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55 TTHHII TTHH VV PP NNDi y l 5 thi th i hc ca LAISAC c tp ch Ton Hc v Tui tr ng trong
4 s t S1 n S 4. s 5 hon ton mi, thay th mt b mt file ngun .Nhng cu hi trong cc trn, ban u hon ton mi, nhng thi gian sau ny thy c ri
rc trong nhng quyn sch luyn thi i hc hay thi th ca mt s trng...
S 372.6/2008 S 1
(Thi gian lm bi:180 pht)
PHN CHUNG CHO TT C CC TH SINH.Cu I.(2 im).Cho ng cong c hm s ( )3 22 1y x x m x m= - - - + (1).
1. Kho st s bin thin v v th ca hm s khi m = 1 .2. Trong trng hp hm s (1) ng bin trong tp s thc R, tnh m din tch hnh phnggii hn bi th ca hm s (1) v hai trc Ox,Oy c din tch bng 1 n v din tch.
Cu II.( 2 im).1. Gii phng trnh nghim thc: 1 tan .tan 2 cos 3 .x x x- =
2. Tm tt c cc gi tr ca tham s k phng trnh : xxx kk 2124)1( -=+-+ c nghim.Cu III.( 2 im)
1 .Trong mt phng vi h to Oxy cho elp (E) :x2 + 4y2 = 4. Qua im M(1 2) k hai ngthng ln lt tip xc vi (E) ti A v B. Lp phng trnh ng thng i qua hai im A v B.
2. Cho tam gic ABC tha mn : ( )
5
os2 3 os2 os2 02c A c B c C + + + = . Tnh ba gc ca tam gic.Cu IV.( 2 im).
1. Tnh tch phn : dxexxx
I xsin2
0
2 .cos2
cos2
+=
p
.
2. Cho ba s thc dng x, y, z tha mn iu kin: .1.2 =+ xzxy
Tm gi tr nh nht ca biu thc: .543
z
xy
y
zx
x
yzS ++=
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PHN T CHN:Th sinh chn cu V.a hoc cu V.b.Cu V.a. Theo chng trnh THPT khng phn ban . ( 2 im)
1. Trong khng gian vi h trc to Oxyz cho hai ng thng
(d1) :
=-
=-+
03
042
z
yx (d2):
=-
=+
01
0
x
zy.
Lp phng trnh mt cu c bn knh nh nht tip xc vi c hai ng thng trn.
2. C tt c bao nhiu s t nhin chn c 4 ch s, sao cho trong mi s ch s ng sau lnhn ch s ng lin trc n.Cu 5.b. Theo chng trnh THPT phn ban th im . ( 2 im)
1. Cho hnh chp t gic S.ABCD. y ABCD l hnh vung cnh bng a, SA vung gc vi mtphng(ABCD) v SA = a. Tnh din tch ca thit din to bi hnh chp vi mt phng qua A vunggc vi cnh SC.
2. Gii bt phng trnh : ( ) 2log3log 12 xx - ( )Rx .
Ht
HNG DN GII.Cu I.1.Bn c t gii.
2. Ta c y = 3x2 4x m + 1.
hm s ng bin trong tp s thc R khi1
' 03
y x R m " - (2).
Phng trnh honh giao im ca th (1) vi trc Ox: ( )3 22 1x x m x m- - - + = 0(x 1)(x2x m ) = 0 th (1) lun ct trc honh ti im c nh
(1 0 ). Mt khc v hm s l hm bc ba c h s cao nht a = 1 > 0 , li ng bin trong R nn th lun ct trc tung c tung m.
Hay khi 31
-m ( ) [ ]
3 2
2 1 0 01y x x m x m x= - - - + " .Do , din tch hnh phng gii hn bi th (1) v hai trc ta l:
( )
212
1)1(2
1
0
23 mdxmxmxxS --=+----= .
M S = 16
13-=m (tha iu kin (2)).
Cu II. 1.iu kin :
2
1cos
0cos
02cos
0cos2x
x
x
x
Phng trnh tng ng :cos3x = cos3x.cosx.cos2x.
Hoc :
=
=
=-=
4
3cos
)(0cos0cos3cos403cos 2
3
x
loaxxxx p
pkx +=
6.
Hoc:cosx.cos2x=1 01coscos2 3 =-- xx 0)1cos2cos2)(1(cos 2 =++- xxx
=++
=-
0)1cos2cos2(
0)1(cos
2 xx
xp
pmx
vnxx
mx2
).(01cos2cos2
.2
2=
=++
=
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Vy phng trnh c nghim l : pp
kx +=6
pmx 2= . ),( Zmk .
2. t t = 2x k 10
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MN l ng vung gc chung ca hai ng thng (d1) v (d2) ,ta c
-=
=
=--
=--
=---+
=++--
^
^
1'
1
03'2
06'5
0'3'0
00'46
t
t
tt
tt
ttt
ttt
bMN
aMN.
T suy ra phng trnh mt cu cn tm l :4
9)2()
2
3( 222 =-++- zyx .
2. Gi s s l 4321 aaaax= .Theo yu cu bi ton cc ch s a1, a2, a3, a4 khc nhau tng i mt
v khc khng , v x l s chn nn ta c cc trng hp sau :TH1: a4 = 4 , t yu cu ton s l x = 1234. Do c mt cch chn .TH2: a4 = 6 , t yu cu ton ba s hng a1, , a2 , a3 ch c ly trong tp{ }5,4,3,2,1 v cc ch s
tng dn nn c 35 10C = s cho trng hp ny .
TH3 : a4= 8 ,tng t ba s hng a1, , a2 , a3 cn li ch c ly trong tp{ }7,6,5,4,3,2,1 nn c37 35C = s cho trng hp ny.
Vy c 1+10 + 35 = 46 s c chn theo yu cu ton .Cu Vb.1. Bng phng php ta ,chn A(0,0,0) ,B(a 0 0) D(0 a 0) C(aa 0) S(0 0 a).Gi s mt phng (P) cho ct SB, SC SD ln lt ti E, G , F. Mt phng (P) i qua A v vung
gc SC nn nhn vect )( aaaSC -= lm VTPTphng trnh (P) l :x + y z = 0 .(5)
Ta lp phng trnh ng thng SD
-=
=
=
taz
ty
x 0
(6) . F l giao im ca SD v (P) nn n l nghim
h phng trnh ( 5) v (6) )2
2
0(aa
F . Tng t G l giao im ca (P) v SC )3
2
3
3(
aaaG .
