phân loại và phương pháp giải đại số 10

NGUYEN
ANH
TRUdNG - NGUYEN
PHU
KHANH
DAU
THANH KY - NGUYEN
MINH
NHIEN
NGUYEN TAN
SIENG
-
DO
NGOC
THUY
(Nhdm
gi£o
viSn
chuy§n
todn
THPT)
PHAN
LOAI
&
PH
V0N6
PHAP
6IAI
10
1/'
(Tai han On nb^t c6

cbinh
If
vk
ho
sung)
Danh cho hoc sinh Idp 10 on tap
va nang cao kien thufc
Bien
soan
theo
noi dung, chifofng
trinh
sach
gi^o khoa Bo GD & DT
Cac em hoc sinh than men!
'Than loai va phuong phap giai Dai so 10" la mot trong nhung cuon
thuoc bo
sach
"Phdn
loai
vd
phuong
phdp
gidi:
lap 10, 11, 12 ", do nhom tac
gia chuyen toan THPT bien
soan.
V6i
each
viet khoa hoc va sinh dong giup ban doc tiep can v6i mon Toan

mpt
each
tu nhien, khong ap luc, ban doc tro nen tu tin va nang dong hon;
hieu ro ban
chat,
biet
each
phan tich de tim ra trong tam cua van de va biet giai
thich,
lap luan cho tung bai toan. Sir da dang ciia he thong bai tap va
tinh
Huong
giup ban doc iuon hiVng thu khi giai toan.
Tac gia chii trong bien
soan
nhCing cau hoi mo, no! dung co ban bam sat
sach
giao
khoa va cau true de thi dai hoe, dong thoi phan bai tap thanh cac
dang toan co loi giai chi tiet. Hien nay de thi dai hoc khong kho, to hop cua
nhieu van de don gian, nhung chiia nhieu cau hoi mo ne'u khong nam
chae
ly
thuyet se lung
tiing
trong
viee
tim loi giai bai toan. Vol mot bai toan, khong
nen thoa man ngay voi mot loi giai minh vua tim
duoc

ma phai co gang tim
nhieu
each
giai nha't cho bai toan do, moi mot
each
giai se co them phan kien
thuc moi on tap.
Mon
Toan la mot mon rat ua phong
each
tai tu, nhung phai la tai tu mot
each
sang
tao va thong minh. Khi giai mot bai toan, thay vi
diing
thoi gian de
luc Ipi tri nho, thi ta can phai suy nghl phan tich de tim ra phuong phap giai
quyet bai toan do. Doi voi Toan hoc, khong co trang
sach
nao la thua. Tung
trang^ tung dong deu phai hieu. Mon Toan doi hoi phai kien nhan va ben bi
ngay tu nhCrng bai tap don gian nha't, nhirng kien thuc ca ban nha't. Vi chinh
nh&ng
kien thuc co ban moi giup ban doc hieu
duoc
nhi>ng kien thuc nang cao
sau nay.
Ludwig
Van
Beethoven:

"Giot
nuoc
co the lam mon tang da, khong
phai
vi giot nude co sue manh, ma do
nuoc
ehay
lien tuc ngay dem. Chi co su
phan da'u khong met moi moi dem lai tai nang. Do do ta eo the khSng
dinh,
khong nhich tung bude thi khong bao gid cd the'di xa ngan dam".
Mae dii tac gia da danh nhieu tam huye't cho cudn
sach,
song
su sai sdt la
dieu
khd tranh
khdi.
Chung toi rat mong nhan
duoc
su phan bien va gdp y quy
bau eiia quy doe gia de nhu-ng Ian tai ban sau eud'n
sach
duoc
hoan thien hon.
Thay mat nhdm bien
soan
Nguyen Phu Khanh.
Cty TNHH MTV DWH
Khang

Vi.'(
l^huon^
1 MENH DE - TAP HOOP
§1.
MENH DE VA MENH DE CHUA BIEN
A. TOM
TATLY
THUYET
<i)inhnghia:
^
r;,^ ; ,! ; '
Mln/j
(fe la mot eau khang
dinh
Diin^hoac Sfli. ,, ,-, ^
Mot
menh de khong the vua
diing
hoae
vua sai ^ r
2.
M$nh
de phu
dinh:
Cho menh de P. Menh de "Khong phai P " goi la
menh
dephu
dinh
cua P.
Ky

hieu la P. Ne'u P dung thi P sai, ne'u P sai thi P dung
3.
Menh
de keo
theo
vd
menh
de ddo
Cho hai menh de P va Q.
NJenh
de "ne'u P thi Q" goi la
menh
dekeo
theo
Ky
hieu la P => Q. Menh de P => Q chi sai khi P dung Q sai
Cho menh de P => Q. Khi dd menh de Q => P goi la
menh
de ddo eua
P=*Q
4.
Menh
de
tuorng
duong
Cho hai menh de P va Q. Menh de "P ne'u va chi ne'u Q" goi la
menh
de
tuong
duang

Ky
hieu la Pc^Q. ;.S
M^nh
de P
<=>
Q dung khi ca P => Q va Q => P ciing
diing
Chu y: "Tuong duong" con duge goi bang cac thuat ngu
khae
nhu "dieu
kien
can va
dii",
"khi va chi khi", "ne'u va chi ne'u".
5.
Menh
de
chiia
bien
Menh de ehiia bien la mot cau khMng djnh
chua
bien nhan gia trj trong mpt
tap X nao dd ma vdi moi gia tn cua bien thuoc X ta
duoc
mgt m^nh de.
Vi
dy: P(n): "n
chia
het cho 5" vdi n la sd'tu nhien
P{x; y):" 2x + y = 5" vdi X, y la so thuc

6. Cac ki hi^u V , 3
menh
de phu
dinh
cua
menh
de cd
chua
ki
hieu
V ,3 .
Ki
hieu V: doc la vdi moi, 3: doc la ton tai
Phii
djnh cua
m^nhde"VxeX,P(x)
" la menh de " 3x e
X,P(x)"
Phu djnh ciia menh
de"3x€X,P(x)
" la menh de " Vx e
X,P(x)
" ^ ,
Phdn
loai
va
phuamg
phdpgiai
Dai so'10
B.

CAC
DANG
TOAN
VA
PHl/QNG
PHAP
GIAI.
|DANG
TOAN
l. XAC DINH MENH DE VA TJNHDUNG SAI CIJA MENHDE
Ca 1. CAC Vf DU MINH HOA
Vi
d\ 1: Cac cau sau day, cau nao la menh de, cau nao khong phai la menh
de? Ne'u la menh de hay cho biet menh de do
diing
hay sai.
(1) O day dep qua!
(2) Phuong
trinh
- 3x +1 = 0 v6 nghiem
(3) 16 khong la so nguyen to
(4; .ai phuong
trinh
- 4x + 3 = 0 va x^ - Vx + 3 + 1 = 0 c6 nghiem chung.
(5) So 7t
CO
Ion hon 3 hay
khong?
(6) Italia v6 dich Worldcup
2006

(7) Hai tam
giac
bang nhau khi va chi khi chiing c6 dien tich bang nhau.
(8) Mot tu
giac
la
hinh
thoi
khi va chi khi no c6 hai duong
cheo
vuong goc
v6i
nhau.
Lai
gidi
Cau (1) va (5) khong la menh de(vi la cau cam than, cau hoi)
Cac cau (3), (4), (6), (8) la nhung menh de dung
Cau (2) va (7) la nhifng menh de sai.
Vi
du 2: Cho ba menh de sau, voi n la so tu nhien
(1) n + 8 la so chinh phuong
(2) Chu so tan ciing ciia n la 4
(3) n-l la so chinh phuong
Biet
rang c6 hai menh de
diing
va mot menh de sai. Hay xac djnh menh
de nao dung, menh de nao sai.
Lcn
gidi

Ta
CO
so chinh phuong c6 cac dm so tan cung laO, 1, 4, 5, 6, 9 . Vi vay
- Nhan thay giua menh de (1) va (2) c6 mau thuan. Boi vi, gia sir 2 menh de
nay dong thai la dung thi n + 8 c6 chir so tan ciing la 2 nen khong the la so
chinh
phuong. Vay trong hai menh de nay phai c6 mot
mf
nh
de la
diing
va
mpt
m^nh de la sai.
- Tuong tu, nhan thay giiia menh de (2) va (3) cung c6 mau thuan. Boi vi, gia
sir m?nh de nay dong
thoi
la
diing
thi n - 1 c6 ch\x so tan cung la 3 nen
khong
the la so chinh phuong.
Vay trong ba menh de tren thi menh de (1) va (3) la
diing,
con menh de (2)
la sai.
IT
Cty
TNHH
Ml v t>\ Kliang

\
ga 2. BAI TAP
LUY|N
TAP
Bai 1.0: Cac cau sau day, cau nao la menh de, cau nao khong phai la m^nh de?
Ne'u
la menh de hay cho biet menh de do dung hay sai.
a) Khong
di\xgc
di loi nay! . , ? '• .
b) Bay gio la may
gio?
;n:|r'c<Ho y
c) Chieh tranh the
gioi
Ian
thii
hai ket thiic nam
1946.Ir
„:; v!
=\
d)
16chia3dul.
e)
2003
khong la so nguyen to. ,.
£) v'5 la so v6 ti.
g) Hai duong
tron
phan biet c6 nhieu nhat la hai diem chung. ^

Huang
dan
gidi
Cau khong phai m^nh de la a), b) d,^ -
Cau d), f) la menh de dung. Cau e) sai. Cau g)
diing
Bai 1.1: Tai Tiger Cup 98 c6 bon doi lot vao vong ban ket: Vi^t Nam,
Singapor,
Thai Lan va Indonexia.
Truoc
khi thi dau vong ban ket, ba ban Dung,
Quang, Trung du doan nhu sau:
Dung:
Singapor
nhi, con Thai Lan ba.
Quang:
Viet Nam nhi, con Thai Lan tu. .
Trung:
Singapor
nhat va Indonexia nhi.
Ket qua, moi ban du doan dung mot doi va sai mpt doi. Hoi moi dpi da dat
giai
may?
Huang
dan
gidi
,, -
Ta
xet du doan ciia ban Dung
+ Ne'u

Singgapor
nhi thi
Singapor
nhat la sai do do Indonexia nhi la
dung(mau thuan)
+ Nhu vay Thai lan
thii
ba la dung suy ra Vi?t Nam nhi,
Singapor
nhat va
Indonexia
thii
tu
DANG
TOAN
2: CAC
PHEP
TOAN
VE
MENH
DE.
Cac
phep
todn
menh
deduce
sit
dung
nham
muc

dich
ket not cac
menh
delai
vox
nhau
tao ra mot
menh
de moi. Mot so cac
phep
todn
menh
de la :
Menh
de phu
dinhiphep
phu djnh),
menh
dekeo
theoiphep
keo
theo),
menh
deddo,
menh
detitang
duongiphep
tuvngducmg).
Phan
lo^i vA

phucmg
phdp
gidi
Dai sd'lO
Ca 1. CAC Vf DU
MINH
HOA
Vi
1: Neu m?nh de
phii
djnh cua cac m^nh de sau, cho bie't
mf
nh
de nay
diing
hay
sai?
P: "
Hinh
thoi
c6 hai duong
cheo
vuong goc vai nhau"
Q: " 6 la so nguyen to"
R: " Tong hai canh cua mpt tarn
giac
Ion han canh con lai"
S:
"5>-3"
K:

"Phuong
trinh
x'* - 2x^ + 2 = 0 c6 nghi^m "
H:
" (x/3 - x/l2)^ - 3 "
Lai
gidi
Ta
CO
cac m^nh de
phii
dinh
la
P: "Hai duong
cheo
ciia
hinh
thoi
khong vuong goc voi nhau", menh de
nay sai
Q
: "6 khong phai la so'nguyen to", menh de nay dung
R: "Tong hai canh ciia mot tarn
giac
nho han
hoac
bang canh con lai",
mf
nh
de nay sai

S: " 5 < -3", menh de nay sai
K:
" phuang
trinh
- 2x^ + 2 = 0 v6 nghi^m", m|nh de nay dung vi
x'^-2x^ +2 = (x^ if
+l>0
H:
" (V3-Vi2) =3 ", minhde nay sai
Vi
dv 2:
Phat
bieu m^nh de P => Q va phat bieu m^nh de dao, xet
tinh
diing
sai ciia no.
a) P: "Tu
giac
ABCD la
hinh
thoi"
va Q:" Tu
giac
ABCD, AC va BD cat nhau
tai
trung
diem moi duong"
b) P: "2>9" va
Q:"4<3"
c) P: "Tam

giac
ABC vuong can tai A" va Q:" Tam
giac
ABC c6 A = 2B "
d)
P: "Ngay 2 thang 9 la ngay
Quoc
Khanh ciia
nuac
Viet Nam" va Q: "Ngay
27 thang 7 la ngay thuong binh
lift
si"
Lai
gidi
a) Menh de P Q la "Neu tu
giac
ABCD la
hinh
thoi
thi AC va BD cat nhau
tai
trung
diem moi duang", menh de nay dung.
M?nh de dao la Q
ri>
P: "Neu tu
giac
ABCD c6 AC va BD c^t nhau tai
trung

diem
moi duang thi ABCD la
hinh
thoi",
menh de nay sai.
b) Menh de P => Q la "Neu 2 > 9 thi 4 < 3", menh de nay
diing
vi menh de P sai.
Menh
de dao la Q => P: "Neu 4 < 3 thi 2 > 9", menh de nay
diing
vi m^nh de
c) Menh de P => Q la "Neu tam
giac
ABC vuong can tai A thi A = 2B ", menh
de nay
diing
Menh
de dao la Q => P: "Neu tam
giac
ABC c6 A = 2B thi no vuong can tai
A",
menh de nay sai
d)
Menh de P => Q la "Neu ngay 2 thang 9 la ngay
Quoc
Khanh ciia nuoc Viet
Nam
thi ngay 27 thang 7 la ngay thuong binh liet sT" ' '
Menh

de dao la Q => P: "Neu ngay 27 thang 7 la ngay thuong binh liet sT thi
ngay 2 thang 9 la ngay
Quoc
Khanh ciia nuoc Viet Nam"
Hai
menh de tren deu
diing
vi menh de P, Q deu
diing
Vi du 3:
Phat
bieu menh de P o Q bang hai
each
va xet
ti'nh
diing
sai ciia no
a) P: "Tii
giac
ABCD la
hinh
thoi"
va Q: "Tu
giac
ABCD la
hinh
binh hanh
CO
hai duong
cheo

vuong goc voi nhau"
b) P: "Ba't phuong
trinh
Vx^ -3x > 1 c6 nghiem" va Q: " ^(-1)^ -3.(-l) > 1 "
Lai
gidi
a) Ta
CO
menh de P <=> Q
diing
vi menh de P => Q, Q => P deu
diing
va duoc
phat bieu bang hai
each
nhu sau:
"Tii
giac
ABCD la
hinh
thoi
khi va chi khi tii
giac
ABCD la
hinh
binh hanh
CO
hai duong
cheo
vuong goc voi nhau" va

