bKng xay ra o x = y.
Tdrdotaco:
,
ww
2sin--cos—
2
2
B = 60".
Tifdng ty A = C = 6O" => ABC la tam giac deu.
Neu x, y e
fa
4sin^cos^^—^ + 2 - 4 s i n 2 2„, ,,,,,2„.,::„,.,,.,„f
2
MBP = 120" o
Ta CO nhan x6t hien nhien sau:
bo
2
o
Xet each giai khdc nffa nhU'sau:
ce
2
^v,
o BMNP la hinh thoi c6 B M = BN = BP = 1
28>y3
Nhir the minP =
^
o ABC la lam giac deu.
(1)
^,
IT
IT
. ,
„
cosC4tosA,n
C+ cosA + cosB ,
A+ B ,
3
A+B+C+
--H
^
cos
l-cos - - - 2
2
<
2
2 — <- ^.Qj.
(2)
2
• A-B „
sm
= 0
T .
C
A-B
2 .
. 2
2sin — = cos
Vay maxS = -
,|cos(e,,e3) + 2|t
Do
X = 3yf3
2 ,
cos(e|,e2)
Do cos (c|, 62) = cos (18()" - B j = - cos B . -
A= B=C
Taco
+ 2
Ti/dng tiT cos(e|,e3) = - c o s A va cos(e2,Cj) = - c o s C .
Tirdo m i n = f ( 3 ^ / 3 ) =
x>>/3 ,
Ket hcJp lai suy r a P >
+
+ 62
+ 2
+
>0
iL
ie
uO
nT
hi
Da
iH
oc
01
/
X
+63)^
=>
o A = B = C.
cosA + cosB + cosC + IT 1
^
—2- < cos— = — => cosA + cosB + cosC < —.
4
~
3 2
2
Dafu bkng xay r a < » A = B = C = ^ .
ABC la tarn giac deu.
379
--
V a y maxS = -
«
ABC
o — ..,
I nqii i i u y
iMim
r\ndl
la tarn giac deu. Ta cung thu lai ket qua tren.
M U C
L U C
3. L a i x6t each g i a i khac nffa nhu'sau:
Ta c6 S = cosA + cosB + cosC
A + B
= 2cos-
C O S - — —
2,.
D a u b^ng trong (*)
hMng
A-B
xay ra o
.
A-B
= 1 o A
3
§1. V a i bai loan m 6 dau
+ cosC < 2sin — + cosC
2
cos
1. M('/ d i u v6 gia tri l«tn nha't va nho nhS't cua ham s6'
3
§2. N h i n l a i cac bai loan ve gia t r i Idn nha't va nho nha't cua h a m so trong
cac k l t h i t u y e n sinh vao dai hoc, cao dS^ng
= B
13
• '^i;'
iL
ie
uO
nT
hi
Da
iH
oc
01
/
§3. C d sd l i thuye't cua b a i toan t i m gia trj Idn nha't va nho nha't cua h a m so
X e t h a m s c r f ( x ) = 2 s i n - + cosx v d i O < x < 7 t
cua h a m s6'
T a c 6 f'(x)
IT
X = J
= cos--sinx=
2
c o s - 1 - 2 s i n —,
2
2j
§ 1. Phifdng phap sit dung bat d^ng thufc Cosi
(do 0 < X < j i )
V d i chu y k h i 0 < X < j i t h i cos J > 0, ta c6 bang b i e n thien sau:
3
om
suy ra S < - , va S =
2
2
la tam giac deu.
<^
1.3. PhiTdng phap t h e m bdt hang so
63
1.4. PhiTdng phap t h e m bdt hang tuT
73
1.5. PhiTdng phap n h d m cac so hang
90
1.6. PhiTdng phap siJ d u n g k i thuat ngiTdc dau trong bat dSng thiJc C o s i
111
§2. PhiTdng phap stir dung bat dang thiirc Bunhiacopski t i m gia t r i Idn nha't
va nho nha't h a m so
119
§3. Cac phifdng phap thong dung khac siir dung ba't dSng thiJc
de t i m gia t r i Idn nha't va nho nha't ciia ham so
133
1. PhUdng phap xuat phap tiif mot ba't ddhg thiJc da biet tClT tri/dc
133
va nho nhS't cua h a m s5'
ww
Nhi/ the maxS = ^
50
141
191
ChiTcTng 4. Phi/dng phap c h i ^ u bien thidn h a m s6' tim gia trj Idn nhd't
w.