Do din tch thit din AEGF : [ ] .32
)(22a
AFAGAGFdtS ===
2. iu kin : x>1 , 2x .
Ta c ( ) 2log3log 12 xx - xx 22
3 log1
)1(log1
-.
Khi 21 t v tx 2= . Bt phng trnh vit li 14
1
4
3143
+
-
tt
tt (7)
ttt
tf
+
=
4
1
4
3)( l hm s lin tc trong )
2
1( +
Ta c
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S 418. 4/2012
S 2(Thi gian lm bi:180 pht)
PHN CHUNG CHO TT C CC TH SINH( 7, 0 im):Cu I. (2, 0 im).Cho th ( C ) c phng trnh 3 23 2y x x= - + .
1. Kho st v v th ( C ).2. Qua im un I ca th ( C ) vit phng trnh ng thng ( d ) ct th ( C ) ti hai im
A, B khc I sao cho tam gic MAB vung ti M, trong M l im cc i ca th ( C ).Cu II. (2,0 im).
1. Gii phng trnh :22cos 3
tan cot .sin2
xx x
x+ =
2 nh tham s m h phng trnh( )
( )
3 19
3 21
x y x m
y x y m
+ + = -
+ + = +
c nghim.
Cu III.(1, 0 im).Tnh tch phn : I = ++++
1
0
12 2)12( dxexx xx .
Cu IV ( 1,0 im).Cho hnh chp t gic S.ABCD, y ABCD l hnh vung cnh a, mt bn SABl tam gic u v vung gc vi y ABCD. Tnh th tch khi nn c ng trn y ngoi tiptam gic ABC v nh ca khi nn nm trn mt phng (SDC).
Cu V.( 1, 0 im).Tm gi tr nh nht ca biu thc: P=3 3 3
3 3 3
a c b a c b
b a bc c b ac a c ab+ +
+ + +,
trong a, b, c l ba s thc dng ty .PHN RING( 3, 0 im): Th sinh ch c lm mt trong hai phn (phn A hoc B)
A.Theo chng trnh Chun.Cu VI a.(2, 0 im)
1. Trong mt phng, vi h trc ta Oxy lp phng trnh ng trn c bn knhR = 2, c tm I nm trn ng thng ( )
1 : 3 0d x y+ - = v ng trn ct ng thng
( )2 : 3 4 6 0d x y+ - = ti hai im A, B sao cho gc
0120AIB = .
2. Trong khng gian, vi h trc ta Oxyz cho ba im ( ) ( ) ( )
123 , 010 , 10 2 A B C - .
Tm trn mt phng (P): 2 0x y z+ + + =
im M sao cho tng
2 2 2
2 3MA MB MC+ +
c gi tr nhnht.
Cu VII. a ( 1,0 im).Gii phng trnh :os
4tanx=2012c x
p +
.B. Theo chng trnh Nng Cao.Cu VI b.(2, 0 im).
1. Trong mt phng, vi h trc ta Oxy cho hai ng thng ( )1 : 3 3 2 0d x y- - + = v
( )2 : 3 3 2 0d x y+ - - = .
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Lp phng trnh ng thngD ct hai ng thng ( ) ( )1 2,d d ln lt ti B, C sao cho tam gic
ABC u c din tch bng 3 3 (vdt), trong nh A l giao im ca ( ) ( )1 2,d d .
2.Trong khng gian, vi h trc ta Oxyz cho hai ng thng cho nhau
( )11 2 3
:1 2 3
x y zd
- - -= = v ( )2
1:
3 2 1
x y zd
-= = .Lp phng trnh mt phng (P) sao cho khong cch
t ( )1d n (P) gp hai ln khong cch t ( )2d n (P).
Cu VII b ( 1, 0 im).Gii phng trnh:os2x
t anx=2012c
.Ht
HNG DN GIICu I.1.T kho st .
2.Theo cu trn hai im ( )02M , ( )1 0I ln lt im cc i v im un ca th ( C ).
Gi k l h s gc ca ng thng qua I nn c phng trnh ( )
1y k x= - . Phng trnh
honh giao im ca ( C ) v ( d ) l ( ) ( )( )
3 2 23 2 1 1 2 2 0x x k x x x x k- + = - - - - - = .
( d ) ct ( C ) ti hai im A,B khc M th phng trnh 2( ) 2 2 0g x x x k= - - - = (*)
c hai nghim khc 1 0 3(1) 0
gk
gD > > -
.
Gi s hai giao im l ( ) ( )1 1 2 2 , A x y B x y , trong 1 2,x x l hai nghim ca phng trnh (*)
v ( ) ( )1 1 2 21 , 1y k x y k x= - = - .V I l tm i xng ca th ( C ) nn tam gic MAB vung ti M
th ( ) ( )2 2
1 2 1 22 2 5 2 5AB MI x x y y= = - + - =
( ) ( ) ( ) ( )( )
2 22 21 2 1 2 1 21 20 1 4 20k x x k x x x x + - = + + - = .
3 2 1 5 1 53 2 0 2, ,
2 2
k k k k k k - + - -
+ + - = = - = = . So vi iu kin k > -3 ta c ba ng thng
( ) ( ) ( )1 5 1 5
2 1 , 1 12 2
y x y x y x- + - -
= - - = - = -
Cu II.1. K 02sin x . Phng trnh tng ng2 21 os6xos 3 os2x os2x 4cos 2x-5cos2x+1=0
2
cc x c c
+= = 2(cos 2 1)(4 cos 2 4 cos 2 1) 0x x x - + - =
os2x=1c (loi),1 2
cos22
x- -
= (loi),1 2 1 1 2
cos 2 arccos .2 2 2
x x kp - + - +
= = +
2.K 0 , 0x y . Cng v theo v ca hai phng trnh ca h ta c :
( ) ( ) ( )
2
2 3 40 0 3 40 0x y xy x y x y x y+ + + + - = + + + - = .
Gii phng trnh bc hai ny ta c 8x y+ = - (loi), 5x y+ =
Th 5y x= - vo phng th nht ca h ta c4
2
mx
-= . tn ti nghim x th 4m .
V4 6
52 2
m my
- += - = . tn ti nghim y th 6m - .
Vy h c nghim th 6 4m- .