"Tu
giac
ABCD la
hinh
thoi
neu va chi neu tii
giac
ABCD la
hinh
binh hanh
CO
hai duang
cheo
vuong goc voi nhau"
b) Ta
CO
menh de P
<=>
Q
diing
vi menh de P, Q deu dung(do do m^nh de P => Q,
Q
=> P deu diing) va dugc phat bieu
bSng
hai
each
nhu sau:
"Bat phuong
trinh
\/x^ -3x > 1 c6 nghiem khi va chi khi -3.(-\) > 1"

va "Bat phuong
trinh
-3x >1 c6 nghiem neu va chi neu ^(-1)^ -3.(-l) > 1"
ea
2.
BAI
TAP
LUYIN TAP
Bai 1.2: Neu menh de phu
dinh
ciia cac menh de sau, cho bie't menh de nay
diing
hay
sai?
P: "Trong tam
giac
tong ba goc bang 180""
Q:"(V3-V27)^ la so
nguyen"
R: "Viet Nam v6 djch Worldcup
2020"
' '
Phdn
loai
va
phuong
phdpgidi
Dai so 10
S:"-^>-2"
2

K:
"Bat phuong
trinh
x^°" >
2030
v6 nghiem "
Huong
dan
gidi
'
'^Mat*
Ta
CO
cac menh de phu
dinh
la
P: " Trong tam
giac
tong ba goc khong bang 180°", menh de nay sai
Q
: " (N/3 - \/27) khong phai la so nguyen ", menh de nay sai
R:" Viet Nam khong v6 djch Worldcup
2020",
menh de nay chua xac djnh
dirge
diing
hay sai
S: "<-2", menh de nay dung
K
:" Bat phuong

trinh
x^"^'^ >
2030
c6 nghiem ", menh de nay dung
Bai 1.3:
Phat
bieu menh de P => Q va phat bieu menh de dao, xet
tinh
diing
sai
ciia no.
a) P: "Tu
giac
ABCD la
hinh
chii
nhat" va Q: "Tu
giac
ABCD c6 hai duong
thang AC va BD vuong goc v6i nhau"
b) P: "-V3>-V2" vaQ:
"{-^fsf
>
{-^f
"
c) P: "Tam
giac
ABC c6 A = B + C " va Q: "Tam
giac
ABC c6 BC^ = AB^ + AC^ "

d)
P: "To
Hiru
la nha Toan hoc Ion cua Viet Nam" va Q: "Evariste Galois la nhk
Tha loi lac cua The'gioi"
Huang
dan
gidi
a) P => Q: " Neu tu
giac
ABCD la
hinh
chu nhat thi tu
giac
ABCD c6 hai
duong
thang AC va BD vuong goc voi nhau", menh de sai
Q
=> P: " Neu tu
giac
ABCD hai duong thang AC va BD vuong goc vol
nhau thi tu
giac
ABCD c6 la
hinh
chCr nhat", m?nh de sai
b) P ^ Q:" Neu -73 > -y/2 thi [-^f > [-^f ", menh de dung
P ^ Q:" Neu (-^f
>
(-^flf

thi - > -72 ", menh de sai
c) P=>Q: "Neu tam
giac
ABC c6 A = B + C thi tam
giac
ABC c6
BC?=AB?+AC?"
Q
=> P: "Neu tam
giac
ABC c6 BC^ = AB^ + AC^ thi A = B + C "
Ca hai menh de deu
diing.
d)
P^Q:" Neu To Huu la nha Toan hoc Ion cua Viet Nam thi Evariste Galois
la nha Tho loi lac cua The'gioi", Q => P :" Neu Evariste Galois la nha Tho loi
lac ciia The'gioi thi To
Hiru
la nha Toan hpc Ion ciia Viet Nam ". Hai m^nh
de
diing.
Cty TNHHMTV
nvv)l
Khang
Vict
Bai 1.4:
Phat
bieu m^nh de P
<=>
Q bang hai

each
va va xet
tinh
diing
sai ciia no
a) Cho tu
giac
ABDC. Xet hai menh de |
P: "Tu
giac
ABCD la
hinh
vuong".
Q: "Tii
giac
ABCD la
hinh
chu nhat c6 hai duong
cheo
vuong goc voi nhau".
b) P: "Bat phuong
trinh
x^ -3x + l >0 c6 nghiem" va Q: " Bat phuong
trinh
x^ - 3x +1 < 0 v6 nghiem"
Huang
dan
gidi
a) Ta
CO

menh de P
<=>
Q
diing
vi menh de P Q, Q => P deu
diing
va
dugc
phat bieu bang hai
each
nhu sau:
"Tu
giac
ABCD la
hinh
vuong khi va chi khi tii
giac
ABCD la
hinh
chu
nhat
CO
hai duong
eheo
bang vuong goc voi nhau " va
"Tu
giac
ABCD la
hinh
vuong neu va chi neu tu

giac
ABCD la
hinh
chii
nhat
CO
hai duong
cheo
vuong goc voi nhau "
b) Ta
CO
menh de P o Q sai vi menh de P
diing
con Q sai.
Phat
bieu menh de P
<=>
Q b3ng hai
each
"Bat phuong
trinh
x^ - 3x +1 > 0 c6 nghiem khi va chi khi ba't phuong
trinh
x^ - 3x +1 < 0 v6 nghiem" va "Bat phuong
trinh
x^ - 3x +1 > 0 c6 nghiem
neu va chi neu ba't phuong
trinh
x^ - 3x +1 < 0 v6 nghiem"
Bai 1.5: Cho cac menh de : < .

a-\/3
A
: "Neu AABC deu c6 canh bang a, duong cao la h thi h = " ;
B : "Tii
giac
CO
bon canh
bSng
nhau la
hinh
vuong" ;
C: "15 la so nguyen to" ;
D
: "
>/T25
la mpt so nguyen".
a) Hay cho bie't trong cac menh de sau, menh de nao
diing,
menh de nao sai :
A
=^ B, A ^ D, B =^ C.
b) Hay cho bie't trong cac menh de sau,
mf
nh
de nao
diing,
menh de nao sai :
A<=>
B, B C, B o D.
Huang

dan
gidi
Ta
CO
A va D la cac menh de
diing,
B va C la cac m^nh de sai. Do do ;
a) Menh de A=>B sai vi A
diing
va B sai. ' >
Menh
de A D
diing
vi A va D deu
diing.
. ,
Menh
de B
=:>
C
diing
vi B sai.
b) Menh de A B sai vi menh de A B sai (Hoac A
diing
va B sai). Menh de
B
<=>
C
diing
vi hai menh de B va C deu sai.

Menh
de A
Ci>
D
diing
vi hai m^nh de A va D deu
diing.
Phiht
loai
va
pltucnig
phiip
giiii
Dai so'10
Bai 1.6: Hay phat bieu menh de keo
theo
P => Q, Q => P va xet
tinh
diing
sai
cua menh de nay-
a) Cho tiV
giac
ABCD va hai menh de: ' ' •
P: "Tong 2 goc doi cua tu
giac
loi bang 180" " va Q: " Tu
giac
noi tiep duoc
duong

tron
".
b) P :" >/2 - >
-1"
va Q: "(^/2 - Vsf > (-1)' "
Htt&ttg dan
gidi
a) P => Q : " Neu tong 2 goc doi cua tiV
giac
loi bang 180" thi tiV
giac
do npi tiep
du(?c
duong
tron
".
Q
P: "Neu tu
giac
khong noi tiep duong
tron
thi tong 2 goc doi cua tu
giac
do bang 180""
Menh
de P => Q dung, menh de Q => P sai.
b) P Q: " Neu V2 - V3 > -1 thi (V2 - Sf > {-if "
Q=>P:"Neu (V2 - Vs)^ < (-1)^ thi V2->/3>-l "
Menh
de P => Q sai vi P dung, Q sai, menh de Q => P

diing
vi P va Q deu
dung.
DANG
TOAN
3: MENH DE CHUA BIEN VA MENH DE CHUA
KiHIEU
V, 3.
ffll.CAC
Vl DU
MINH
HOA
Vi
du 1: Cho m^nh de
chua
bien "P(x): x > x"'", xet
tinh
dung sai ciia cac
menh de sau:
a) PO)
c) VxeN, P(x)
b) P
d)
3xeN, P(x)
Lai
gidi
a) Ta
CO
P(l): 1 >
1^

day la menh de sai
b)
Taco
p'''
1
3J
1
:
—
>
3
day la menh de
diing
c) Ta
CO
Vx € N, X > x'"* la m^nh de sai vi P(l) la menh de sai
d)
Ta
CO
3x € N, x > x'' la mfnh de
diing
vi x - x"' = x(l - x)(l + x) < 0 voi moi
so' ty nhien.
Cty TNHH MTV DWH
Khang
Viet
Vi
du 2: Dung cac ki hieu de vie't cac cau sau va vie't menh de phu djnh ciia no.
a) Tich ciia ba so'tu nhien lien tiep chia he't cho sau
b) Voi mpi so thuc binh phuong ciia la mot so khong am. i

c) Co mot so nguyen ma binh phuong ciia no bang chi'nh no.
d)
Co mot so hCru ti ma nghich dao ciia no Ion hon chinh no.
Loi
gidi
i-uij \
a) Ta
CO
P: Vn e N, n(n + l)(n + 2):6, menh de
phii
djnh la r
P:3neN, n(n + l)(n + 2)/6. ^'/^
b) Ta
CO
Q
:
Vx e M, x^ > 0, menh de
phii
djnh la Q :3x 6 K, x^ < 0
c) Ta
CO
R : 3n
G
Z, n^ = n , menh de
phii
djnh ia R : Vn e Z, n^ ;t n .'' '
d)
3q € Q, — > q , menh de phu
dinh
la Vq e Q, i < q .

q M
Vi
du 3: Xac djnh
tinh
diing
sai ciia menh de sau va tim
phii
dinh
ciia no :
a) A:
"VxeR,
x^ > 0 "
b) B: " Ton tai so tu nhien deu la so nguyen to".
c) C: " 3x6N, X chia he't cho x +
1
"
d)
D:"Vn£N, n'^-n^+l la hop so"
e) E: "Ton tai
hinh
thang la
hinh
vuong ".
f)
F: "Ton tai so thuc a sao cho a +1 + —< 2"
a + 1
Lai
gidi
a) Menh de A
diing

va A : 3x € R, x' < 0
b) Menh de B
diing
va B : "Voi moi sotu nhien deu khong phai la so nguyen to"
c) Menh deC sai va C: " Vx e N,x/(x + 1)"
d)
Menh de D sai vi voi n = 2 ta c6 n'* - n^ +1 = 13 khong phai la hop so
M^nh
de
phii
djnh la D: " 3n e N, n"* - n^ +1 la so so nguyen to"
e) Menh de E
diing
va E : " Voi mpi
hinh
thang deu khong la
hinh
vuong ".
0 Menh de F
diing
va menh de
phii
djnh la F: "Voi moi so thuc a thi
a + l + >2 .
a + 1
Phdtt
loai
va
phucnig
phdp

gM Dai so 10
Ca 2. BAI TAP
LUYfN
TAP
Bai 1.7: Xet
diing
(sai) menh de va phu
dinh
cac menh de sau :
a) VxeM, -x^ +1 >0
b) Vx € K, x"* - x^ +
1
= (x^ + 73x + l)(x^ - N/3X +1)
c) 3x e N, n +3 chia het cho 4
d)
3qeQ,
2q2-l=0
e) 3nGN,n(n + l) la mpt so chinh phuong
Huang
dan
gidi
a) Menh de Vx
G
M, x'' - x^ +1 > 0 sai ch^ng han khi x = -1 ta c6
(-1)'-(-!)'+1
= -1<0.
M^nh
de phu
dinh
la 3x e R, x'' - x^ +1 < 0 .

b) Menh de Vx e x"" - x^ +1 = (x^ + V2x + l)(x^ -
N/2X
+1) dung vi
x'' - x^ +1 = (x^ +
])^
- 3x^ = (x^ + x/3x + l)(x2 - >/3x + l)
Menh
de phu djnh la 3x e
IR,
x"* - x^ +1
9^
(x^ + VSx + l)(x^ - VSx +1).
c) Menh de 3x
G
N, n^ + 3 chia het cho 4
diing
vi n = 1 e N va n^ + 3 = 4:4
M^nh
de
phii
dinh
la " Vx e N, n + 3 khong chia het cho 4"
d)
Men de 3q e Q, 2q^ -1 = 0 sai. Menh de phu
dinh
la Vq e Q, 2q^ -1 0
e) Menh de "3nGN,n(n + l) la mot so chinh phuong"
diing.
M|nh de
phii

dinh
la " Vn e N, n (n +1) khong phai la mot so chinh phuong"
Bai 1.8:
a) Voi neN, cho menh de chua bien P(n):"n^ +2 chia het cho 4". Xet
tinh
diing
sai ciia menh de
P(2007).
b) Xet
tinh
diing
sai cua menh de P{n) : " 3n e N*, ^n(n +1) chia het cho 11".
^
Huang
dan
gidi
a) Ta
CO
: Voi n =
2007
thi n^ + 2 =
2007^
+ 2 la so le nen khong chia het cho 4.
Vay
P(2007)
la menh de sai.
b) Xet bieu thiVc ^ , voi n G N*. ta c6 :
2
Voi
n = 10 thi = 55 : chia het cho 11. Vay menh de da cho la menh

de
diing.
|
Cty TNHH MTV DWH
Khang
Vjc-i
Bai 1.9:
a) Cho menh de P : "Voi moi so thuc x, neu x la so
hiiu
ti thi 2x la sd'hiru
ti".
Dung
ki hieu viet P, P va xac djnh
tinh
diing
- sai ciia no.
b)
Phat
bieu MD dao ciia P va chiing to MD do la
diing.
Phat
bieu MD
duoi
dang MD tuong duong
Huang
dan
gidi
'f i
' sr
a) Menh deP"VxG

R,XGQ=>2X
GQ"
.MD
diing.
P: "3x
G
R,x 6 Q 2x g Q". MD sai
b) MD dao ciia P la " Voi moi so thuc x, XGQ khi va chi khi
2XGQ".
Hay
" Vx e R,x
G
Q <:> 2x
G
Q". 'O
riff'
Bai 1.10: Cho sotu nhien n. Xet hai menh de chiia bien: •'•
'i'
A(n):
"n la
sochan",
B(n) : "n^ la so ch§n".
a) Hay phat bieu menh de A(n) B(n). Cho biet menh de nay
diing
hay sai ?
b) Hay phat bieu m?nh de " Vn
G
N,
B(n) A(n) ".
c) Hay phat bieu m^nh de "Vn e N, A(n) <r> B(n)".

