Tnr(mg
THPT
Chuyen
Le
Quy
Don
DE
THI
THU
D~I
HOC
LAN
3 -
MON
TOAN
KHOI
A
Tinh
Ba
Ria
Viing
Tau
Thai
gian him bai: 180 phut.
I.
pHAN
CHUNG
CHO
TAT
CATHi
SINH:

Cau
II
(2 di8m):
1.
Giai
phuong
trinh x
3
-6x
2
+ 12x
-7
={j_x
3
+9x
2
-19x
+
11
.
2sinx
+ 1
cos2x
+
2cosx
7
sinx
+5
2.
Gi1ii

III
bc
p an sau: =
x
Cau
IV
(l
di~m):
Cho
hlnh chop SABCD co
ABCD
1a
hinh binh hanh tam
0,
AB = 2a,
AD
2aJ3
,
c~c
Cl;l.llh
ben
b~ng
nhau
va
bk
g
3a,
gOi
M
1a

Tim
GTNN
va
GTLN
ella
P =
X4
+ l x
2
y2.
n.
pHAN
RIENG
(3
diim)
Thi
sinh
chi
dU'Q'c
lam
m9t
trong
hai
phAn (phAn 1 ho,"c phAn 2).
1.
Theo
ChU'01l~
trinh
chuAn:
Cau

Tim
toa
dQ
cac dlnh A, B, C bi8t
~g
C co hoanh
dQ
1a
m¢t
s6
nguyen.
2.
Trong
khong gian
Oxyz
cho hai
duang
th~ng
Cd
1
):
~
-1
72=Z~:,
(d
2
):
{;=~::
2
Z

MN
=
3.J6
.
Cau
VII.a
(1
di~m):
Tim
t~p
hQ'P
cac diam
biau
di€n cho
s6
phuc
Z th6a man M
tMc:
Iz
+ 3 - 2il
12z
+
1-
2il
2.
Theo
chU'O'D~
tdnh
nang
cao:

ABC
bi~t
XB <
Xc
.
.
x-I
Y+2 Z 2
{X
2 - t .
~
2.
Trong
khong gian
Oxyz
cho hai
duang
thang
(d!):

;;;
=
,
(d
2
):
y = 3
+t
va
m~t

,{eX
e
Y
(In y
In
x)(l
+xy)
G
1a1
he phuong trmh .
.
21nH2lny
_ 3.4
ln
:<
;;;
4.iny
Thi sinh
kMng
(/l.f9c
sa
dl,mg
tai
lif?u
Can b9 coi thi
kMng
gial thlch gi them.
HQ
va
ten thf

NQidung
Cho ham so y =x
j
-
3x" + 3.
Khao sat
S\l bien thien va ve do
thiJC)cua
ham so.
T~p
xd va
Gi61
h~
y'
=3x"
6x
y'
::;
0
~
x =0 hay x =2
Bang
bien thien:
y"
va diem uon
Gia tri
d~c
bi~t
Do thi
va

-6x
k
=>
x
3
3x
2
+ 3 =
3x
3
-
6x
2
+
3x
2
-
6x
1
~
2x
3
-
6x - 4 = 0
~
X = 2 hay x =
-1.
,.
x = 2
=>

6 { Y
~
x
3
-
6x'
+
12x-7
Iota
c
y3=_x
3
+9x
2
19x+ll
=>
y3
+ 2y =x
3
-
3x
2
+ 5x - 3
Q
y3
+ 2y = (x
_1)3
+ 2(x
-1)
(1)

._.
-
Di~m
2:=2d
"L=.1.25d
0.25
0.25
0.25
0.25
0.25
'L= 0.75d
0,25
0.25
0.25
2:=24
:L=ld
I
!
0.25
i
0.5
0.25
www.MATHVN.com
www.mathvn.com
lx
~j
¢::>
x=2.
x=3
I

<=>
(cosx+l)(2cosx y3):;tO
<=>
cosx:;e
_~
-
0.25
(1)
<=>
(2sinx + 1)(cosx +
1)
==
cos2x + 2cosx
-7sinx
+ 5
<=>
2sinxcosx + 2sinx + cosx + 1 = 1 - 2sin
2
x + 2cosx - 7sinx + 5
0.25
<=>
2sinxcosx - cosx +
~
= 0
<=>
cosx(2sinx -
1)
+ (2sinx - 1)( sinx + 5) = 0
<=>
(2sinx - 1)( cosx + sinx + 5) = 0

~:
+
k21t
(k E Z).
0.25
Call
III
rnh
'h
han
I
fX3~X3+8+(6X3+4X2)lnxd
I
1
tIC
P sau:
==
x
2.:
= Itt
,
'x
f
X2~X3
+ 8 dx
~
12
(6x
2
+4x)ln

D6i
c~n:
x
1
=>t=3
x = 2
=>
t =4,
[ r
0.25
2 2
t
3
2
h
do
II
rt.3"tdt =3"'
3"
3
==
g(
64-27)
==
-,
$
).
Tinh h
D~t
u = lnx

I
241n2-
-3-+
x2
1 = 241n2
-
3
J
A 77 23 2
.J t"
S-
0.25
I
!
VayI=
241n2+ =241n
+-
I
J
. 9
39
www.MATHVN.com
www.mathvn.com

-

~"""

r
-~

di~n
tich cua hinh
cAu
ngo~i
ti~p
tii'
di~n
SOCD.
~
Ta
co SA = SB =SC =SO nen SO 1. (ABC D).
,
~
/),.
SOA = .=
/),.
SOD nen
OA
::;;
OB
OC =
00
=>
ABCD
1ft
hinh
chu
nhat.
0.25
•