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IO
FE
D
CB
A
S
Cu III.Ta c ...)2(.)12(1
0
11
0
121
0
12 222
++++++ ++=++ dxedxexxdxexx xxxxxx
Dng phng php tng phn ta tnh tch phn ++
1
0
12 dxe xx .
t
=
+=
=
= ++++
xv
dxexdu
dxdv
eu xxxx 1122
).12(
Suy ra dxexxxedxe xxxxxx 11
0
21
0
11
0
1 222 )2()( ++++++ +-= .
Do : 31
0
11
0
12 )()12(22
exedxexx xxxx ==++ ++++Cu IV. Gi E, F ln lt trung im AB v CD suy ra ( )
EF SAB^ .
Gi O l tm ca tam gic u ABC. Trong mt phng (SEF) t O dngng thng song song EF ct SF ti I , suy ra I l nh hnh nn
Ta c:2 2 2
EF= .
EF 3 3 3
OI SOOI a
SE
= = =
Bn knh ng trn y2 2 3 3
OS= .3 3 2 3
R SE a a= = =
Vy th tch ca hnh nn l3
2 21 1 1 2 2.3 3 3 3 27
aV R h a a
p
p p= = =
Cu V. Ta c :
.)(
2
3
3
a
c
c
b
b
a
cbab
aca
bcab
ca
+
=+
=+
Tng t:
2
3
3
bcb a
c ac b ac
a b
=
+ +
2
3
3.
cac b
a ba c ab
b c
=
+ +
t X=
b
a Y=
c
b Z=
a
cX,Y,Z >0 v X.Y.Z=1
2 2 2X Y ZP
Y Z Z X X Y = + +
+ + +.
M4
2 ZY
ZY
X ++
++
4
2 ZX
ZX
Y ++
++ .
4
2
ZYXYX
YX
Z++
++
+
P=2
3
2
222
++
+
++
++
ZYX
YX
Z
XZ
Y
ZY
X.Vy Max(P)=
2
3khi a = b = c.
Cu VI a.1.Ta c tm ( ) ( )1 3I x x d- + .Gi H l hnh chiu ca I xung ng thng ( )2d suy ra
tam gic HIA l na tam gic u c cnh IH = 1.Do ta c
2
3 4( 3) 6( , ) 1 6 5 1 11
5
x xd I d IH x x x
+ - + -= = - + = = =
Vy c hai ng trn ( ) ( )2 2
1 2 4x y- + - = , ( ) ( )2 2
11 8 4x y- + + =
2.Gi ( )
x y z l ta im I sao cho :2 2 1
2 3 0 3 3 2
IA IB IC I
+ + = -
uur uur uur r
Ta c ( ) ( ) ( )2 2 2
2 2 2 2 2 2 22 3 2 3 6 2 3MA MB MC MI IA MI IB MI IC MI IA IB IC+ + = + + + + + = + + +uuur uur uuur uur uuur uur
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2 2 22 3MA MB MC+ + c gi tr nh nht khi v ch khi MI t gi tr nh nht, lc M l hnhchiu ca I xung mt phng (P).
ng thng IM vung gc vi (P) nn c phng trnh tham s2 2 1
, ,3 3 2
x t y t z t= + = + = - +
Giao im M l nghim ca h to bi phng trnh (P) v ng thng (d).
Gii h ta c5 5 13
18 18 9
M - - -
.
Cu VIIa.K: cosx 0, t anx>0 sinx 0 . Phng trnh tng ng1
osx 1 12 sinx osx2 2
1sinx
2
sinx 2012sinx.2012 osx.2012
osx2012
cc
cc
= =
V hm s tanx c chu k kp , tanx>0 ta ch xt min nghim sao cho s inx>0, cosx>0 t suy ramin nghim s inx0,cosx>0 )
tanx>0 vi s inx
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2 2 2 2cos sin sin cossinx =2012 s inx.2012 osx.2012osx
x x x xcc
- =
V hm s tanx c chu k kp , tanx>0 ta ch xt min nghim sao cho s inx>0, cosx>0 t suy ramin nghim s inx
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2. Trong khng gian vi h ta Descartes Oxyz , cho tam gic ABCc ( ) ( ) ( )
111 , 0 1 1 , 3 5 3A B C- - - .
Lp phng trnh ng phn gic trong gc A ca tam gic ABC.
Cu VII.a (1 im)Cho cc s phcztha mn iu kin: 1 2 5z i- + = . Tm s phc w c mun
ln nht, bit rng: w = z+1+i .B. Theo chng trnh Nng Cao
Cu VI.b (2 im)1. Trong mt phng vi h ta Descartes Oxy , cho hai im ( ) ( )
A 34 , 5 3B . Xc nh im M
trn ng Elip ( )2 2
: 18 2
x yH + = sao cho din tch tam gic MAB c gi tr nh nht.
2. Trong khng gian vi h ta Descartes Oxyz, cho hai ng thng ( )11 1 1
:1 2 2
x y zd
- - -= = v
( )21 3
:1 2 2
x y yd
+ -= =
-ct nhau nm trong mt phng ( )P . Lp phng trnh ng phn gic ca gc
nhn to bi ( )1d , ( )2d nm trong mt phng ( )P .
Cu VII.b (1 im) Gii h phng trnh
2
1 20121
2 4 .log 0
x y
y
xy
x
-+ = + + =
, vi n thc 0, 0x y> > .
..Ht
HNG DN GIICu I.a) T gii.
b) tip tuyn ti ( ) ( )0 0 mM x y C song song vi ng thng 3 1y x= - + th h s gc ca tip
tuyn
( )
0' 3y x = -
( )
20
2
31
m
x
- - = -
-
2
0 03 6 1 0x x m - + - =
, 2m -
(1)
Gi thit: 0 03 1
( ) 1010
x yd M d
+ -= =
20 0
20 0
3 12 11 0
3 8 9 0
x x m
x x m
- + + =
+ - + =, (2)
Gii (1) v (2) ta c 1m = hoc43
3m = .
Cu I a)iu kinosx 0
cos4x 0
c
. Phng trnh tng ng ( ) ( )8 2 s inxcos2x= tanx+tan4x + tanxtan4x-1
sin5x cos5x8 2 s inxcos2x=cosxcos4x cosxcos4x
- sin 8 sin 54
x x p = -
2.