Huang
dan
gidi
a) A(n) => B(n) : "Neu n la so chan thi n^ la so chan". Day la menh de
diing,
vi
khi
do n = 2k (k
G
N) => n^ =
4k2
la so chan.
b) " Vn G N, B(n) => A(n) " : Voi moi so tu nhien n, neu n^ la so chan thi n la so
chan.
c) " Vn € N, A(n)
<=>
B(n) " : Voi moi so tu nhien n, n la so chin khi va chi khi n^
la
sochan.
Bai 1,11: Xet
tinh
diing
sai ciia cac m^nh de sau:
a) P:"V\GR,VyGR:x + y =
l"
b) Q :"3x
G
R,3y
G
R : x + y = 2"

c) R:"T GR,VyGR:x + y=3" d) S :" Vx
G
R,3y
G
R : x + y = 4"
Huang
dan
gidi
' '
a) Menh de P sai vi ch^ng han X = 1GIR, y = 2GK nhung x + y ^ 1
b) Menh de Q
diing
vi X = y = 1 => x + y = 2 •>
c) Vi X + y = 3 nen voi moi y
G
M thi
luon
ton tai x = 3 - y do do
mf
nh
de R
diing
'
d)
Menh de S
diing
"
Phatt loai va
fhucntg
phdp

giai
D(it
so
§2:
AP
DUNG MENH DE VAO SUY LUAN TOAN HOC
A.
TOM
TAT LY
THUYET
/. 't)inh
li
va.
chung
minh
dinh
U.
• Trong toan
hoc
djnh
ly la
mot menh
de
diing.
Nhieu
dinh
ly
dugc phat bieu
duoidang "VxeX, P(x)=:>Q(x)",
P(X),Q(X)

la cac
menh
de
chiia bien
•
Co
hai
each
de
chiing minh
dinh
li duoi dang tren
Cach
1:
Chung minh true tie'p gom
cac
budc
sau:
-
Lay
xeXba'tkyma
P(x)
dung
- Chung minh
Q(x)
dung(bang suy luan
va
kien thuc toan
hoc da
biet)

Cach
2:
Chung minh bang phan djnh li gom
cac
buoc
sau:
- Gia
su
ton tai
XQ e X sao
cho P(x„) dung
va
Q(X(,)
sai
- DiJng
suy
luan
va cac
kien thuc toan
hoc de
di de'n mau thuan.
2. Djnh
li dao,
dieu ki?n
can,
dieu ki?n dii, dieu ki^n
can va du.
•
Cho
djnh li duoi dang "Vx

6 X,
P(x)
=:>
Q(x)" (1). Khi
do
P(x)
la
dieti
kien
dii deed Q(x)
Q(x) \a
dieu
kien
can de
CO
P(x)
• Menh
de
Vx€X,
Q(x)=>
P(x) dung
thi
duoc
goi
dinh
li dao
ciia djnh
li
dang
(1)

Liic
do
(1) duoc goi
la
dinh
Ixj
thuan
va
khi
do c6 the
gop lai thanh mot djnh li
Vx
G
X,
Q(x)
o
P(x),
ta
goi
la "
P(x)
la
dieu
kien
can va dii de c6
Q(x)"
Ngoai
ra
con noi" P(x) ne'u
va

chi neu Q(x)"," P(x) khi
va chi
khi Q(x)",
B.
CAC
DANG
TOAN
VA
PHl/QNG
PHAP
GIAI.
DANG
TOAN 1: PHUONG PHAP CHUNG MINH BANG PHAN CHUNG.
QLCAC
V(
DUMINH
HOA
Vi
du 1:
Chung minh
rSng
voi moi
so tu
nhien
n, n''
chia
het cho 3
thi
n
chia

he't
cho 3.
Lai
giai
Gia
su n
khong
chia
he't
cho 3
khi
do n = 3k +1
hoac
n = 3k + 2, keZ
;
Voi n = 3k +
1
ta c6 n^ = (3k + if =
27k^
+
27k^
+ 9k +1
khong
chia
he't
cho
3 (mau thuan)
Voi
n = 3k + 2 ta c6 n^ =(3k +
2)^

=27k^+54k^+36k
+ 4
khong
chia
het
cho
ba
(mau thuan)
Vay
n
chia
he't
cho 3.
1L
Chf TNHH
MTV
DVVH Khang Vi^t
Vi
du
2:
Cho tarn thuc f (x)
= ax^ + bx + c, a 0 .
Chung minh rang ne'u ton
tai
so thuc
a sao cho a.f (a) < 0
thi phuong
trinh
f (x)
= 0

luon
c6
nghiem.
Ta CO
f(x) = a
X
+
•
2a
Lai
gidi
^-,A
=
b2-4ac.
4a
Gia
su
phuong
trinh
da cho v6
nghiem, nghTa
la
A
< 0.
Khi
do
ta co:
af(x)
= a'
^

b^2
X
+
•
2a
\ itiv.tn
>0,
VxeR
4
*,»vwi'
Suy
ra
khong ton
tai a de af (a) < 0 ,
trai
voi
gia
thiet.
Vay dieu
ta gia su 6
tren
la sai, hay
phuong
trinh
da cho
luon
c6
nghiem.
Vi
du 3:

Chung minh rang mot tarn
giac
c6
duong trung tuye'n
vua la
phan
giac
xua't phat
tu
mot
dinh
la
tam
giac
can tai
dinh
do.
Lai
gidi
Gia
su
tam
giac
ABC co AH vua la
duong trung tuyen
vua la
duong phan
giac
va
khong

can
tai
A.
Khong
mat
tinh
tong quat xem nhu AC
> AB .
Tren
AC lay D sao cho
AB
=
AD
.
Goi
L la
giao diem
cua BD va
AH
.
Khi
do
AB
=
AD, BAL
=
LAD
va AL
chung
nen AABL

=
AADL
Do
do
AL
=
LD
hay L la
trung diem
cua BD B
Suy
ra LH la
duong trung binh
cua
tam
giac
CBD
=> LH//DC dieu nay mau thuan vi LH,DC Ccit nhau tai
A
Vay tam
giac
ABC can tai A. ;
Q
2. BAI
TAP
LUYEN
TAP
Bai
1.12:
Chung minh b^lng phuong phap phan chung: Ne'u phuong

trinh
bac
hai
ax^ + bx + c = 0 v6
nghiem thi
a va c
cung da'u.
Huong
dan
gidi
Gia
su
phuong
trinh
v6
nghiem
va a, c
trai
da'u. Voi dieu ki^n
a, c
trai
da'u
CO
a.c<0 suy ra
A
= b^ - 4ac =
b^
+
4(-ac)
> 0

Nen
phuong
trinh
co hai
nghiem phan biet, dieu
nay mau
thuan
voi gia
thiet
phuong
trinh
vo
nghiem.
Vay phuong
trinh
v6
nghiem thi
a, c
phai cung da'u.
Bai
1.13:
Chung minh bang phuong phap phan chung: Ne'u hai so'nguyen duong
CO
tong binh phuong
chia
he't cho
3
thi
ca
hai

so
do phai
chia
he't cho
3.
Phdn lo^i va
phuang
phdp
gi&i
Dai so 10
Huang dan
gidi
Gia su
trong
hai so' nguyen duang a va b c6 it nhat mot so khong chia het
cho 3, chang han a khong chia het cho 3. The thi a c6 dang: a = 3k + 1
hoac
a = 3k+2. Luc do a^ = 3m + 1 , nen neu b chia het cho 3
hoac
b khong chia het
cho 3 thi a^ + b^ cung co dang: 3n + 1
hoac
3n + 2, tuc la a^ + b^ khong chia
het cho 3,
trai
gia thie't. Vay neu a^ + b^ chia het cho 3 thi ca a va b deu chia
het cho 3.
Bai
1.14: Chung
minh

rang : Neu do dai cac canh cua tam
giac
thoa man bat
ding
thuc a^ + b^ > 5c^ thi c la do dai canh nho nhat cua tam giac.
Huong dan
gidi
Gia su c khong phai la canh nho nhat cua tam giac.
Khong
mat
tinh
tong quat, gia sua < c => a^ < c^ (1)
Theo bat
ding
thuc
trong
tam giac, ta co b < a + c b^ < (a + c)^ (2).
Doa<c^(a + c)^
<4c2
(3)
Tu
(2) va (3) suy ra b^ < 4c^ (4).
Cong ve voi ve'(l) va (4) ta co a^ + b^ < 5c^ mau thuan voi gia thie't
Vay c la canh nho nhat cua tam giac.
Bai
1.15; Cho a, b, c duong nho hon I. Chung
minh
rang it nhat mot
trong
ba

ba't
ding
thuc sau sai a(l -b) > i, b(l - c) > c(l - a) >i
Huong dan
gidi
Gia sir ca ba bat
ding
thuc deu
dung.
Khi
do, nhan theo ve'ciia cac bat
ding
thuc tren ta duoc:
4;
3
1
a(l-b).b(l-c).c(l-a)> - hay a(l-a).b(l - b).c(l-c) > — (*)
64
Mat
khac
a(l-a)
= -a^+a = ^-
1
1
a —
N2
1
< —
4
DoO<a<l=>0<a(l-a)<^

Tuong
tu thi 0 < b(l - b) < i, 0 < c(l - c) < 1
Nhan
theo ve ta dugc a(l-a).b(l - b).c(l - c) < — (**)
64
Bat
ding
thuc (**) mau thuan (*)
Vay
CO
ft nhat mot
trong
cac bat dang thuc da cho la sai. (dpcm)
Bai
1.16: Neu a]a2>2(b5+ b2) thi it nhat mot
trong
hai phuang
trinh
+
ajx + bj = 0, +
a2X
+ b2 = 0 co nghiem.
Huong dan gia
i
Gia su ca hai phuang
trinh
tren v6 nghiem
Khi
do D, = ai^ - 4bi < 0, D] = - 4b2 < 0
_^a/-4bi+a2^-4b2

<Oc:>a,^+32^ <4(bi+b2) (1) '
Ma (a, - a2 f > 0
<=>
a? + a^ >
23^32
(2)
Tu
(1) va (2) suy ra 23^32 < 4(bj + b2) hay 3^32 < 2{b] + bj)
trai
gia thie't.
Vay phai co it nhat 1
trong
hai so Ai, A2 Ion han 0 do do it nhst 1
trong
2
phuang
trinh
x + ajx + b^ =0, x + a2X + b2 = 0 co nghiem.
Bai
1.17: Chung
minh
rang sfl la so v6 ti.
Huong dan
gidi
,^
De dang chung
minh
dugc neu n^ lasochSnthi n laso'chan
Gia sir V2 la so hCru ti, tuc la J2= — ,
trong

do m, n e N*, (m, n) = 1
n
Tit
N/2 = — => m^ = 2n^ => la so chan
n
=>
m la so chan => m = 2k, k e N
Tir
m^ = 2n^ =:> 4k^ = 2n^ :=> - 2k^ => n^ la so chin => n la so chan
Do
do m chan, n chan, mau thuan vai (m, n) = 1 .
Vay N/2 la so v6
ti.
a + b + oO (1)
Bai
1.18: Cho cac so a, b, c thoa cac dieu ki§n :
•
ab + be + ca > 0 (2)
abc > 0 (3)
Chirng
minh
rang ca ba so a, b, c deu duong.
Huong dan
gidi
Gia sir ba so a, b, c khong dong
thai
la so duang. Vay co it nhat mot so
khong
duang.
Do

a, b, c
CO
vai tro
binh
ding
nen ta co the gia sir: a < 0
Neu
a = 0 thi mau thuan voi (3)
Ne'u
a < 0 thi tir (3) => be < 0
Ta CO (2)
<=>
a(b + c) > -be => 3(b
^
b
+
c<0 a
+
b
+
c<0 mauth|ulli%y»P
TiWH
BiNH
THUAfJ
Vav
ca b3 so a, b, c deu duone.
Than
loai
va
phumtg

phapgiai
Dai so 10
Bai 1.19: Chung minh
bang
phan chiing dinh li sau : "Neu tarn
giac
ABC c6 cac
duang phan
giac
trong BE, CF
bang
nhau, thi tarn
giac
ABC can".
Huang
dan
gidi
• Neu B > C thi ta dung hinh binh hanh
BEDF
nhu hinh ve .
Ta CO : B > C => B^ > Q =i> > Q (1)
Ngoai ra, BE = CF => DF = CE
=> Di +D2 =C2 +C3 (2).
Tu
(1) va (2) suy ra
< Q => EC < ED => EC < FB .
Xet cac tarn
giac
BCE va CBF, ta thay : ^
BC

Chung,
BE = CF, BF > CE nen Q > B^ =^ C > B. Mau thuan.
• Truong hop C > B, chung minh hoan toan tuong tu nhu tren.
Do do B = C. Vay tarn
giac
ABC can tai A.
Bai 1.20: Cho 7 doan
thSng
c6 do dai Ion hon 10 va nho hon 100. Chung minh
rang luon tim dugc 3 doan de c6 the
ghep
thanh mgt tarn
giac.
Huang
dan
gidi
Truoc
het sap xep cac doan da cho
theo
thu tu tang dan cua do dai aj, ,87
va chung minh rang trong day da xep luon tim dugc 3 doan lien tiep sao cho
tong cua 2 doan dau Ion hon doan cuoi (vi dieu kien de 3 doan c6 the
ghep
thanh mgt tarn
giac
la tong cua 2 doan Ion hon doan thu ba).
Gia su dieu can chung minh la khong xay ra, nghia la dong thoi xay ra cac
bat dang thuc sau: aj <a3; a2 +33
<a^; ;
+a(,<ay