0.25
A_I
4a
3
J15
, _ 3 3
c:
V~y
VSABCD
-
"3SABCD'SO
==
3 . Do do
VSABMD
-
"4
VSABCD
= a ;15.
~
GQi
G la
trQng
tam
/),.
OCD, vi
/),.
OCD deu nen G cling
1ft
tam
dUOng

tT\Ic
cua SO
=>
KO::;;
KS
rna
KO =
KC
=
KD
nen K
18.
tam
mi},t
cAu
n
o~i
tiS
ill
di~n
SOCD.
Taco:
GO::;;
CD
=
2a
fj
fj
0.25
R=KO=

=
-4t
+ 2.
1
f(t)
= 0
¢::>
t = .
2
f(-,,!,) =
!
fO) = 1
va
f(
!)'::;;~.
3
9'
2 2
Yay MaxP =
maxf(t)
= l
va
Min
P
=::
minf(t)
= 1

[-1;1]
2 [-1;1) 9

Dinh A thuoc tia Oy, duemg cao ve tir C nfun tren duemg thfutg (d): X + 5y
O.
I=ld
i
Tim
tQa
do
cac
dinh A, B, C
bi~t
rfutg C co
hO<lnh
do h\ mot
s6
nguyen
» A thuoc
tia
Oy nen
A(O;
a)
(a>
0).
, Vi A E (T) nen
a'
- 2a - 8
~
0
""
[a
4

~
:::::>
x = _ 20
:::::>
C(5;
-1)
(Do
Xc
E Z)
13
13
» (AB)
1-
(d)
nen
(AB): 5x - y + m = 0 rna (AB)
qua
A nen 5.0
4+m
0
:::::>m==4.
V~y
(AB): 5x
-y
+ 4
O.
B E (AB)
:::::>
B(b; 5b + 4).
[b-O

:::::>
B(-I;
-1)
(nh~).
0.25
i
V~yA(O;
4),
B(-I;
-1)
va
C(5;
-1).
2
r2
t
Trang
kg
Oxyz cho (d}):
x
1
+2
z 2 , _
,:
=
1 =
-2
,(d
2): y=:03+t
vam~tphang

-2
+ m; 2
2m)
I
NE(d
2
):::::>N(2
n; 3 + n; 4 + n)
:::::>
NM
==(2m+n-l;m-n
-
5'-2m-n
-2)'
n
=(1'-1'1)
, ,
a ' ,
0.25
-
MN
II
(a)
:::::>
na .NM =0
<=>
2m
+ n - 1
-em
n - 5)

2m
+ 2 = 6
<=>
m
2
m-2
o
<=>
m
-1
hay m =
2.
0.25
i
~-
x+l
y+3
z-4
I
» m =
-1:
M(-I;;Zj
4)
va
NM
-3(Ui::-1)
~
Ed):·

-""

VIl.a
Tim
t~p
hqp
cac dibm
bi~u
dien cho
s6
phuc z th6a:
Iz
+ 3
2il
=
12z
+
1-
2il
I=ld
!
G9i M(x; y)
la
di~m
bi~u di~n
cho
s6
phuc z =x + yi (x; y E R).
Taco:
Iz+ 3 -
2i
l=

V~y
t?P
hqp
cac diem
_~
ia
duang
tron (T): 3x- +
3l-
2x - 4y 8 =
O.
0.25
VI.b
I,=2d
I
Trong
m,t
phing
Oxy. cho tam giac ABC
cO
dinb A(
0.4
ttrong
tim
G (
~;
~
)
I=ld
,

2
Y
-4
~
~
4
YI
1
I 2 3
BC qua I
va
c6 VTPT
la
OA
= (0;4) = 4(0;1)
:=;>.
BC: y
=-1.
GQi
B(b;
-1),
vi I
la
trung
di~m
BC nen C(4 b;
-1).
-
-
Ta

b=-l
:=;>B(-1;-l)vaC(5;-I)(nh~)
I * b =
5:=;>
B(5;
-1)
va
C(-I;
-1)
(lo~i)
0.25
»
BC::::
(
6;
0)
:=;>
BC
6; d(A; BC)
=5
:=;>
SABC = 15.
I 0.25
2
r
2
-
t
x-I
+2

v6i.
(d
l
),
c~t
(a)
4ti N sao cho
MN
3.
M
E (d
2
):=;>
M(2
m; 3 + m; 4 + m).
r=2-m+2t
(d) qua M
va
II (dJ)
nen
(d):
y:::::
3+m+t
0.25
z
4+m-2t
N (d) n
(a)
nen
tQa

(6
+ 2m; 3 + m;
-2m
6)
0.25
I I
:=;>
NM2
(2m +
6)2
+
(m
+
3)2
+
(-2m-
6i
I
Do
d6
MN
==
3
~
9(m +
3i
= 9
~
m + 3 = ± 1
~

ki~n:
x, y >
o.
I
Ta
c6: 1 + xy >
O.
*
x>
y:
VT
(1) > 0 va
VP(I)
<
O:=;>
VT(I)
>
VP(l)
(voU)
* x <
y:
VT(1) < 0 <
VP(l)
(vo
Ii)
0.25
Do d6 tir (1)
:=;>
x =
y.

~
0.25
i
2
1nx
==
4
'-:~
V~y
Mc6
nghi~m
la
x y
e.l.
,

_

_1
0.25 I
BET
www.MATHVN.com
www.mathvn.com