12 3 ( ). (*)5 2
.52 13
x k
k Zx k
p p
p p
= - +
= +
So vi iu kin (*) chnh l nghim ca phng trnhb) Cng hai v ca hai phng trnh ca h, ta c:
( ) ( ) ( ) ( )
22 1 3 1 2 1 10 0 1 3 1 10 0x y x y x y x y x y+ + + + + + + - = + + + + + - =
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Gii phng trnh bc hai ny ta c1 5
1 2
x y
x y
+ + = - + + =
thay1 5
1 2
y x
y x
+ = - - + = -
vo mt trong hai phng trnh ca h v tip tc gii ta c nghim ca h
cho l ( ) ( ) ( )
: 0 3 , 1 0x y
Cu IIITch phn vit li ( )2
1
ln 1 ln ln.
e x xI dx
x
+ +=
t lndx
t x dtx
= = khi 1 0 1x t x e t= = = =
Do :
( ) ( )1 11
2 2
200 0
ln 1 ln 11
tI t t dt t t t dt
t = + + = + + - +
1 2
20
1 (1 )ln(1 2) ln(1 2) 1 2
2 1
d t
t
+= + - = + + -
+ .
Cu IV. Dng ( )SH ABCD^ , vSA SB SD= = HA HB HD= = H l tm ng trn ngoi tip tamgic ABD, m tam gic ABD u H va l trng tm, trc tmtam gic ABD.Do HD AB^ , m //AB CD HD DC ^ (1), theo nh l bang vung gc ta cng c SD DC ^ (2). T (1) v (2) gc nhn 030SDH= chnh l gc ca hai mt phng ( )SDC v ( )ABCD .Xt tam gic SHD vung ti H c
0 1 3 3, 30 2 2 2
SD a SDH SH a HD a AB a= = = = = .
Vy th tch hnh chp SABCD l 31 3 3. .3 16ABCD
aV S SH = =
Cu V.Cch 1.t1 1 1 1 1 1
1a b cx y z x y z
= = = + + = v 0, 0, 0x y z> > > .
Biu thc cho vit li2 4 3 9 6 36
x y zP
x y z= + +
+ + +
Ta c1 1 1
1 .2 4 2 3 9 3 6 36 6
x y zP
x y z
- = - + - + -
+ + +
1 1 11
2 3 6P
x y z
- = - + +
+ + + .
Ta li c:1 1 4
2 2x x
+
+
ng thc xy ra khi 2x =
Tng t v cng cc bt ng thc li suy ra ,ta c:1 1 1 1
2 3 4 2x y z
- + + -
+ + +
Do :1 1 1 1 1
12 3 6 2 2
P Px y z
- = - + + -
+ + +
Vy gi tr nh nht1
2P= khi v ch khi
1 1 1, ,
2 3 6a b c= = =
H
D
CB
A
S
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Cch 2. Chn im ri:1 2 4 1 1 3 9 1 1 6 36 1
, , .2 4 16 2 3 9 36 3 6 36 144 6
a b c
a b c
+ + ++ + +
+ + +
Suy ra2 4 3 9 6 36 1
116 36 144 2
a b cP P
+ + ++ + +
Cu VI a1) Phng trnh ng trn vit li2 2
1 21
2 2
x y- + + =
.
t1
sin 2 sin 12
xxa a
-
= = + 2
os = 2 os 22
yc y ca a
+
= - , trong [ ]
0 2 .a p
Do ta im ( ) ( )2 sin 1 2 os 2M c Ca a+ - .Phng trnh ng thng : 8 0AB x y+ - = .
Ta c ( )
2 cos 94
2M ABd
pa
- -
= . ax cos - 1 0 3
4 4MABS M x y
p p
a a p
= - = + = = -
.
Vy im ( ) ( )
0 3M C- th din tch tam gic MAB c gi tr ln nht.
2) Ta c ( )1 2 2 3AB AB= - - - =uuur
, ( )2 4 4 6AC AC= - =uuur
Ly im D l trung im AC ( )
23 1D AC AB AD - =
Nn phn gic trong gc A ca tam gic ABC cng l ng trung tuyn ca tam gic cn ABD
Gi H trung im ( )1 11 1BD H - , do ng phn gic cn tm c phng trnh
1
: 1
1 2
x
AH y
z t
=
= = -
.
Cu VIIa. Xt s phcz x yi= + . T gi thit suy ra ( ) ( )2 2
1 2 5x y- + + = . Suy ra tp hp im
( )M x y biu din s phc z l ng trn tm ( )1 2I - , bn knh 5R = . D dng c c
( )5 sin 1 5 cos 2M a a+ - vi [ ]0 2 .a p Mt khc ( ) ( )w = z+1+i= x+1 1y i+ +
( ) ( ) ( ) ( ) ( )2 22 22
w x+1 1 5 sin 2 5 os 1 10 2 5 2sin osy c ca a a a = + + = + + - = + -
t 2 sin cost a a= - , tn tiath ( ) ( )2 2 21 2 5 5 2sin cos 5t t a a+ - -
Do Max
w 20= khi v ch khi2 1
sin , os = 3 35 5
c x ya a= - = = - . Vy s phc l w 4 2i= -
Cu VIb. 1) t sin 2 2.sin2 2
xxa a= = os = 2 os ,
2
yc y ca a = , trong [ ]0 2 .a p
Do ta im ( ) ( )2 2 sin 2 osM c Ha a .Phng trnh ng thng : 2 11 0AB x y+ - = .
Ta c: ( )
4 os 11
45
M AB
c
d
pa
- -
= .. in cos 1 2 14 4MAB
S M x yp pa a - = = = =
Vy im ( ) ( )
21M H th din tch tam gic MAB c gi tr nh nht
2) D dng nhn thy hai ng thng ( )1d , ( )2d ct nhau ti giao im ( )111I .
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Chn trn ( )1d im ( )233 3M IM = .Phng trnh tham s ( )2d
( ) ( )21 2 1 2 3 23 2
x t
y t N t t t d
z t
=
= - + - + - = -
.
( ) ( )2 1 20
3 9 0 13 , 23 12
tIN IM IN N N
t
== = = - - =
.
Ta c: ( ) ( ) ( )
1 2122 , 1 2 2 , 1 2 2IM IN IN= = - - = -uuur uuur uuur .M
02 2. 1 0 90IM IN MIN= > < uuur uuur gc nhn ca
( )1d , ( )2d chnh l gc 2MIN .Gi ( )231K trung im ca 2MN nn ng thng qua hai im I, K
l ng phn gic ca gc nhn to bi ( )1d , ( )2d c phng trnh
1
: 1 2
1
x t
IK y t
z
= +
= + =
.