Tu
gia thie't a^, aj c6 gia tri Ian hon 10, ta nhan dugc 83 > 20 . Tu 82 > 10
va 83 > 20 ta nhan dugc a^ > 30, 35 > 50, ag > 80 va ay > 130 . Dieu
ay > 130 la mau thuan voi gia thiet cac do dai nho hon 100. Co mau thuan
nay la do gia sir dieu can chung minh khong xay ra.
Vay, luon ton tai 3 doan lien tiep sao cho tong cua 2 doan dau Ion hon doan
cuoi. Hay noi
each
khac
la 3 doan nay c6 the
ghep
thanh mgt tam
giac.
DANG TOAN 2: SUDUNG
THUAT
NGU
DIEU
KIEN
CAN,
BIEU
KIEN
DU,
DIEU
KIEN
CAN VADU. ||
Ca 1. CAC Vf DU
MINH
HOA "
*
'' '

Vi
du 1: Cho djnh li: "Cho so tu nhien n. Neu n'^
chia
het cho 5 thi n
chia
het
cho 5". Dinh li nay dugc viet duoi dang P => Q.
a) Hay xac dinh cac menh de P va Q.
b)
Phat
bieu dinh li tren
bang
each
dung thuat
ngii
"dieu kien can".
c)
Phat
bieu djnh li tren b^ng
each
diing
thuat
ngii
"dieu kien du".
d) Hay phat bieu djnh li dao (neu c6) cua djnh li tren roi dung cac thuat
ngii
"dieu kien can va du" phat bieu gop ca hai dinh li thuan va dao.
Lai
gidi.
a) P : "n la so tu nhien va n^

chia
het cho 5", Q : "n
chia
het cho 5".
b) Voi n la so tu nhien, n
chia
het cho 5 la dieu kien can de n'^
chia
het cho 5 ;
hoac
phat bieu
each
khac
: Voi n la so tu nhien, dieu kien can de
chia
het
cho 5 la n
chia
het cho 5.
c) Voi n la so tu nhien, n"^
chia
het cho 5 la diiFu
kien
du de n
chia
het cho 5.
d) Dinh li dao : "Cho so tu nhien n, neu n
chia
het cho 5 thi n"^
chia

het cho 5".
That
vay, neu n = 5k thi n'^ = 5\k^: So nay
chia
het cho 5.
Dieu
kien
can va du de n
chia
he't cho 5 la n'^
chia
het cho 5.
Vi
du 2:
Phat
bieu cac menh de sau voi thuat
ngii
"Dieu kien can", "Dieu
kien du" » f'
a) Neu hai tam
giac
bang
nhau thi chung c6 dien tich
bang
nhau
b) Neu
so'nguyen
duong
chia
het cho 6 thi

chia
het cho 3
c) Neu hinh thang c6 hai duong
cheo
bang
nhau thi no la hinh thang can
d) Neu tam
giac
ABC vuong tai A va AH la duong cao thi AB^ =
BC.BH
Lai
gidi
a) Hai tam
giac
bang
nhau la dieu kien du de chiing c6 dien tich
bang
nhau
Hai
tam
giac
c6 dien tich
bSng
nhau la dieu kien can de chiing
bang
nhau
b)
So'nguyen
duong
chia

het cho 6 la dieu kien dii de no
chia
het cho 3
So nguyen duong
chia
het cho 3 la dieu kien can de no
chia
het cho 6
c)
Hinh
thang c6 hai duong
cheo
bang
nhau la dieu ki|n du de no la hinh
thang can
Hinh
thang can la dieu kien can de no c6 hai duong
cheo
bang
nhau
d) Tam
giac
ABC vuong tai A va AH la duong cao la dieu ki?n dii de
AB^-BC.BH ^.R
Tam
giac
ABC c6 AB^ = BC.BH la dieu i<ien can de no vuong tai A va AH
la duong cao. 1^ ' . ^ '
Ca 2. BAI TAP LUYEN TAP | •
Bai 1.21:

Phat
bieu cac djnh ly sau day
bSng
each
sir dung khai niem " Dieu
kien
can","
Dieu kien du "
a) Neu trong mat phang, hai duong
thSng
cijng vuong goc voi duong thang
thii
3 thi hai duong thang do
song song
voi nhau
b) Neu
so'nguyen
duong c6 chu tan cung
bang
5 thi
chia
het cho 5
c) Neu tu
giac
la hinh thoi thi 2 duong
cheo
vuong goc voi nhau
d) Neu 2 tam
giac
bang

nhau thi chung c6 cac goc tuong ung
bang
nhau
e) Neu
so'nguyen
duong a
chia
het cho 24 thi
chia
het cho 4 va 6
i,
Huang
dan
giai
a) Trong mat phang, hai duong thang citing vuong goc voi duong thang thu 3
la dieu kien du de hai duong thang do
song song
voi nhau.
Trong mat phang, hai duong thang do
song song
voi nhau la dieu kien can
de hai duang thang cung vuong goc voi duong thang
thii
3.
b)
So'nguyen
duong co chu so tan ci^mg
bang
5 la dieu kien du de
chia

he't cho 5.
So nguyen duong
chia
he't cho 5 la dieu kien can de no co
c\\\x
so tan ciing
bang
5.
c) Tu
giac
la hinh thoi la dieu kien du de no co 2 duong
cheo
vuong goc voi nhau
Tu
giac
CO
hai duong
cheo
vuong gck voi nhau la dieu kien can de no la
hinh
thoi
d) Hai tam
giac
bang
nhau la dieu kien du de chung co cac goc tuong ung
bang
nhau
Hai
tam
giac

co cac goc tirong ung
bang
nhau la dieu kien can de chiing
bang
nhau
e)
So'nguyen
duong a
chia
he't cho 24 Ih dieu kien du de no
chia
he't cho 4 va 6
So'nguyen
duong a
chia
he't cho 4 va 6 la dieu
kifn
can de no
chia
he't cho 24.
Bai 1.22. Dung thuat ngij' dieu kien can va ctu de phat bieu
dinh
li sau
a) Mot tam
giac
la tam
giac
can, neu va chi neu no co hai goc
bang
nhau

b) Tu
giac
la hinh binh hanh khi va chi khi tu
giac
co hai duong
cheo
cat nhau
tai
trung diem ciia moi duong
c) x>yo^>3/^ • « ^^-^
d) Tu
giac
MNPQ la hinh binh hanh khi va chi khi MN = QP. ^V*^
Huong
dan
gidi
a) Mpt tam
giac
la tam
giac
can la dieu kien can va du de no co hai goc
bang
nhau
CtyTNHH MTV DWH
KHaWg
VieT
b) Tu
giac
la hinh binh hanh la dieu kien can va du de tu
giac

co hai dirong
cheo
cat nhau tai trung diem cua moi duong g
c) X > y la dieu kien can va
dude
;
d) Dieu kien can va du de tu
giac
MNPQ la hinh binh hanh la MN = QP .
Bai 1.23: Su dung thuat ngCf "dieu kien can", "dieu kien dii" de phat bieu
dinh
li
sau:
a) "Neu mot tu
giac
la hinh vuong thi no co bon
canh
bang
nhau".
Co
dinh
li dao cua djnh li tren khong , vi
sao?
M i i
b) "Neu mot tu
giac
la hinh thoi thi no co hai duong
cheo
vuong goc".
Co djnh li dao cua djnh li tren khong , vi

sao?
^
Hu&ng dan
gidi
^,
a) Mot tu
giac
la hinh vuong la dieu kien du de no co 4
canh
bang
nhau.
Mot
tu
giac
CO
4
canh
b3ng
nhau la dieu kien can de no la hinh vuong.
Khong
CO
djnh li dao vi tu
giac
co 4
canh
bang
nhau co
thela
hinh thoi
b) Mot tu

giac
la hinh thoi la dieu kien du de no co hai duong
cheo
vuong goc.
Mot
tu
giac
CO
hai duong
cheo
vuong goc la dieu kien can de no la hinh thoi
Khong
CO
djnh li dao vi tu
giac
co hai duong
cheo
vuong goc co the la hinh
vuong
hoac
mot da
giac
bat ki co hai duong
cheo
vuong goc.
i li
V
§3: TAP HOP VA
CAC
PHEP

TOAN
TREN
TAP HOP
.A,:S;:
A.TOM
TAT LY
THUYET
1.
r^p hap
• Tap hop la mot khai niem co ban cua toan hoc, khong djnh nghia.
• Cach xac dinh tap hop: -
+ Liet ke cac phan tir: vie't cac phan tir cua tap hop trong hai dau moc
+ Chi ra
tinh
chat
dac trung cho cac phan tu cua tap hop.
• Tap
rong:
la tap hop khong
chua
phan tu nao, ki hieu 0. ' \^'
2. Tap hap con - Tap hap
bdng
nhau
^^J'"
•
AcBo(Vx6A=^xeB)
,^
Cac
tinh

chat:
+ AcA, VA
+0cA,
VA + A c B,B c C=> A c C
• A =
Bo(AcBvaBcA)o
(Vx,^ ^ A « x £ B)
3.
Mot so tap con cua tap hap sdthuc
Ten
gQJ,
ky
hi^u
Tap
hgp
Hinh
bieu
dien
Tap
so'
thyc''
(-00;+00)
0
a
Doan
[a ; b]
{x€Rla<x<b)
/////[
]-H-H-
Khoang

(a ; b)
Khoang
(-00;
a)
Khoang
(a ;
+00)
{xgIRIa<x<bl
|x6Klx<a
xeKIa
<x
a
/////(
)////
•
a
)//////
a
/////(
Nira
khoang
[a ; b)
Nua
khoang
(a ; b]
Nua
khoang
(-co; a]
Nua
khoang

[a ;+co)
|xeKla<x<b|
{x€Rla<x<b}
|x€Klx<a}
(x6Rlx>a|
a h I
/////[
)//// ^,
a
iin/{
]////
>•
a
)ni/iii
•
a
III II III
4.
Cac
phep todn
tap
hap
•
Giao
ciia
hai tap
hop:
AnBofxIxeA
va xeB)
•

Hop
cua hai tap
hop:
AuB<=>|xlx6A hoac x e
B)
•
Hi^u cua hai tap hgp: A\B<=>(xlx6A va xiB}
Phan
bii:
Cho
B
c A thi
CAB
= A
\.
B.
CAC
DANG
TOAN
VA
PHl/aNG
PHAP
GIAI.
DANG
TOAN
1:
XAC
DINH
TAP
HOP

VA
PHEP
TOAN
TREN
TAP
HOP
£ai.CAC
Vf DU MINH HOA
Vi
d\
1: Xac
dinh
cac tap hop sau
bang
each neu
tinh
chat dac
trung
A
= {0 ; 1; 2; 3; 4} ; B = {0 ; 4; 8;
12;16}
; C = {1;2;4;8;16}
Ta
CO
cac tap hop
A,
B, C
duoc
viet
duoi

dang
neu cac
ti'nh
chat dac
trung
la
A
=
{xeNlx<4};
B = (x e N
I
x;4 va x < 16)
C
= (2"
I
n < 4 va n G
Nl
' ' '
Vid\i2:Cho
A = {-4;-2;-l;2;3;4} va B = {x e Z
l|x|
< 4} .
Tim
tap
hop
X sao cho
a)XcB\
b)AeXcB
c) A u X = B voi X
CO

dung
bon
phan
tu
Lai
gidi
'|x|<4 f-4<x<4
Taco''
<^ oxG -4;-3;-2;-l;0;l;2;3;4
X
e Z [
X
e Z
Suyra
B = {-4;-3;-2;-l;0;l;2;3;4}
a) Taco B\ =
{-3;0;1}
Suy
ra X c
B
\
thi
cac tap hop X la
•
0, {-3},{0}, {!}, {-3;0}, {-3;!}, {0;1},
{-3;0;1}
b)
Taco {-4;-2;-l;2;3;4}cXc{-4;-3;-2;-l;0;l;2;3;4}suy
ratlphgp
X la

{-4;-2;-1;
2;
3; 4}, {-4;-2;-3;-1;
2;
3; 4}, {-4;-2;-1;0;2; 3; 4}
{-4;-2;-l;l;2;3;4},
{-4;-2;-3;-l;0;2;3;4}, {-4;-2;-3;-l;l;2;3;4}
{-4; -2;
-1;
0;
1;
2;
3;
4},
{-4; -3; -2; -1;0;
1;
2;
3;4}
c) Ta
CO
A u X = B voi X c6
diing
bon
phan
tu
khi
do tap hop X la
{-4;-3;0;l},{-3;-2;0;l},
{-3;-l;0;l},
{-3;0;1;2}, {-3;0;1;3},

{-3;0;1;4}
Vi du 4: Cho cac tap
hop:
'
A
=
{XER|(X2+7x
+
6)(x^-4)
=
o};
B =
{XGNI2X<8}
C
=
|2X
+
1IXGZ
va -2<x<4)
a)
Hay
viet
lai
cac tap hop
A,
B,
C
duoi
dang
liet

ke cac
phan
tu
2.3
b) Tim
c) Tim
A
B, A n B, B
\, C^^^B (B \) .
(ALJC)\B.
Laigidi
x'^
+ 7x + 6 = 0
-
4 = 0
x =
-l
x = -6
hoac
a) • Tac6:(x^+7x + 6)(x2-4) = 0
i
Vay A = {-6;-2;-l;2} ; ,
'x
e N fx e N
oxe
{0,1,2,3,4}.
Vay B =
{0;1;2;3;4|
(xeZ
x = -2

x = 2
•
Ta CO •
<=>
i
2x<8 x<4
•
Ta
CO
<
X
< 4^ ^
{-2,-1,0,1,2,3,4}.
Suy ra C = {-3;-l;l;3;5;7;9}
b) Tac6: A u B = {-6;-2;-l;0;l;2;3;4}, A n B = {2}, B
\ =
{0;2;4}
S",
CAUB
(B
\) = (A u B) \B \) =
{-6;-2;-l;l;3}
c) Taco: AuC = {-6;-3;-2;-l;l;2;3;5;7;9}
Suyra
(AuC)\
= {-6;-3;-2;-l;5;7;9}
£•
2.
BAI TAP
LUYIN

TAP
Bai
1.24: Xac
dinh
cac tap hg'p sau bSng each neu
tinh
chat dac
trung
^,
A
= {-4;-3;-2;-l;0 ; 1; 2; 3; 4}, B =
{l;
3; 5; 7; 9}, C =
{0;1;4;9;16;25}
ia»
Huong dan gidi
Ta
CO
cac tap
hgp
A,B,C duoc
viet
duoi
dang
neu cac
tinh
chat dac
trung
la
A =