~ 3
1.2. Phirdng phap suT diing trifc tiep bat dang thiJc Cosi
Fhifrfng phap Ivhfng giac h6a tim gia trj Idn nhS't va nho nhfl't c u a h a m s6'
fa
ABC
42
Chrfdng 3.
ce
A = B
42
1.1. Sijrdung ba't ding thuTc Cosi c d ban
v^ nho nha't cua h a m so
ok
Tir do k e t hdp v d i (*)
33
2. Cac bai toan khac su-dung ba't ding thiJcde t i m gia t r i I d n nha't
.c
2
bo
13J
^ _ TV o
s/
1
-
3
IT
Vay maxf(x) = f
0
ro
1
f(x)
+
•'
/g
1
f'(x)
n
up
0
Ta
IT
X
30
hiidng 2. Phrfefng phap suT dyng bS't dang thrfc de tim gia trj Idn nhSft va nho nhS't
' ''^^^
206
§1. SuT diing triTc tiep chieu bien thien ham so de t i m gia trj Idn nha't, nho nha't ... 206
la lam gidc deu.
§2. SU diing chieu bien thien h a m so c6 k e t hdp t h e m cac phiTdng phap khac
B a i t o a n mof d ^ u c u 6 n s a c h n a y d a chtfng to srf p h o n g p h i i v a d a d a n g c u a
c a c phif(/ng p h a p d u n g de t i m g i a t r j I d n nhfi't v a n h o nhS't c u a n i g t b i 6 u
thuTc h a y m o t h a m so' c h o trxidc.
phiTcTng p h a p t i m g i a t r j Idn nM't
b a y k l lufdng t r o n g c u d n s a c h n a y mdi
216
ChiYdng 5.
Phrfdng phap mien gia trj ham so'tim gia trj Idn nhS't, nho nhS't ciia h a m s6' ....227
B a i t o a n n e u t r f i n k h e p l a i c u 6 n s a c h n a y c u n g vdi m p t t h d n g d i $ p
Cac
de t i m gia t r i Idn nha't, nho nha't
r^ng:
va n h o nhS't m a c h u n g t d i d a t r i n h
that phong phii l a m sao!
Chifdng 6. Phrfdng phap siJ dung 66 thj hoSc hinh hqc
d4 tim gia trj idn nhS't va nho nhS't cua h a m so'
243
C h i f r f n g 7 . L f n g d u n g c u a g i a t r f l«?n nha't v a g i a t r j n h o n h a t t r o n g b a i t o a n
g i a i phi/(fng t r i n h v a bS't phi/«/ng t r i n h c o t h a m .s(Y
263
§ 1 . M o ' i l i e n h e giiJa g i a t r i i d n n l i a l , n h o nha't c u a h a m so
•
v a sU C O n g h i p m ci5a p h i f i t n g t r i n h v a b a t phU(
263
§ 2 . C a c b a i t o a n b i p n l u a n p h i f ( * n g t r i n h v a b a t phiTttng t r i n h c(') i h a m so
CO m 3 t trong cac k i t h i t u y c n sinh v a o D a i hoc, Cao ddng
264
d e b i p n l u a n p h i f d n g t r i n h v a b a t phiTcing t r i n h c 6 t h a m so
272
A . B i e n l u a n p h i / d n g t r i n h c 6 t h a m so
272
B . B i e n l u a n b a t p h U d n g t r i n h c d t h a m so'
285
Chi^(/ng 8 . V a i b a i t o a n k h a c v e g i a t r j I d n nha't v a n h o nha't c i i a h a m so'
§ 1 . Cfng d u n g g i a t r j I d n nha't v i i n h o nha't d e g i a i phU'dng t r i n h
*!
iL
ie
uO
nT
hi
Da
iH
oc
01
/
§ 3 . Sic d u n g g i a t r j I d n nha't v a n h o nha't c u a h a m so'
296
>•! t ^
v a b a t p h U d n g t r i n h k h d n g c 6 t h a m so
296
§ 2 . G i a t r i I d n nha't v a n h o nha't c u a h a m so p h u t h u p c t h a m so
308
A . B i e n l u a n t h c o t h a m s o ' g i a t r i I d n nha't v a n h o n h a t c u a h o h a m so
314
up
c u a c a c h a m so p h u t h u o c t h a m so'
s/
B . C a c iJng d u n g c u a v i § c k h a o s d t g i a t r i I d n nha't v a n h o nha'l
ro
§ 3 . G i d i t h i O u m O t s o b a i t o a n g i a t r j I d n nha't, n h o nha't t r o n g so h o c ,
330
/g
h i n h h o c , li/dng giac
.c
B . D i e m q u a b a i t o a n g i a t r i I d n nha't, n h o n h a t t r o n g h i n h h o c t o h d p
om
A . V a i b a i t o a n v e g i a t r i I d n nha't, n h o nha't t r o n g so h o c
ok
C . V a i b a i t o a n v e g i a t r j I d n nhaft v a n h o nha't t r o n g h i n h h o c k h o n g g i a n
bo
D . D i e m q u a v a i b a i t o a n v e g i a t r i I d n nha't, n h o n h a t t r o n g h i n h h o c p h d n g
330
342
348
355
w.
fa
ce
E . D i e m q u a m o t so' b a i t o a n t i m g i a t r j I d n nha't, n h o nha't t r o n g l i f d n g g i a c . . . . 3 7 0
ww
'
308
Ta
p h u t h u o c t h a m so'