Cu VII b, V : x > 0 , y > 0 T (1)
2012 2012 2012
1log log (1 ) log (1 ) (*)
1
xx y y y x x
y
+= - - + = - +
+
t ( ) ( )2012log 1f t t t= - + , xc nh ( )0t" + Ta c :
( )
1'( ) 1 01 ln 2012
f tt
= - >+
vi ( )0t" +
f(t) lun lun ng bin trong ( )0+
Do phng trnh (*) ( ) ( ) ( ) ( )2012 2012log 1 log 1x x y y f x f y x y- + = - + = =
T phng trnh (2) : 2 2 14
12 4 .log 0 4 .log 2 log
4
yy yx x x
+ = = - =
(3)
Th x = y vo phng trnh (3) ta gii phng trnh 14
1log
4
x
x
=
bng phng php v hai th
14
logz x= v 14
xz =
trn cng mt h trc ta vung gc xOz ta thy phng trnh c nghim
duy nht1
2x = . Th li th h phng trnh c nghim
1 1,
2 2x y= =
..
S 440 2/2014
S 4(Thi gian lm bi:180 pht)PHN CHUNGCu 1( 2 im). Cho ng cong ( )mC c phng trnh ( )
3 3 2 5,y mx m x m= - - + - ( m l tham s)
1) Kho st v v th hm s trn khi m = 1.2) Chng minh rng vi mi tham s m ng cong ( )
mC lun lun ct mt ng thng c nh
ti ba im c nh.Cu 2( 1 im).Gii phng trnh 4sin .sin 2 . os3 t anx. tan 2 . os6x x c x x c x= .
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Cu 3 ( 1 im).Gii h phng trnh4 3 1 0
4 (1 )(1 ) 6 1 1 0
x y y
x y x
- + + =
+ + - + + =.
Cu 4( 1 im). Tnh tch phn sau ( )4
2
0
ln(cos )xI e ta nx x dx
p
= + .
Cu 5( 1 im). Cho t din ABCD c , ,AB CD a AC BD b AD BC c= = = = = = v 2 2 2 3a b c+ + = (
n v chiu di ). Tnh th tch nh nht ca t din ABCD .Cu 6( 1 im). Cho a, b, c l ba cnh ca mt tam gic tha mn iu kin 2c b abc+ = .
Tm gi tr nh nht ca biu thc3 4 5
Sb c a a c b a b c
= + ++ - + - + -
.
PHN RING( Th sinh ch c chn mt trong hai phn A hoc B)A. Theo chng trnh Chun
Cu 7a.( 1 im). Trong mt phng, vi h trc ta Oxy cho tam gic ABC c nh (12)A , trng
tm (11)G v trc tm2 10
3 3
H
. Hy xc nh ta hai nh B v C ca tam gic.
Cu 8a( 1 im). Trong khng gian vi h trc ta Oxyz cho tam gic ABC vi
( ) ( ) ( )
1 0 0 , 3 2 2 , 2 0 3A B C - . Lp phng trnh mt cu ( )
C c bn knh nh nht i qua A v lnlt ct hai cnh AB, AC ca tam gic ABC ti hai im E, F sao cho 3, AF 2AE= = .
Cu 9a( 1 im). Gi E l tp cc s t nhin c ba ch s abc ( )0a sao cho ba s a,b,c khc
nhau theo th t tng dn. Tnh xc sut ly ra trong tp E mt phn t l s chn.B. Theo chng trnh Nng Cao
Cu 7b.( 1 im).Trong mt phng vi h trc ta Oxy, cho hai ng thng
1 2( ) : 1 0, ( ) : 1 0d x y d x y+ - = - + = . Lp phng trnh ng trn ( )C ct 1( )d ti A v 2( )d ti hai
im B, C sao cho tam gic ABC l tam gic u c din tch bng 24 3 n v din tch.
Cu 8b.( 1 im). Trong khng gian vi h trc ta Oxyz cho tam gic ABC c nh (120)A ,trng tm (110)G v trc tm
2 10 0
3 3H
. Hy xc nh ta hai nh B v C ca tam gic.
Cu 9b( 1 im). Gii phng trnh ( ) ( ) 12 3 2 3 2x x
x++ + - = .
,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,
HNG DN V GII.
Cu 1. 1.Khi m = 1 hm s vit li . 3 3 4,y x x= + - T kho st v v
2. Hm s vit li ( )
3
3 1 6 5y m x x x= - + + - .
im c nh ( nu c) m ng cong i qua tha nghim h:3 3 1 0
6 5
x x
y x
- + =
= -.
t 3 2( ) 3 1 0 '( ) 3 3f x x x f x x= - + = = - , khi '( ) 0 1, 1f x x x= = = -V th ca hm s f(x) c ax . ( 1). (1) 3.( 1) 0m minf f f f= - = -
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Cu 2. K cos 0.cos 2 0x x 4sin . sin 2 . os3 tanx. tan 2 . os6 4sin . sin 2 . os2 . os3 tan x.sin 2 . os6x x c x x c x x x c x c x x c x= = . Xy ra:
+ sin 2 02
kx x
p= = , so vi iu kin phng trnh c nghimx mp= .(1)
+ 4 sin . os2 . os3 t anx. os6 sin 4 . os3 s inx. os6 sin x=-sin5x ,3 4 2
k kx c x c x c x x c x c x x x
p p p= = = = - - (2)
T (1) v (2) so vi iu kin phng trnh c nghim ( )3
kx k Z
p= .
Cu 3. K 1, 1x y - - . t 2 1 , 1 0, 0a x b y a b= + = + th h tr thnh2 2 3 3 0
2 3 1 0
a b b
ab a
- + - =
- + =T h phng trnh trn ta bin i v dng s phc sau
( ) ( ) ( ) ( )22 2 23 3 2 3 1 0 3 3 0 3 3 0 ( )a b b ab a i a bi i a bi i z iz i a bi z C- + - + - + = + - + + - = - + - = + =
Gii phng trnh ny, ta c 1 21 , 1 2z i z i= + = - +
Suy ra ( )1 1a b= = do nghim ca h phng trnh trn l3
04
x y
= - =
Cu 4.V4 42
0 0
ln(cos )x xI e ta nxdx e x dx
p p
= + .