{xeNI|x|<4},
B =
{xeNlx
la so le nho han 10}, C = {n^l n lasotu
nhien
nho hon 6)
14
Bai
1.25 Cho tap hop A = |x e M
I
^ e Z
a) Hay xac
dinh
tap A bang each
liet
ke cac
phan
tu
b) Tim tat ca cac tap con
ciia
tap hop A . ''^
Huang
dan gidi
a) Ta CO Vx > 0 suy ra 0 <
14 ^14 .
3^ + 6 " 6
14 14 14
Mat
khac
—7=

e Z nen — =
1
hoac — = 2
3Vx+6
1
64
Hay
^ = ~ hoac x = —. Vay A =
3Vx
+ 6 ' 3\fx + 6
i
9' 9
b) Ta't ca cac tap con
ciia
tap hgp A la 0, jij, , •
T
!
B^i
1.26: Cho A = |x
G
iR
I
(x"* -16)(x^ = B = {x e N
I
2x - 9 < 0} .
Tim
tap hgp X sao cho
a) XcB\ . ^
A\
= XnA vol X c6

dung
hai
phan
tu
Huong
dan gidi
Ta
CO
A = {-2;-l;l;2} va B =
{0;1;2;3;4}
^
a) Taco B\ =
{0;3;4}
Suy ra X c B
\ thi cac tap hop X la
0,{O},|3},{4},{O;3},{O;4},{3;4MO;3;4}
b) Ta
CO
A
\ = {-2;-l} v6i X c6
dung
hai
phan
tu khi do X = {-2;-l}.
Bai
1.27: Cho tap A =
{-1;1;5;8},
B ="G6m cac uoc so
nguyen
duong

cua 16"
a)
Viet
tap A
duoi
dang
chi ra
tinh
chat dac
trung
ciia
cac
phan
tu.
Viet
tap B
dual
dang
liet
ke cac
phan
tu.
b) Xac
dinh
cac phep
toan
A n B, A u B, A
\ .
Huong
dan gidi

a) Taco A = {xeR|(x + l)(x-l)(x-5)(x-8) = 0}
B
=
{1;
2; 4; 8; 16}
b) Ta
CO
A n B = (1;8), A u B = {
-1;
1; 2; 4; 5; 8; 16}, A
\ - {
-1;
5)
Bail.28:Chocactaphgp
E =
|
XGNM
<x<7| .
A
= {
XGN|(X2-9){X^-5X-6)
= 01
va
B = {x 6 N
IX
la so
nguyen
to nho hon 6}
a)
Chung

minh
rang
A c E va B c E
b) Tim CgA ;
CEB;CE(AUB)
c)
Chung
minh
rang
:
E\(A
n B) = (E
\( E\B)
Huong
dan gidi
a) Taco E =
{1;2;3;4;5;6}
A = {3;6} va B =
{2;3;5}
Suy ra A (z E va B c E
b) Taco
CEA
=
E\
= {1;2;4;5};
C^B
= E
\ =
{1;4;6}
,

A^B
= {2;3;5;6}^CE(AUB) =
E\(AUB)
= {1;4}
c) Taco AnB = {3}=>CE(AnB) =
E\(AnB)
= {l;2;4;5;6}
E\
= {1;2;4;5};E\ =
{1;4;6}=>(E\A)U(E\B)
=
{1;2;4;5;6}
Suy ra
E\(AnB)
=
(E\A)u(
E\B).
DANG
TO
AN
2: SUDUNG
BIEU
DO VEN DE
GIAI
TOAN.
.>"X
a IK
'Phucmg
phdp
gidi.

•
Chuyen bai toan ve ngon ngi> tap hop
•
Su*
dung bieu do ven de minh hoa cac tap hop
•
Dua vao bieu do ven ta thie't lap
dugc
dang
thiic
(hoac
phuang
trinh
he
phuang
trinh) tu do tim
du'oc
ke't qua bai toan ' .i.j i
Trong
dang
toan nay ta ki hieu n
(X)
la so phan tu cua tap X .
£•
1. CAC Vl DU
MINH
HOA ^ qv" q(>!
nil
••\ (t :.,) A
i:>i

VUr
Vi
du 1: Moi hoc sinh cua lop
lOAi
deu biet choi da cau
hoac
cau long, bie't
rang c6 25 em bie't chai da cau, 30 em biet choi cau long , 15 em biet choi
ca hai. Hoi lap
lOAi
c6 bao nhieu em chi bie't da
cau?
bao nhieu em chi
bie't danh cau long?
ST
so lop la bao
nhieu?
Lai
gidi
Dua vao bieu do ven ta suy ra so hoc
sinh chi bie't da cau la 25 -15 = 10
So hoc sinh chi bie't danh cau long la
30-15 = 15
Do
do ta
CO
sT
so'hoc
sinh cua lap
lOAi

la
10
+
15
+
15 = 40
Vi
du 2: Trong lop IOC c6 45 hoc sinh trong do c6 25 em thich mon Van, 20
em thich mon Toan, 18 em thich mon Su, 6 em khong thich mon nao, 5
em thich ca ba mon. Hoi so em thich chi mot mon trong ba mon tren.
Loi
gidi
Goi
a,b,c
theo
thu tu la so hoc sinh chi thich mon Van, Su, Toan;
X
la
so'hoc
sjnh chi thich hai mon la van va toan
y
la so hoc sjnh chi thich hai mon la Su va toan
z
la so hoc sjnh chi thich hai mon la van va Sir
Ta
CO
so em thich it
nhat
mot mon la 45 - 6 = 39
Sua vao bieu do ven ta c6 he phuong

trinh
a
+
x + z
+
5
= 25 (1)
b
+
y + z
+
5 = 18 (2) 25(V)
c
+
X + y
+
5 = 20 (3)
x
+ y + z
+
a
+
b
+
c
+
5 = 39 (4)
Cong ve'voi
ve'(l),
(2), (3) ta c6

20(T)
a + b
+
c
+
2(x + y + z)
+
15
=
63 (5)
Tu
(4) va (5) ta CO
a + b + c + 2(39-5-a-b-c) + 15 = 63 <=>a + b + c = 20 "" .
';
[(I.I''
••
Vay
chi
CO
20 em thich chi mot mon trong ba mon tren. ,,. , , ,
P
2. BAI TAP
LUYEN
TAP
Bai
1.29: Mpt nhom hoc sinh
gioi
cac bp mon: Anh, Toan, Van. Co 8 em
gioi
Van,

10 em
gioi
Anh, 12 em
gioi
Toan, 3 em
gioi
Van va Toan, 4 em
gioi
Toan va Anh, 5 em
gioi
Van va Anh , 2 em
gioi
ca ba mon. Hoi nhom do c6
bao nhieu em?
Huong dan
gidi
Ky
hi?u A la tap hop nhiing hoc sinh
gioi
Anh, T la tap hop nhCrng hoc sinh
gioi
toan, V la tap hop nhijng hoc sinh
gioi
Van.
Theogiathie'ttac6:n(V) = 8, n(A) = 10, n(T) = 12,
n(VnT)=
3, n(TnA) = 4,
n(VnA)
= 5, n(AnBnC) = 2.
n(VuAuT)

= n(V) + n(A) +
n(T)-n(VnA)-n(AnT)
-n(Tn
V)
+ n(VnAnT)
8 + 10 +
12-3-4-5
+ 2 = 20 .
Vay
nhom do c6 20 em.
Bai
1.30: Co 40 hoc sinh
gioi,
moi em
gioi
it
nha't
mot mon. Co 22 em
gioi
Van,
25 em
gioi
Toan, 20 em
gioi
Anh. Co 8 em
gioi
diing
hai mon Van, Toan; Co
7 em
gioi

dung hai mon Toan, Anh; Co 6 em
gioi
diing
hai mon Anh, Van.
Hoi:
Co bao nhieu em
gioi
ca ba mon Van, Toan, Anh?
(^f^
, i
Huang dan
gidi
*t. ,
Ky
hieu A la tap hop nhijng hoc sinh
gioi
Anh, T la tap hop nhCrng hoc sinh
gioi
toan, V la tap hop nhirng hoc sinh
gioi
Van.
Theo gia thie't tac6:n(V) = 22, n(T) = 25, n(A) = 20,
n(VnT)=
8, n(TnA) = 7,
n(VnA)
= 6, n(AuBuC) = 40.
n(VuAuT)
= n(V) + n(A) + n(T)-n(VnA)
-n(AnT)-n(TnV)
+ n(VnAnT)

=>n(VnAnT)
=
n(VuAuT)-n(V)-n(A)
-n(T)
+
n(VnA)
+ n(AnT) + n(TnV)
40-22-25-20 + 8
+
7 + 6 = 14 .
Vay
CO
14 em hoc
gioi
ca ba mon
Bai 1.31: Trong Ky thi tot nghiep pho thong, o mot
trirong
i<e't qua so thi sinh
dat danh hieu xuat sac nhir sau; Vo mon Toan: 48 thi sinh; Ve mon Vat ly: 37
thi
sinh; Ve mon Van: 42 thi sinh; Ve mon Toan
hoac
mon Vat ly: 75 thi sinh;
Ve mon Toan
hoac
mon Van: 76 thi sinh; Ve mon Vat ly
hoSc
mon Van: 6(i
thi
sinh; Ve ca 3 mon: 4 thi sinh. Vay co bao nhieu hoc sinh nhan

duoc
danh
hieu xuat sac ve:
a) Mot
mon?
b) Hai
mon?
c) it nhat mot
mon?
if
•s**'
Huong
dan
gidi
Ggi A, B, C Ian
lugt
la tap hop nhi>ng hoc sinh xuat sac ve mon Toan, mon
Vat Ly, mon Van.
Goi a, b, c Ian
lugt
la so hoc sinh chi dat danh hieu xuat sac mot mon ve
mon Toan, mon Vat Ly, mon Van.
Goi X, y, z Ian
lugt
la so hoc sinh dat danh hieu xuat sac hai mon ve mon
Toan va mon Vat Ly, mon Vat Ly va mon Van, mon Van va mon Toan.
Diing
bieu do Ven dua ve he 6 phuong
trinh
6 an sau:

a + X + z + 4 = 48 a = 28
b
+ X + y + 4 = 37
b
= 18
c + y + z + 4 = 42 c = 19
a + b + X + y
<=> •
a + b + X + y + z =
71
X
= 6
a + c + X + y
+ z = 72
y
= 9
b
+ c + X + y + z = 62
z
= 10
DS: a) 65 thi sinh dat danh hieu xuat sac 1 mon
b) 25 thi sinh dat danh hieu xuat sac 2 mon
c) 94 thi sinh dat danh hieu xuat sac it nha't 1 mon.
DANG
TOAN
3:
CHUNG
MINH
TAP HOP
BANG

NHAU,
TAP HOP CON.
'
'Phuang
phdp
gidi.
• De chung
minh
A c B
Lay Vx, x e A ta di chung
minh
x G B
• De chung
minh
A = B ta di chung
minh
+
AcBvaBcA
hoac
Vx, x e A o x e B
P
1. CAC Vf DU
MINH
HOA
Vi
du 1: Cho cac tap hop
71
A
=
— + k7i, k e Z

3
2Tt
, B = {-y+ kn,keZ
a) Chung
minh
rang A = B.
b) AcC
, „ I 271 kTl
va C = + —,
3 2
keZ
Led
gidi
a) • Ta CO Vx G A => Bk,, G Z : x =
—
+ k(,7i suv ra
x = |-7r + (k(, +l)7r = -^ +
(ko+l)7t.
Vi
ko G Z
=i>
k() +1 € Z do do x G B suy ra A c B (1).
• Vx € B 3k() G Z : x = - — + k()7t suy ra
3
x = -y+ '^ + (ko - '>)'^ = ^ + (k()
-l)7t.
Vi
k() G Z =::> k() -1 6 Z do do X G A suy ra B c A (2).
Tu(l)va(2)suyra A = B.
b) Ta CO Vx G A => 3k() G Z

:
x = + k,)7i suy ra
_n_
2(ko+l)7t_
27T 2(k() +1)71
3 2 3 2
Vi
k() e Z => 2(k(, +1) G Z do do x G C
Suy ra A c C .
Vi
du 2: Cho A va B la hai tap hop. Chung
minh
rang
a)(A\B)cA- b)An(B\A) = 0 c) Au(B\A) = AwB
Lai
gidi
a) Ta CO Vx, X G A
\
<=>
<!
=> x e A
" X«B
uy ra (A
\) c A
hi'
X G A
•5)
Taco
xG An(B\;
^ IXG(B\A)

Suy ra An(B\) = 0
X G A
X G B <r> X G 0
c) Ta CO X e A >>J (B
\) <=>
X e A
xe(B\A)
x e A
xeB »
X e A
xeB
« X e A B
Vi
du 3: Cho cac tap hop A, B va C . Chung
minh
rang
a) An(BuC) = (AnB)u(AnC)
b) Au(BnC) = (AuB)n(AuC)
Lm
gidi
XG A
a) Ta
CO X
e A
n(BuC)
<=> j ^ ^
<=>
X
G
A

'XGB
xeC
x 6 A
XGB
X G A
XGC
1T :X
XG AnB
XG AnC
<»xG(AnB)u(AnC)
Suy ra An(BuC) = (AnB)u(AnC) .
b)
Taco
xGAu(BnC)o
X G A
XGBOC
X G A
XGB
XGC
X
G
A
XGB [XGAUB
'XGA
[XGAUC
XGC
•r
Suy ra Au(BnC) = (AuB)n(AuC)
Ca 2. BAI TAP
LUVeN

TAP
«>xe(AuB)n(AuC)
Bai 1.32: Cho cac tap hop
A
=
— + k2n, k
G
Z
6
, B =
ll^
+
k27r,kGz}
vaC = {| +
^,kGz}
a) Chung
minh
rang A = B. b) A c C
Huang
dan
gidi
a) • Ta CO Vx G A => 3k() G Z: x = +
k()27r
suy ra
6
X =
+27t
+ (ko
-1)271
= — +(ko - 1)271 .

6 6
Vi
ko G Z => kg -1 G Z do do X G B suy ra A c B (1).
CiyTNHH MTV DWH
Khang
Viet
• Vx G B => 3ko G Z: X = +
kQ27t
suy ra
,,,,,,,
1,
6
X =
i^-27t
+ (ko
+1)271
= + (k„+1)271.
6 V >' ' 6
Vi
ko
G
Z => ko +1
G
Z do do x
G
A suy ra B c A (2).
Tu
(1) va (2) suy ra A = B. -
b) Ta CO Vx G A => 3ko GZ:x = + ko2TC suy ra
71 71 71 , - 71 (4ko-l)7t

x = -
—+
—- —+
ko27r
= —+ - '
6 2 2 " 3 2
Vi
ko G Z => 4kQ - IGZ do do XGC. Suy ra A c C .
Bai 1.33: Cho cac tap hop A c B, C c D . Chung
minh
rang '
a)AuCc:BuD
b)AnCcB
C)CBAUA
= B
Huang
dan
gidi
^XGA
XGC
a) Ta CO VX, X G A u C <=>
VoixGA
viAcB=>xGB=>xGBuD.
Suy raAuCcBuD.
X G A
b) Ta CO Vx, X G A n C <=>
XGC
X G A
Vi
A c B => X G B. Suy ra A n C c B.

XGB
c) VX, X G CgA u A <=>
x G CBA
X G A
<=>
X^AOXGB
X
G
A
Suy ra
CBAUA
= B
Bai 1.34: Cho cac tap hop A, B va C . Chung
minh
rang
a) (A\B)u(B\A) = (AuB)\(AnB)
b) A\(BnC) = (A\B)u(A\C)
c) A\(BuC) = (A\B)n(A\C)
Huang
dan
gidi
fx
G A
'XGA\ ^
I,,.II
a)
Taco
Vx, XG(A\B)U(B\A).
XGB\
XgB

XGB
[xiA
no it''
l^ltan
loai vd
phumig
phiipgiiii
Uai so W
<=>
X e A
xeB [xe
ALJB
x^A
[xgAnB
xgB
<=i>(AvjB)\(AnB)
Suy ra (A
\B \) = (A u B)\A n B).
X € A
b) Vx,x6A\(BnC)c>^ ""l^^c^,
X g BnC
X e A
X e A
x^C
xeB
xgC
xe A\
xe A\
«x6(A\B)u(A\C),
c) Vx, XG A\(BuC)<=>

. (J
X G A
X g B w C
<=> <
X G A
xgB
X g C
X G A
xgB fxeA\
. ,^«xG(A\B)n(A\C)
XGA
XGA\
x^C
DANG TOAN 4: P//£P TOAiV
TiJEAT
TAP CON CUA TAP SO
THUC.
(Phuang
phdp
gidi.
• De tim AnB talamnhirsau
- Sap xep
theo
thu tu tang dan cac diem dau
miit
ciia cac tap hgpA, B len
true so
- Bieu dien cac tap A, B tren true so (phan nao khong thuoe cac tap do thi
gaeh
bo)

- Phan khong bj
gach
bo chinh la giao ciia hai tap h^p A, B
• Detim A^^B taiamnhusau ,^ t
- Sap xep
theo
thu tu tang dan cac diem dau mut cua cac tap hgpA, B len
true so ' ^ ^
- To dam cac tap A, B tren true so
- Phan to dam chinh la hop cua hai tap hop A, B ^
Lfy
INHH MI V UVVH KHang
Vtei
• De tim A
\ ta lam nhu sau
- Sap xep
theo
thu tu tang dan cac diem dau
miit
cua cac tap hgp A, B len
true so ,^
- Bieu dien tap A tren true so
(gach
bo phan khong thuoe tap A),
gaeh
be
phan thuoe tap B tren true so
- Phan khong bi
gaeh
bo chinh la A

\ . ir nt -r- •
ea 1. cAc VI Du
MINH
HOA
Vi
du 1: Cho cac tap hop:
A =
{x6Rlx<3}
B = {XGRI1<X<5} C = {X
G
R I-2 <
X
< 4}
a) Hay viet lai cac tap hgp A, B, C duai ki hieu khoang, nira khoang, doan.
b) Tim A u B, A n B, A
\ .
c) Tim (BuC)\(AnC)
Len
gidi
a)
Tac6:
A = (-oo;3) B = (];5] C =
[-2;4].
b) • Bieu dien tren true so
Suy ra A w B = (-«; 5]
1
3 5
i
^—t
• Bieu dien tren true so ] 3 '5

H-H-i
)\/^/v\]\/^AA>
Suy ra AnB = (l;3)
• Bieu dien tren true so
13 5 '
(////)WM]\\\\
Suy ra A\ = (-oo;l]
c)
Bang
each
bieu dien tren true so ta c6: AnC =
[-2;3)
va BuC =
[-2;5]
Suy
rataco
(BuC)\(AnC) = [3;5]
Nhan
xet:
Viec
bieu
diln
tren true so de tim cac phep toan tap hop ta larr
tren
giay nhap va
trinh
bay ke't qua vao.
Vi
dv 3: Cho cac tap hgp A = (-oo;m) va B = [3m - l;3m + 3]. Tim m de
a)AnB

= 0 v v , 4
b)BcA
C)ACCRB '
d)CRAnB5t0
Lai
gidi
Ta
CO
bieu dien tren true
so'cac
tap A va B tren hinh ve
, , ,. ,
y///////.
Vay m > 1 la gia tri can tim. ' '
bijirM:>'^vr!
.i::^;;,
.^rrvn;!
ru^dT :"
uxT'DAoi
3 3A77-]3W + 3
b) Ta CO B c A <=> 3m + 3<m<=>m< — , , , , ,^ ,,,,,
2
/////[
]////»
'3 ^ ' •'
Vay m < - — la gia trj can tim. ^
c)
Taco
CRB =
(-«);3m-l)u(3m

+
3;+K)
Suy ra A c CRB «> m < 3m - ]
<=>
m > -
2 , „(
Vay m > i la gia trj can tim. ,
d) Ta CO CjiA =
[m;+oo)
suy ra C.pAn B
?^0c=>m<3m
+ 3<=>m>-^
Vay m > - ^ la gia trj can tim.
ca 2. BAI TAP
LUYEN
TAP
Bai 1.35: Xac djnh cac tap hgp AuB,A\C,AnBnCva bieu dien tren tryc so
cac tap hgp tim dugc biet:
a) A =
{xeR|-l<x<3},B
= {xeR|x>l}, C =
(-oo;l)
b) A =
{xeR|-2<x<2},B
=
{x€R|x>3},
C = (-oo;0) '8 *
Huang
dan
gidi

a) Co A =
[-l;3]
va B =
[l;+oo)
AwB
=
[-l;+c»),
A\ = [1;3], AnBnC = (l)
b) Co A = [-2; 2] va B = [3;+oo)
AuB
=
[-2;2]u[3;+«),
A\ = [0;2], AnBnC = ^
Bai 1.36: Cho hai tap hgp A = [0; 4), B = {x e
IR
/|x|
< 2}.
Hay xac djnh cac tap hgp AuB,AnB,A\
Humg
dan
gidi
A
=
[0;4),B
=
[-2;2],
AuB =
[-2;4),
AnB = [0;2], A\ = (2;4)
Bai 1.37: a) Cho A=ixeRI-l<x<5), B =

{xeRI-2<x<0
hoac
1 < x < 6 )
C=( X e RI X > 2
1
Tim
A n B, A o C, B
\ va bieu dien
each
lay ket qua tren true so
b) Cho A = (-00,-2), B = [2m +1,
+co).
Tim m de A u B = R . ' /
Huang
dan
gidi
• '
•'
' « " ' '
a) AnB = [-l;0)u(l;5) AuC =
[-!;+«))
B\ =
(-2;0)u(l;2)
b) AuB = R<=>2m+l<-2om< —
Bai 1.38: a) Tim m de
(1;
m] n (2; +») * 0 .
b) Viet tap A gom cac phan tir x thoa man dieu kien
Huang
dan

gidi
a) De (1; m]n(2; +oo)^0 thi m<2.
b) Viet tap A gom cac phan tix x thoa man dieu ki^n
SO.
x<3
X
+1
> 0 duoi dang tap so.
x<0
X
< 3
X
+ 1 > 0 duoi dang tap
x<0
X
< 3
•x<3
X
e
(-00;
3]
Co •
X
+ 1 > 0 o
•
X
> -1 «
•
X
€ [ -1; +

oo)
(bieu dien tren true so)
X
< 0
x < 0
X€(-oo;0)
<=>
X6(-oo;
3]n[-l;
+ao)n(-oo; 0)<=> x€[-l; 0).
Vgy A = [-l; 0).
Bai 1.39: Cho tap hgp A ^
a) AcB
m
-1;
m
+1
b) AnB = 0
Huang
dan
gidi
va B = (-oo;-2)u[2;+co). Tim m de
m
+1
Dieu
kien de ton tai tap hgp A la m -1 < —^ m < 3 (*)
a) AcB<=>
Ac(-«;-2)
Ae[2;+oo) '
^<-2

2 <^
m-l>2
m
< -5
m>3
Ket hgp voi dieu kien (*) ta c6 m < -5 la gia trj can tim
35
ury ii\nttMl V
UVVH
Khang vtft
b) AnB = 0«
-2 < m -1 ,,
m
+ 1
^»-l<m<3
<2 m<3
Ket hop voi dieu ki^n (*) ta c6 -1 < m< 3 la gia tri can tim
Bai 1.40: Cho hai tap khac rong
:
A = (m -1;4], B = (-2 ;2m + 2), voi meR. Xac
dinhmde: j":, '
a)
AnB^0;
• b) AcB; 1*
Be A; , ., d)
(AnB)c(-l;3).
' Huang dan gidi
\6i
A = (m -1;4], B = (-2 ;2m + 2) khac tap rdng, ta c6 dieu ki^n
m-1 <4

<=>
i
m< 5
, o -2 < m < 5 (*).
[2m + 2>-2 [m>-2
Voi dieu kien (*), ta c6 :
a)
AnB^0
« m-1 <2m + 2« m >-3 . So sanh voi (*) ta thay cac gia trj m
thoa man yeu cau AnB^0 la -2<m<5.
UN
A D fm-l>-2 fm>-l
b)
AeBc^l^^^^^^^l^^^
<^m>l. So sanhnta thay cac gia
trim
thoa man yeu cau A e B la
1
< m < 5.
^ D A fm-l<-2 fm<-l ^ y
c)
BcA<^
o om<-l.
2m + 2<4 [m<J
m
thoa man yeu cau B c A la -2 < m <
-1.
d)
(AnB)c(-l;3)o-
So sanh voi (*) ta thay cac gia trj

m-l>-l 1
2m.2<3^°^'"^2^*°'(*)^-
§4. SO GAN
DUNG.
SAI SO
A. TOM TAT LY THUYET ^ '
1.
S6
gdn dung
Trong nhieu truong hop ta khong the biet
dugc
gia trj dung cua d^i lugng
ma ta chi bie't so gan diing ciia no. . ,
^. Vi du: gia trj gan diing ciia n la 3,14 hay 3,14159; con doi vai 72 la 1,41 hay
1,414;
Nhu vay c6 su sai l?ch giua gia tri chinh xac cua mot dai lugng va gia tri
gan dung ciia no. De danh gia miic dg sai l#ch do, nguoi ta dua ra khai
niem sai so tuyet do'i.
2.
a)
a -a
d
b)
Sai
so tuyet doi:
Sai so tuy^t doi ciia so gan dung
Neu a la so gan diing ciia a thi ^^ =
so'gan
diing a . ., (,,( »*•
Dp chinh xac cua mpt so gan diing

Trong thuc te^ nhieu khi ta khong bie't a nen ta khong
tinh
dugc
Ag
nhien ta c6 the danh gia Ag khong vugt qua mot so duang d nao do.
Neu Aa < d thi a - d < a < a + d, khi do ta viet a = a ± d
Goi la do chinh xac cua so gdn dung. j
Sai so tucmg doi
Sai
sotuang do'i ciia so gan diing a, ki hieu la 63 la ti sogiiia sai so'tuyet
dugc
ggi la sai so'tuyet dot ciia
Tuy
doi va |a|, tuc la 83 =
lal
4.
5.
Nhdn xet: Neu a = a ± d thi Ag < d suy ra 83 < —. Do do — cdng nho thi
lal
lal
chat lugng ciia
phep
do dac hay
tinh
toan cdng cao.
Quy tron so gdn dung
Nguyen
tac quy tron cac so nhu sau:
Neu chii so
ngay

sau hang quy tron nho han 5 thi ta chi
viec
thay chir so' do
va cac chir
so'ben
phai no boi 0.
Neu chu so
ngay
sau hang quy tron Ion hon hay bing 5 thi ta thay chij so do
va cac chii
soben
phai no boi 0 va
cong
them mot don vj vao sohang lam
tron.
Nhdn xet: Khi thay so diing boi so qui tron den mot hang nao do thi sai so'
tuyet doi ciia so' qui tron khong vugt qua nua don vi ciia hang qui
tron.
Nhu vay, do chinh xac ciia so qui tron bang nua don vj ciia hang qui
tron.
Chii y: Cac viet so quy tron oia so gan dung can cii vao dp chinh xac cho truoc:
Cho so' gan diing a voi do chinh xac d. Khi
dugc
yeu cau quy tron a ma
khong noi ro quy tron den hang nao thi ta quy tron a den hang cao nha't ma
d
nho hem mpt dan vi ciia hang do.
Chu sochdc (dang tin)
Cho so gan diing a ciia so a vai do chinh xac d. Trong so' a, mgt chii so'
dugc

ggi la chu so chic (hay dang tin) neu d khong vugt qua nua dan vj
ciia hang c6 chir so' do.
:
„
Nhdn xet: Tat ca cac chCr so diing ben
trai
chii
so'chac
deu la chii so
chac.
Tat
ca cac chu so diing ben phai chii so khong
chac
deu la chii so khong
chSc.
©ang
chudn cua so gdn diing
Neu so gan diing la sothap phan khong nguyen thi dang chuan la dang ma
mgi chii so'ciia no deu la chii
chac
chan. /j.:,
- Neu
so'gan
diing
la so
nguyen thi dang chuan ciia
no la
A.10*^ trong
do
A

la
so
nguyen,
k la
hang thap nha't
c6
chu
sochSc
(k e
N ).
(suy
ra
moi chii
so
cua
A
deu
la
chu
so'chic
chan)
,y
Khi
do do
chinh
xac d
=
CS.IO"".
6.
JCi

hieu
khoa
hoc cua mot so
MQ\O thap phan
khac
0
deu
vie't dugc
dudi
dang a.lO",
1 <
|a|
< 10,
n e Z
(Quy uoc
10"" = )
dang nhu vay dugc goi
la
kihieu
khoa
HQC
cua
so
do.
B.
CAC
DANG
TOAN
VA
PHlTaNG PHAP GIAI.