Ta tnh4 4 4 4 4
2 42 2
0 0 0 0 0
11 (t anx) 1
o sx o sx
xx x x xe dxJ e ta nxdx e dx e dx e d e
c c
p p p p p
p
= = - = - = + -
M4 4 4
44
00 0 0
(t anx) t anx t anx t anxx x x xe d e e dx e e dx
p p p
pp
= - = - . Nn4
0
1 t anxxJ e dx
p
= -
Ta li c
( )
4 4 4 444
00 0 0 0t anx ln( osx) ln(cos ) ln(cos ) ln 2 ln(cos )
x x x x x
K e dx e d c e x e x dx e e x dx
p p p p
pp
= = - = - + = + .
Do 4
4
0
1 ln 2 ln(cos )xJ e e x dx
p
p
= - - . Vy tch phn 41 ln 2I ep
= - .
Cu 5.
Qua cc nh ca tam gic BCD ta k cc ng thng song song vicc cnh i din, chng ln lt ct nhau tng i mt to thnh tamgic BCD c din tch gp 4 ln din tch tam gic BCD.Ta c CD = 2CD = 2AB suy ra tam gic CAD vung ti A, tng
t ta cng chng minh c hai tam gic CAB, BAD ln ltvung ti A.t ' , ' , 'AB x AC y AD z= = = th ta c
( )
2 2 2
2 2 2 2 2 2 2 2 2
2 2 2
4
4 2
4
x y c
y z a x y z a b c
z x b
+ =
+ = + + = + + + =
D'
C'
B'
D
C
B
A
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( )
( )
( )
2 2 2 2
2 2 2 2
2 2 2 2
2
2
2
x b c a
y a c b
z a b c
= + -
= + -
= + -
( )( )( )2 2 2 2 2 2 2 2 2' ' '1 1 2
4 24 12ABCD AB C DV V xyz b c a a c b a b c= = = + - + - + - .
m ( )( )( )32 2 2 2 2 2 2 2 2
2 2 2 2 2 2 2 2 2 27 13 27
b c a a c b a b cb c a a c b a b c
+ - + + - + + -+ - + - + - = =
( C-si)
212ABCD
V .Vy th tch ca t din ln nht 212ABCD
V = khi v ch khi 1a b c= = = .
Cu 6. V a, b, c l ba cnh tam gic nn 0, 0, 0a b c a c b b c a+ - > + - > + - > v p dng BT
( )
1 1 4, 0x y
x y x y+ >
+
Ta c1 1 1 1 1 1
2 3Sb c a a c b b c a a b c a c b a b c
= + + + + +
+ - + - + - + - + - + -
2 4 6
c b a + +
Mc khc t gi thit2 1
2c b abc a
b c
+ = + =
Nn ta c2 4 6 1 2 3 3
2 2 4 3ac b a c b a a
+ + = + + = +
.Vy
3 4 54 3S
b c a a c b a b c= + +
+ - + - + -.
Do gi tr nh nht ca 4 3S= , khi v ch khi 3a b c= = = .T CHNA. Theo chng trnh chun
Cu 7a.Gi ( )M x y l ta trung im cnh BC, ta c3 1
12 2
AM AG M
=
uuuur uuur
ng thng BC qua1
12
M
v c VTPT1 4
3 3
AH -
uuurnn c phng trnh BC: 4 1 0x y- + = (1)
Gi ( )O x y l ta tm ng trn ngoi tip tam gic ABC, ta c7 1
3 6 6
OH OG O
= -
uuur uuur
Suy ra ng trn ngoi tip tam gic ABC c tm7 1
6 6
O
-
, bn knh170
36R OA= = c phng
trnh l ( )2 2
7 1 170:
6 6 36C x y
- + + =
(2)
Vy ta im B, C l giao im ca ng thng BC v ng trn ( )C chnh l nghim ca h
hai phng trnh (1) v (2) . Gii h ny ta c ( ) ( )1 0 , 3 :1B C-
Cu 8a. Ta c ( ) ( )
222 , 3,0,3AB AC -uuur uuur m . 0AB AC= uuur uuur tam gic ABC vung ti APhng trnh tham s ng thng AB l 1 , ,x t y t z t= + = = suy ra ta im
( )
2 2 21 3 1, 1E t t t AE t t t t t+ = + + = = = -
Khi ( )1 (211) 111t E AE = uuur
cng hng vi vec t ( )
222ABuuur
v 3 2 2AE AB= < = im E
nm trn cnh AB ca tam gic ABCPhng trnh tham s ng thng AC l 1 ', ', 'x t y t z t= - = = suy ra ta im
( )
2 21 '0 ' ' 0 ' 2 ' 1, ' 1F t t AF t t t t- = + + = = = -
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Khi ( )' 1 (001) 101t F AE = -uuur
cng hng vi vec t ( )3,0,3AC -uuur
v
2 3 2AF AC= < = im F nm trn cnh AC ca tam gic ABCDo mt cu i qua ba im A, E, F c AE v AF vung gc nhau th mt cu c bn knh nh nht
EF 6
2 2R = = v tm I l trung im EF
1 11
2 2I
Vy phng trnh mt cu cn tm l
( )2 2
2 1 1 31
2 2 2x y z
- + - + - =
Cu 9a. Gi s ta c mt s abc tha iu kin cho th r rng a, b, c c ly trong tp
{ }123456789X = v ta c mt cch duy nht sp xp theo ba s c th t tng dn, nn ba s
c chn l mt t hp chp ba ca 9 phn t trong tp X, vy s phn t trong tp E l 39E C=
Chn abc l s chn , ta c cc trng hp sau:+ 8c = th a, b c ly t cc s 1,2,3,4,5,6,7 nn ta c s cch chn trong trng hp ny l 27C .
+ 6c = th a, b c ly t cc s 1,2,3,4,5 nn ta c s cch chn trong trng hp ny l 25C .
+ 4c = th a, b c ly t cc s 1,2,3 nn ta c s cch chn trong trng hp ny l 23C .
Gi A l tp cc bin c s chn ca tp E th s cc phn t ca A l2 2 2
7 5 3A C C C= + + .Do xc su cn tm l :
2 2 27 5 3
39
17
42
A C C C
E C
+ += =
B. Theo chng trnh Nng Cao
Cu 7b.Ta nhn thy hai ng thng 1( )d v 2( )d vung gc nhau ti H(01) nn theo yu cu biton th tm I ca ng trn nm trn ng 1( )d v hai im B, C i xng nhau qua giao im H.