DANG
TOAN 1: TINH
SAI SO
TUYETDOI,
SAI SO
TUONG
DOI CUA SO
GANDUNG
.
VIET
SO QUY
TRON.
ea 1. cAc
vf Du
MINH
HOA
Vi
1: Do
dai
cua cai cau ben
thuy
hai
(Nghe An) nguai
ta do
dugc
la
996m
±
0,5m
•

Sai so
tuang doi to!
da
trong phep
do
la
bao
nhieu.
Loi
gidi
.
Ta
CO
do
dai gan dung
cua cau
la a
= 996
vol
do
chinh
xac d
=
0,5
Vi
sai
so
tuy^t doi
Ag <
d

=
0,5
nen
sai
so
tuong doi
|a|
a| 996
Vay
sai
so
tuang doi toi
da
trong phep
do
tren
la
0,05%.
Vi
dv 2:
Hay
xac
dinh
sai so
tuyet doi
cua cac so
gan
dung
a, b
biet

sai so
tuong
doi
cua
chung.
a)
a
=
123456,
5,
= 0,2%
b) a
= 1,24358,
8,
=
0,5%
Lai
gidi
Taco
8,=^oA,=|a|5,
a)
Vol
a
-123456,
8^,
0,2% ta c6
sai
so
tuyet doi
la

Aa
=123456.0,2% = 146,912
b)
Voi
a
= 1,24358, 8^ = 0,5%
ta c6
sai
so
tuy?t doi
la
Ag
=
1,24358.0,5% = 0,0062179
.
Vi
dv 3:
Lam tron
cac so
sau
voi
do
chinh
xac cho
truac.
a)
a
= 2,235
voi
do

chinh
xacd
= 0,002
b)
a =
23748023
vai
do
chinh
xacd
=
101
Lai
gidi
a)
Ta
CO
0,001
< 0,002
<
0,01
nen
hang
cao
nhat
ma
d
nho hon mot don
vj
cua

hang
do
la
hang phan tram
Do
do
ta
phai quy tron
so a
= 2,235
den hang phan tram
suy
ra a * 2,24 .
b)
Ta
CO
100
< 101 < 1000 nen
hang
cao
nhat
ma d
nho hon mot don
vi
ciia
hang
do
la
hang nghin
:» it • , - :

Do
do
ta
phai quy tron
so a
= 23748023 den
hang nghin
suy
ra a
a;
23748000
Vi
du
4: a)
Hay vie't
gia trj gan
diing
ciia
\/8
chinh
xac
den hang phan tram
va hang phan nghin biet
Ts
= 2,8284 l/oc
lugng
sai so
tuy^t doi trong
moi
truang hgp.

b) Hay vie't
gia trj gan
diing
ciia
\/2015^
chinh
xac
den
hang
chuc
va
hang
tram
biet
V2015^
=
25450,71
. l/oc
lugng
sai so
tuyet
doi
trong
moi
truang hgp.
Lai
gidi
a)
Ta
CO

Vs
=2,8284
do
do gia trj gan
diing
ciia
N/8
den
hang phan tram
la
2,83
Taco
78-2,83 =2,83-78
<
2,83-2,8284
=
0,0016
Suy
ra
sai
so
tuyet doi ciia
so
gan
diing
2,83
khong vugt qua
0,0016 .
Gia trj
gan

diing
ciia
78
den
hang phan nghin
la
2,828
Ta CO
178 -
2,828|
=
78 -
2,828 < 2,8284
-
2,828 =
0,0004
Suy
ra sai so
tuyet doi ciia
so
gan
diing
2,828
khong vugt qua
0,0004 .
b)
Su
dung may
tinh
bo

tiii
ta c6
\/2015^ = 25450,71966
>:\i
Do
do
gia trj
gan
diing
ciia
N/2015'' den
hang
chuc
la 25450
=
72015''
-
25450
<
25450,72
-
25450
= 0,72
Ta
CO
72015^
-25450
Suy
ra sai
so

tuyet doi ciia
so
gan
diing
25450
khong vugt qua
0,72.
Gia trj gan
diing
ciia
72015'*
den hang tram
la 25500.
Ta
CO 72015" -
25500 = 25500
-
^2015"
<
25500
-
25450,71 = 49,29
Suy
ra
sai
so
tuy^t doi ciia
so
gan
diing

25500
khong vugt qua
49,29
.
ea
2.
BAI TAP
LUVeN
TAP
Bai
1.41. Su
dung
may
tinh
bo
tiii,
hay
vie't
gia trj gan
diing
ciia moi
so
sau,
chinh
xac
den hang phan tram
va
hang phan nghin
:
a)

73 ; b) .
Huong
dan
gidi
a)
Sit
dung may
tinh
bo
tiii
ta c6
N/S
= 1,732050808 Do
do:
Gia tn gan
diing
ciia
V3
chinh
xac den
hang phan tram
la 1,73. Gia
tri
gan
dung ciia
-JS
chi'nh
xac
den hang phan nghin
la 1,732.

b)
Sit
dung
may
tinh
bo
tiii
ta c6 gia tn cua la 9,8696044 Do do : Gia trj
gan
diing
ciia chinh
xac den
hang phan tram
la 9,87. Gia
tri
gan
diing
cua chinh
xac
den hang phan nghin
la 9,870.
Bai
1.42:
Hay vie't so'quy
tron
cua so a
voi do chinh
xac d
dugc cho
sau

day;
a)
a = 17658 ± 16 ; b) a =15,318
±0,056
.
Huang
dan
gidi
a) Vi
10 < 16 < 100
nen hang
cao
nha't ma
d
nho han mot don
vj cua
hang do
la
hang tram. Nen
ta
phai quy
tron
so 17638
den hang tram. Vay
so
quy
tron
la
17700
(hay vie't

a* 17700).
b)
Ta
CO
0,01 < 0,056 < 0,1 nen
hang
cao
nha't
ma d nho hon mot
don
vj
ciia
hang
do la
hang phan
chuc.
Do do
phai quy
tron
so 15,318 den
hang phan
chuc.
Vay
so
quy
tron
la 15,3
(hay vie't
a
«15,3).

2
Bai
1.43:
Cho
so x =
Cho
cac
gia trj
gan
diing
ciia
x la :0,28 ; 0,29 ; 0,286 .
Hay
xac
dmh
sai so
tuyet doi trong
tung
truong hop
va cho
biet gia trj
gan
diing
nao
la
tot nha't.
Huang
dan
gidi
Ta

CO
cac sai so
tuyet doi
la :
0,28
7
175
Au
=
0,29
7
700
0,286
3500
Vi
Ac
<
Ab
<
An
nen
c = 0,286 la so
gan
diing
tot nha't.
Bai
1.44:
Mot mie'ng da't
hinh
chii

nhat
c6
chieu
rpng
x =
43m
±
0,5m
va
chieu
daiy
=
63m
±
0,5m.
Chung
minh
ring
chu vi
P
ciia miehg da't
la P =
212m
±
2m
.
Huang
dan
gidi
Gia sir

X =
43
+
u,
y =
63
+ V.
Taco
P = 2x + 2y = 2(43 + 63) +
2u+2v
= 212 +
2(u
+v).
Theo
gia thie't
-0,5<u<0,5va -0,5<v<0,5 nen -2 <
2(u
+
v)
< 2 .
Dodo
P =
212m±2m.
DANG
TOAN
2: XAC
DINH
CAC
CHU
SO

CHAC
CUA MOT
SO
GAN
DUNG, DANG
CHU AN
CUA CHU SO GAN
DUNG
VA Ki
HIEU
KHOA
HOC CUA MOT SO.
PI.CAC
ViDUMINH
HOA
-^^[dyl^'r"
so chac va
vie't dang chuan ciia
so
gan
diing
a
biet
a)
So
nguoi
dan
tinh
Nghe
An la a = 3214056

nguoi
voi do
chinh
xac
d
= 100
nguoi. r
<>"
r
g
_ 1,3462 sai so
tuong doi ciia
a
bang 1%.
Lai
gidi
#
y\^ = 50 < 100 < = 500 nen chu
so
hang tram(so
0)
khong
la so chac,
con
chii
so
hang nghin(so
4) la
chii
so chac.

Vay chu
so chac la 1,2,3,4 .
Cach
vie't
duoi
dang chuan
la 3214.10'^ .
b) Ta CO
6a = ^
Aa
=5a.|a|
=
l%.l,3462
= 0,013462
Suy
ra do
chinh
xac
ciia
so
gan
diing
a
khong vugt qua
0,013462 nen ta c6
the xem do chinh
xac la d = 0,013462.
Ta
CO
= 0,005 < 0,013462 < ^ = 0,05

nen chu
so
hang phan tram(so
4)
khong
la so chac, con
chu
so
hang phan
chiic(so
3) la
chii
so
chac.
i) (
Vay chij'
so chac la 1 va 3 .
Cach
vie't
duoi
dang chuan
la 1,3.
Vi d\i
2:
Viet
cac so
gan
diing
sau
duoi

dang chuan
a)
a = 467346 ±12 b) b = 2,4653245 ± 0,006
Lai
gidi
a)
Ta
CO
— = 5 < 12 < — = 50 nen chu so
hang tram tro di
la
chir
so chSc do
2
2
do
so
gan
diing
viet
duoi
dang chuan
la 4673.10^ . V
b)
Ta
CO
Ml = 0,005 < 0,006 < ^ = 0,05 nen
chu
so
hang phan

chuc
tro di
la
p._£huso
chir
so chac
do do
so
gan
diing
vie't
duoi
dang chuan
la 2,5 .
du
3: Cac
nha khoa
hoc
My dang nghien
ciiu
lieu
mot may
bay c6 the c6
toe
do gap bay Ian toe do
anh
sang.
Voi may
bay do
trong mot nam

(gia
su mot nam
c6 365
ngay)
no
bay dugc bao
nhieu?
Biet
van toe anh
sang
la
^jgo
nghin km/s. Viet ke't qua
duoi
dang ki hieu khoa hoc.
^
Lai
gidi
Ta
CO
mot nam c6 365 ngay, mot ngay c6 24 gio, mpt gio c6 60 phut va mot
phiit
CO
60 giay
Vay mot nam c6
24.365.60.60
=
31536000
giay.
Vi

van toe anh
sang
la 300 nghin km/s nen trong vong mot nam no di dupe
31536000.300
=
9,4608.10'^
km. ; ;
-i;-:::;
ca 2. BAI TAP LUYEN TAP
Bai 1.45: So dan cua mot
tinh
la A =
1034258
± 300 (nguoi). Hay tim cac chi> so
chac
va viet A
duoi
dang chuan.
Hu&ng dan
gidi
Ta
CO
: = 50
<300
< 500 = 1^9^ nen cac chu so 8 (hang don vj), 5 (hang
chuc)
va 2 ( hang tram ) deu la cac chir so khong
chac.
Cac chCr so con lai 1, 0, 3, 4 la chi> so
chac.

Do do
each
viet chuan cua so A la A «
1034.lO''
(nguoi).
Bai 1.46: Do chieu dai cua mot con doc, ta dupe so do a =
192,55
m , voi sai so
tuang
doi khong
vupt
qua 0,2%. Hay tim cac chu so
chac
cua d va neu
each
viet
chuan gia trj gan dung cua a .
Huang
dan
gidi
Ta
CO
sai so
tuyft
doi cua so do chieu dai con doc la :
Aa
=a.5-,
<
192,55.0,2%
=

0,3851
Vi
0,05 < Aa < 0,5 . Do do chu so
chac
cua d la 1, 9, 2.
Vay
each
viet chuan cua a la 193 m (quy
tron
den hang don vi).
Bai 1.47: Cho
3,141592
<
7t
<
3,141593
. Hay viet gia trj gan
diing
cua so TI
duoi
dang chuan va danh gia sai so tuy?t doi ciia gia trj gan dung nay
tron^
moi
truong hop sau :
a) Gia tri gan dung cua
TI
c6 5
chii'
so
chac;

b) Gia trj gan dung cua n c6 6 chir so
chac;
c) Gia trj gan
diing
cua 7t c6 3 ehi>
so'chac.
Huang
dan
gidi
a) Vi
CO
5 chu- so
chac
nen so gan
diing
cua n dupe viet
duoi
dang chuan la
3,1416
(hay TI*3,1416).
Sai sotuyetdoiciia so gan dung la =
|3,1416
-
TI|
<
0,000008.
b) Vi
CO
6 chir so
chac

nen
TI
a
3,14159
va sai so tuyet doi cua
so'gan
dung na\
la A„ =
|3,14159
-
Ti|
<
0,000003.
c) Vic6
3ehiisochacnen
Ti*3,14
va A^ |3,14 -
n|
<
0,001593.
ONTAPCHUONGI
Bai 1-48: Cho Oxy, lap menh de keo theo va nienh de tuong duong ciia hai
menh de sau day va cho biet
tinh
dung, sai ciia chiing:
p
: "Diem M nSm tren phan
giac
cua goc Oxy ".
n

•
"Diem M
each
deu hai canh Ox, Oy". "' " ""'' '
Huang
dangiat
p
^ Q : "Neu diem M nam tren phan
giac
cua goc Oxy thi M
each
deu hai
canh Ox, Oy ": dung.
Q
=> P : "Neu diem M
each
deu hai canh Ox, Oy thi M nam tren phan
giac
ciia goc Oxy " : dung.
p
ci> Q : "Diem M nam tren phan
giac
ciia goc Oxy neu va chi neu (khi va
chi
khi) diem M
each
deu hai canh Ox, Oy" :
diing.
Hay
: P » Q : "Dieu kien can va dii de diem M nam tren phan

giac
ciia goc
Oxy la M
each
deu hai canh Ox, Oy" :
dimg.
. /
Bai 1.49: Cho djnh li : "Cho so tu nhien n. Neu n"^ chia he't cho 5 thi n chia het
cho 5".
Dinh
li nav dupe viet
duoi
dang P => Q.
a) Hay xac
dinh
cac menh de P va Q.
b)
Phat
bieu
dinh
li tren bang
each
diing
thuat ngir "dieu kien can".
c)
Phat
bieu
dinh
li tren bang
each

diing
thuat ngu "dieu kien
dii".
d)
Hay phat bieu djnh li dao (neu c6) cua
dinh
li tren roi
diing
cac thuat ngij
"dieu
ki^n can va
dii"
phat bieu gpp ca hai
dinh
li thuan va dao.
Huong
dan
gidi
a) P : "n la so tu nhien va
n'^
chia he't cho 5", Q : "n chia he't cho 5".
b) Voi n la so tu nhien, n chia he't cho 5 la dieu kien can de n'^ chia he't cho 5 ;
hoac
phat bieu
each
khae
: Voi n la so tu nhien, dieu kien can de chia he't
cho 5 la n chia he't cho 5.
c) Voi n la sotu nhien,
n"*

chia he't cho 5 la
dieu
kien
du de n chia he't cho 5.
Djnh
li dao : "Cho so'tu nhien n, neu n chia he't cho 5 thi n^ chia he't cho 5".
That vay, neu n = 5k thi
n"^
= 5\k'^: So nay chia he't cho 5.
Dieu
kien
can va du de n chia he't cho 5 la
n"*
chia he't eho 5. ' ''
•'
Bai 1.50: Cho tap X = { 1 ; 2 ; 3 ; 4 ; 5 ; 6 ; 7 }.
3) Hay tim tat ca cac tap con ciia X c6 chiia cac phan tir 1, 3, 5, 7.
°) Co bao nhieu tap con ciia X chira
diing
2 phan tir ?
Huang
dan
gidi
^) Cac tap eon ciia X chira c6 cac phan tir 1, 3, 5, 7 dupe thanh lap bang
each
them vao tap {1 ; 3 ; 5 ; 7 } cac phan tir con lai ciia tap X. '
Do do tat ca cac tap con cua X c6 chiia cac phan tu 1, 3, 5, 7 la :
{1;3;5;7},
{l;3;5;7;2},
{];3;5;7;4},

{l;3;5;7;6},
{1;3;5;7;2;4}, {1;3;5;7;2;6},
{1;3;5;7;4;6}
va X.
b) Gia su tap can tim la i a ; b jvoi a b. ., - rv''•
• Vi X
CO
7 phan tu nen c6 7
each
chon phan tir a. Sau khi chon a thi X con 6
phan tu, do do voi moi
each
chon a, ta c6 6
each
chon phan tu b, nhu vay c6
7.6 = 42 cap (a ; b)
theo
each
chon nay.
Nhung voi
each
chon tren thi voi hai phan tu bat k"i a, b ta da chon lap lai
hai
Ian, do la hai cap {a;b) va
(b;a),
nhung chi c6 duy nhat tap{ a;b}.
42
Do do,
CO
—="1^ tap con cua X chua dung hai phan tu.