V tam gic ABC u 23
24 3 4 64ABC
S BC BC = = = .
Ta c 3 26 2 4 22 3
AH BC R IA AH= = = = = .
a ng thng 1( )d vit di dng tham s ta tm ca ng trn l ( 1 )I t t- , m
2 21 2 2 8 2 8 2, 23
IH HA IH t t t= = = = = = - . Vy ta tm ng trn l (2 1)I - hoc l
( 23)I - . Vy ta c hai ng trn cn tm l ( ) ( ) ( ) ( )
2 2 2 22 1 32, 2 3 32x y x y- + + = + + - = .
Cu 8b. Mt phng cha tam gic ABC i qua ba im (120)A , (110)G ,2 10
03 3
H
nn c
phng trnh z = 0. Gi ( )M x y z l ta trung im cnh BC, ta c
3 1
1 02 2AM AG M
=
uuuur uuur
Mt phng (P) cha ng thng BC qua1
1 02
M
v nhn VTPT1 4
03 3
AH -
uuurc phng trnh
l: 4 1 0x y- + = . Do phng trnh ng thng BC l giao tuyn ca hai mt phng 4 1 0x y- + =v z = 0.(1)Gi ( )O x y z l ta tm ng trn ngoi tip tam gic ABC, ta c:
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7 13 0
6 6OH OG O
= -
uuur uuurmt cu ngoi tip tam gic ABC c tm
7 1 0
6 6O
-
, bn knh
170
36R OA= = c phng trnh l ( )
2 2
27 1 170:6 6 36
C x y z
- + + + =
(2)
Vy ta im B, C l giao im ca ng thng BC (1) v ng trn ( )C (2) gii h hai phng
trnh ny ta c ( ) ( )
100 3 :10B C- .
Cu 9b. Phng trnh vit li2 3 2 3
2 02 2
x x + -
+ - =
t2 3 2 3
( ) 22 2
x x
f x + -
= + -
2 3 2 3 2 3 2 3'( ) ln ln
2 2 2 2
x x
f x + + - -
= +
V 2 22 3 2 3 2 3 2 3
''( ) ln ln 02 2 2 2
x x
f x x R + + - -
= + > "
Nn hm s 2 3 2 3 2 3 2 3'( ) ln ln2 2 2 2
x x
f x
+ + - -= +
ng bin trong ( )
- +
Mt khc hm s '( )f x lin tc trong ( )- + v '( ) '( )x f x x f x - - + +
Nn phng trnh '( ) 0f x = c ng mt nghim 0x v f(x) i du qua 0x nn phng trnh f(x) = 0
c nhiu nht hai nghim .Th li ta nhn thy phng trnh cho c ng hai nghim x = 0, x = 1.
.
( Cn mi)
S 5(Thi gian lm bi:180 pht)
Cu 1 (2,0 im).Cho ng cong ( )mC c hm s3 3y x x m= - + ( m l tham s thc).
a) Kho st s bin thin v v th ca hm s khi 2m = .b) nh tham s m qua im un ca th ( )mC k c mt ng thng ( )d to vi
th ( )mC mt hnh phng (H) v ( )d tip tc chn trn hai trc ta mt tam gic (T) sao cho din
tch ca (H) v (T) bng nhau u bng 2 (vdt) .
Cu 2 (1,0 im). Gii phng trnh ( )( )
2
tan .cot 2 1 s inx 4cos 4sin 5 .x x x x= + + -
Cu 3 (1,0 im). Tnh tch phn( )
( )
3
4
ln 4tan
sin2 .ln 2tanx
xI dx
x
p
p
= .
Cu 4 (1,0 im).
a) Trog trng hp khai trin theo nh thc Newton ca biu thc ( )21n
x+ ta c h s cha 8x bng
210. Tnh tng cc h s ca cc s hng c khai trin t biu thc trn theo trng hp .b) Cho cc s phc z tha mn 1 34z - = v 1 2z mi z m i+ + = + + . nh tham s m tn
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ti hai s phc 1 2,z z ng thi tha mn hai iu kin trn sao cho 1 2z z- l ln nht.
Cu 5 (1,0 im). Trong khng gian vi h trc ta Oxyz, qua hai im ( ) ( )1 11 , 0 10M N- - lp
phng trnh mt phnga ct mt cu ( )
2 2 2( ) 2 ( 1) ( 1) 5S x y z+ + + + - = mt thit din ng trn m
din tch hnh trn sinh bi ng trn c din tch S p= .Cu 6 (1,0 im). Cho hnh chp t gic S.ABCD, y ABCD l hnh vung cnh a, cnh bn
( )SA ABCD^ v SA = a. Qua A dng mt phnga vung gc vi SC sao choa ct SC, SB, SD ln
lt ti G, M, N.Tnh theo a th tch khi nn (H), bit rng ng trn y ca (H) ngoi tip t gic AMGN v nhO ca (H) nm trn y ABCD ca hnh chp S.ABCD.Cu 7 (1,0 im). Trong mt phng vi h trc ta Oxy, hy tnh din tch tam gic ABC bitrng hai im (55)H , ( )54I ln lt l trc tm v tm ng trn ngoi tip tam gic ABC v
8 0x y+ - = l phng trnh ng thng cha cnh BC ca tam gic.
Cu 8 (1,0 im). Gii phng trnh nghim thc( ) 2x ln x 2x 2 x 1- + = + .Cu 9 (1,0 im). Cho ba s dng x, y, z tha mn 0 x y z< < - ,
lc 3 nghim ca phng trnh (1) l 0, 3, 3x x k x k= = - + = + .V I l tm i xng ca ng cong ( )mC nn din tch ca hnh phng (H) l:
( )3
23
0
12 3 3
2
k
S kx m x x m dx k +
= + - + - = + ( )21
2 3 2 12
S k k = + = = - (v 3k> - ).
Lc ny ng thng ( )d vit liy x m= - + nn (d) ct hai trc ta ti hai giao im
( ) ( )
0 , 0A m B m . V (T) l tam gic vung cn nn din tch ca (T) l 212
S m=
theo gi thit 2 2, 2S m m= = = - .Vy c hai gi cn tm l 2, 2m m= = - .
Cu 2. iu kin :cos 0
sin 2 0 2
x kx
x
p
.
Ta c ( ) ( )
2 3tan .cot 2 1 s inx 4cos 4sin 5 tan .cot 2 3sin 4sin 1x x x x x x x x= + + - = - -
sin 3 11 tan .cot 2 sin 3 sin 3 sin 3 1 0
cos .sin 2 cos .sin 2
xx x x x x
x x x x
+ = = - =
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Nghim phng trnh xy ra :
hoc sin 3 03
nx x
p= = , so vi iu kin phng trnh c nghim l
2,
3 3x m x m
p p
p p= + = +
hocsin 2 1 sin 2 1
sin 2 . cos 1cos 1 cos 1
x xx x
x x
= = - = "
= = - v nghim
Vy nghim ca phng trnh trn l ( )2
, ,3 3
x m x m m Zp p
p p= + = + .
Cu 3. Ta c:( )
( ) ( )
3 3 3
4 4 4
ln 2 ln 2t anxln 2.
sin 2 .ln 2 t anx sin 2 .ln 2 t anx sin 2
dx dxI dx
x x x
p p p
p p p
+= = +
Tnh( )
( )
( )( )
3 33
44 4
ln 2tanxln 2 ln 2 ln 2 ln 2 3ln 2. . . ln ln(2 tan ) .ln
sin 2 .ln 2 t anx 2 ln 2 t anx 2 2 ln 2
ddxx
x
p p
p
p
p p
= = =
.
Tnh3 3
44
1 1ln(t anx) ln 3
sin 2 2 2
dx
x
pp
pp
= = .
Vyln 2 ln 2 3 1
.ln ln 32 ln 2 2
I
= +
.
Cu 4.a) . Khai trin biu thc trn c s hng th (k+1) l ( )2 ,k knC x k n
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( )( ) ( )
2 2
1 2
1 3463 , 4 3
3 5 3 0
x yM M
x y
- + = - -
- - =.
Vy hai s phc cn tm l 3 46 3 , 4 3z i z i= + = - - .
Cu 5.Mt cu (S) c tm ( 2 11)I - - v bn knh 5R = .Gi r l bn knh ng trn thit din, theo gi thit ta c 2. 1S r rp p p= = = .Gi d l khng cch t I n mt phnga ta c 2 2 2 5 1 2d R r d = - = - = .
Mt phnga qua ( )
0 10N - c dng ( ) ( )
2 2 2Ax 1 0 Ax 0 0B y Cz By Cz B A B C+ + + = + + + = + + .Mt khca qua ( )1 11M - nn tha 0 : Ax 0A C By Az Ba+ = + - + = .
V 2 22 2
3( , ) 2 4 2
2
A Ad d I A B
BA Ba
-= = = = =
+( v 2 2 2 0A B C+ + )
Do c hai mt phnga cn tm l : 2 2 1 0x y z+ - + = , 2 2 1 0x y z- - - = .
Cu 6. Ta c ( )
BC SABC SAB BC AM
BC AB
^ ^ ^
^( v ( )AM SAB ) (1)
Mt khc SC SC AM a^ ^ ( vAM a ) (2)
T (1) v (2) suy ra ( )AM SBC AM MG^ ^ ( v ( )MG SBC )AMG D vung ti M, tng t ta cng c tam gic ANGD vungti N tm H ng trn y ca (H) l trung im AG, c bn
knh2
AGR= . Xt tam gic vung SAC ti A c
. 6 6
3 6
SA ACAG a R a
SC= = = .
V OH l ng cao (H) / /OH OH SC Oa ^ l giao im hai ng cho AC, BD1
2OH CG = . Xt tam gic vung SAC c AG l ng cao , nn
2 2 3
33
ACCG a OH a
SC= = =
Vy th tch hnh nn l( )
2 31 3.3 54H
V R OH ap p= = .
Cu7 Ko di ng cao AH ln lt ct BC v ng trn ngoi tip tam gic ABC ti hai imE v K, ta d dng chng minh c E l trung im HK.ng caoAH BC^ nn c phng trnh 0x y- = , E l giao im ca BC v AH (44)E v H l
trung im HK (33)K , suy ra bn knh ng trn ngoi tip tam gic ABC l 5R IK= =
phng trnh ng trn l ( ) ( )
2 25 4 5, ( )x y C- + - =
Vy hai im B, C l nghim ca h hai phng trnh ng thng BC v ng trn( ) (3 5), (6 2)C B C v nh A l nghim h ca ng cao AH v ng trn ( ) (66)C ADin tch tam gicABCl
( )
6 6 81 1, . .3 2 6
2 2 2ABCS d A BC BC
+ -= = = (vdt).
Cu 8. iu kin 0x > ta c( ) ( )22
x 1x ln x 2x 2 x 1 x ln x
2x 2
+- + = + - =
+
Xt hm s2
x 1f(x)
2x 2
+=
+
/ /
2 2
1 xf (x) f (x) 0 x 1
(x 1) 2x 2
- = = =
+ +
H
N
G
M
O
S
D
CB
A
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Lp bng bin thin ta c ( ) 1, 0f x x " > , ng thc xy ra khi x = 1.
Xt hm s1 1
( ) ln '( ) 1 '( ) 0 1x
g x x x g x g x xx x
-= - = - = = = .
Lp bng bin thin ta c ( ) 1, 0g x x " > , ng thc xy ra khi x = 1.Vy phng trnh c ng mt nghim x = 1.
Cu 9 Ta c
3 3
215
x yy z
zP x y x y zxy z y z x
= + + + + +
. t , , . . 1, 1.x y z
a b c a b c cy z x= = = = >
Biu thc vit li3 3
2 15a bP ca b a b c
= + + ++ +
Ta c ( )
3 33 3 1a ba b ab a b ab
a b a b c+ + + =
+ +( v a, b > 0 ).
Vy ( )2 21 15 16
( ), 1P c c f c cc c c
+ + = + = " +
Ta c2
16'( ) 2 '( ) 0 2f c c f c c
c= - = =
Lp bng bin thin ta c ( ) (2) 12,f c f = khi v ch khi1
2 2 22
c a b z y x= = = = = .
Vy gi tr nh nht 12P= khi v ch khi 2 2z y x= = .
+ chc bn c vui, khe v nht l cc em 12 nm nay 20014-2015 s thnh cng tt p trong ccma thi sp ti.+ cui (S 5), mnh gi cho Ton Hc Tui tr ri, c g sai st mong cc bn gp mnh b sung he he he( Bi vit , thng thng ta son bo phi nhn trc, trc khi ln mng nhng vv hc sinhthn yu he he he )
Nguyn Li