Bai 1.51: Xet
tinh
diing sai ciia menh de sau va neu m?nh de phu djnh ciia no.
a) 3x€Q: 4x^-1 = 0 ;
b)3xGZ,x^=3;
c) Vn G N* : 2" + 3 la mot so nguyen to ; d) Vx e M, x^ + 4x + 5 > 0 .
e) VXGIR,X'*-x^+2x + 2>0
' Huang dan gidi
a) Giai phuong
trinh
: 4x^ -1 = 0 ox =
±ieQ.
Vay menh de da cho dung.
Menh de
phii
djnh
\/x&Q:
Ax^ -
b) Ta
CO
x^ = 3 <=> x = ±V3 . Vi ±N/3 i Z nen menh de da cho sai.
Menh de
phii
djnh Vx G Z, x^ ;^ 3
c) Voi n = 5 thi 2" + 3 = 35, so nay chia het cho 5 (khong nguyen to). Do do
menh de da cho sai.
Menh de
phii
djnh "Bn e N* : 2" + 3 khong phai la mot so nguyen to"
d) Menh de dung vi x^ + 4x + 5 = (x + 2)^ +1 > 0, Vx € K.

Menh de phu djnh 3x e K, x^ + 4x + 5 < 0 'iJiosoa
e) x"* - x^ + 2x + 2 = (x^ -1 + (x + 1)^ nen menh de da cho diing
M?nh de
phii
djnh 3x G K, x^ - x^ + 2x + 2 < 0
Bai 1.52: Xet djnh If: "Neu x la so thuc am thi x + i < -2 ".
x
a) Viet djnh li tren duoi dang ki hieu.
b) Djnh li tren c6 djnh If dao khong ? Giai thich.
c) Su dung thuat
ngi>
"dieu ki^n can" va "dieu kien dii" de phat bieu djnh If tren.
Huang dan gidi
a) xeM, x<0^x + l<-2 .
^
'
b) Menh de dao la x G K, x + - < -2 => x < 0 ".^ •
1
„ x^+2x + l (x + 1)^ 1 n t- - u^-^.
Ta
CO
X
+ - + 2 = = — — do do x +
—
< -2 => x < 0 la m^nh de dung
X X X ^ X °
Vay dinh li tren c6 dinh li dao. ^ ;
g^i 1.53: Chung minh bang phan ehung djnh If sau:
Chung minh rang voi n G N, ta co: n : le o 3n + 1 : chan
Huang dan gidi

Thmn:
Cho n : le, thi n = 2k + l {k eN) V,! ,V, ,
r:> 3« +
1
= 3(2^ + l) + -\=6k + 4 = 2{3k + 2), voi 3^ + 2 G N
=> 3« + 1: chan
Dao
: Cho 3n + 1 : chan, ta chung minh n : le
Dung phuong phap phan chung: ) .
Gia su n : chan, tuc \an = 2k(k e N) * j
=> 3« + 1 = 6A: + 1 =>
3?7
+ 1 : le:
trai
voi gia thie't. Vay 3n + l: le. '^^
Tu hai phan thuan va dao ta
duoe:
«: le o 3« + 1: chan
Bai 1.54: Cho cac tap hop:
A = {xGZi l<x<6),
B
=
{XGQI(1-3X)(X''-3x^+2)
= 0|, C =
(0;l;2;3;4;5;6}
*
a) Viet cac tap hop A, B duoi dang liet ke cac phan tu, tap C duoi dang chi ro
tfnh
dac trung cua phan tu.
b) Tim AnB, AuB, A\B,

Cg^^AAnB.
c) Chung minh rang
An(BuC)
= A.
* Huang dan gidi
a)
A {-l;
0; 1; 2; 3; 4; 5}, B = {-l;i;l VI
•I
V ,
(l-3x)(x'*-3x2+2)
= 0o
x = ±l •
x =
±-j2&Q
• ,
x = l/3
•'::>*o
C
= (XGNIX<6| ' ^'-^ -
b) AnB = |-l;l}; AuB = {-l;0;i;l;2;3;4;5}, A\ =
{0;2;3;4;5),
CBuA(AnB)
=
{0;l/3;2;3;4;5)
BuC = i-l;0;l/3;l;2;3;4;5;6),
An(BuC)
=
{-1;0;1;2;3;4;5)
= A

fiian
loai va pnwimg pnapgiiu
tjiii
so lu
Bai
1.55: Tim quan he bao ham hay bang nhau giua cac tap hop sau day:
a) A =
|X€N|
x<2 }; B =
jxeQ
b) A = |x £ 4x^ - 9 = o|; ' B = |x 6 K
c) A = jx6N|l<x<4}; B = {xeZ
Huang dan gidi
a)
Taco:
A =
{XGN|
x<2 }=> A =
{0;l}.
(1)
(x^ - x)(x^ - 2) = 0 .
(x^
-x)(x2
-2) = 0
X + 4x
-I-
x^ -9 = 0 .
XGQ
B
=

(x2-x)(x2-2)
= 0«
x^ - x = 0
x^ -2 = 0
x = 0 V X = 1
Q
X = ±V2 « O
x = 0
x = l '
B
=
{0;1}(2)
TCr(l)
va (2) cho: A = B.
(3)
(4)
b) Ta c6; 4x^ - 9 = 0 <=> x = ±- g Z A = 0
x^ + 4x = 0 •» X = 0 V X =-4 => B = {0; - 4}.
Tit
(3) va (4) cho: A c B.
c)
Taco:
A = {x e N
|
1 < x < 4} => A = {2 ; 3}.
B
= {x6Z| x^-9 = 0}=^B =
{-3;3}.
Ta thay: 2 e A ma 2 g B nen A cr B; -3 e B ma -3 g A nen B cz A.
Bai

1.56: Cho A = {0; 2; 4; 6}, B = {4; 5; 6}.
a) Hay xac djnh tat ca cac tap con khac rong X, Y ciia A biet ran;
XuY = A va
(AnB)cX;
b) Hay xac djnh tat ca cac tap P biet rang (A n B) c P c (A u B).
Huang dan gidi
a)
Taco
AnB = {4 ; 6}cX.
Do do cac tap X, Y thoa man yeu cau la: X = {4 ; 6} v
Y = {0; 2}, X = {4; 6; 0} va Y = {2}, X = {4; 6 ; 2} va Y = {0}.
b) Ta CO AuB = (0;2;4;6;51, do do cac tap P thoa man dieu kic'
t;
'
(AnB)cPc(AuB)
la:
{4 ; 6}, {4 ; 6 ; 0}, {4 ; 6 ; 2}, {4 ; 6 ; 5}, {4 ; 6 ; 0 ; 2}, {4 ; 6 ; 2 ; 5},
{4;6;5;0}va
{4;6;0;2;5}.
Bai
1.57: Cho ba tap hop:
A =
{xeR|-3<x<l};B
= {x6R|-l<x<5|; C = {x e S
||x|
> 2' .
a) Xac djnh cac tap hop sau day va viet ket qua duoi dang khoang, doan hay
nua
khoang: A o B, A u B, (B \) n C.
b) Chung minh rang:

Cs(AuB)
=
(CsA)n(CsB).
Huang dan gidi
a)
Taco:
AnB = [-3;l)n[-l;5] =
[-l;]);
AuB = [-3;l)u[-l;5] = [-3;5]; B\ = [-1; 5]\[-3; 1) = [l; 5 .
c = {
X e IK.
|x|>2}=(-«;-2]u[2;
+ x); (B
\nC = [2 ; 5
b)Tac6:
C^ (A u B) = CR [-3; 5] = (-x ; - 3)u{5 ; + x). (1)
CgA = CK[-3;l)=(^;-3)u[l; + x); C,B =
C,,[-1;5]
=
;-l)u(5;+x);
(CKA)n(Cj,B)
=
(^;-3)u(5;
+ x). (2)
(1)
va(2)cho:
C^ (Au B) = (CB{A)n(CKB).
Bai
1.58: Cho hai tap hop A, B bat ki.
Chung minh rang: AuB =

AriB<=>A
= B.
Huattg dan gidi
• Thuan:/4 u B =
/4
n B, ta chung minh :/4 = B
Vx, X e A X € A u B (vi A c A u B) => X e A n B
(viAuB = AoB) x€ B (vi AnBc B)
Nhu
the: Vx,x e A => X e B, nen A c B (a) , >
Vx,xeB=>xG AuB(vi BcAuB)=>x€AnB
(viAuB = AnB) => x e A (vi An Bc A)
Nhuthe:
Vx,xeB=>xe
A,nen BcA (b) .i>
Tir (a) va (b) cho/I = B
• Dao:
Cho/4
= B, ta Chung minh : AuB = AnB '
Ta CO A u B = A u A ( vi B = /\ =
/\)
AnB = AnA (viB = /l) =
/4
(d)
Tu
(c) va (d) cho A u B = A n B .
Tu
hai phan thuan va dao ta dugc: AuB =
AnB<=>A
= B

%hu(mq2
HAMSOBACNHATVABACHAI
§l.DAICUaNGVEHAMS6 m-^ ^.i
A. TOM TAT LY
THUYET
* ^
L
©mn ngnia
• Cho D c K, D
7i
0 . Ham so i xac
djnh
tren D la mot qui t3c dat tuong un,
moi
so' X 6 D voi mot va chi mot so y e .
• X dugc goi la bien so (doi so), y dugc gpi la gia tri cua ham so / tai x .
Ki
hieu: v = f (x).
• D dugc goi la tap xdc djnh ciia ham so f. K
2. Cdch cho ham so
• Cho bang bang
•
Cho bang bieu do
•
Cho bang cong thiic y = f(x)
Tap xdc dinh cua ham so y =
f
(x) la tap hop tat ca cac so thuc x sao
chi
bieu

thuc f(x) co nghla.
3.
<^6
thi cua ham so V*>
Do
thi ciia ham so y = f (x) xac
dinh
tren tap D la tap hop tat ca cac
dicn
M(x;
f(x)) tren mat phang toa do voi mgi x e D .
Chu y: Ta thuong gap do thi cua ham so' y =
f
(x) la mot duong. Khi do t
noi
y =
f
(x) la
phuattg
trinh cua duong do.
4. Su
bien
thien
cua ham so ^
Cho ham so f xac
dinh
tren K . ' ^
• Ham so y =
f
(x) dong bien

(tang)
tren K neu
Vxi,X2
e K :
Xj
< X2 => f(xi) < f(x2)
• H^m so y = f(x) nghich bien
(giam)
tren K neu v-
Vxi,X2
GK:XI
<X2
=>f(x^)>f(x2)
^
5. Tinh
chdn
le cua ham so
Cho
ham so y =
f
(x) c6 tap xac
dinh
D .
• Ham so f dugc goi la ham so chin neu voi Vx e D thi -x € D va
f(-x)
=f(x!
• Ham so f dugc ggi la ham so le neu voi Vx e D thi -x
G
D va f(-x) =:-f(x)
•^U

Chii y: + Do thi cua ham
sochan
nhqn
true tung lam true doi xi'mg.
+
Do thi eua ham
sole
nhqn
goe toq do lam tam doi xung.
6: Tinh
tien
do thi
song song
vcri
true toa do ^
Dinh
ly: Cho (G) la do thj cua y =
f
(x) va p > 0, q > 0; ta c6
Tjnh
tien
(G) len tren q don vj thi dugc do thi y =
f
(x) + q
Tinh
tien
(G) xuong
duoi
q don vj thi dugc do thi y =
f

(x) - q
Tinh
tien
(G) sang
trai
p don vi thi dugc do thi y =
f
(x + p)
Tjnh
tien
(G) sang phai p don vi thi dugc do thj y =
f
(x - p)
B. CAC
PANG
TOAN
VA
PHl/ONG
PHAP
GIAI.
DANG
TOAN
1: TJM TAP XAC
DINH
CUA PHUONG
TRJNH.
^hucmgphdpgini.
Tap xac
djnh
cua ham so y = f(x) la tap cac gia tri cua x sao cho bieu thuc

f(x)
CO
nghla
Chu y: Neu P(x) la mot da thuc
thi:
CO
nghla
<=>
P(x) ^ 0
P(x)
7P(X)
CO
nghla o P(x) > 0
CO
nghla
<=>
P(X)
>0
ca 1.
CAC
Vf DU
MINH
HQA
Vi
du 1: Tim tap xac
dinh
cua cac ham so sau
2x'^ + x + l
h) y =
(x^-l)

-2x2
a) DKXD: x^ + x^ - 5x - 2 ^ 0 »
Ldfi
gidi
x^2
-3±S
X
?t
•
Suy ra tap xac dmh cua ham so la D = K
\;
^2
-3-V5 -3
+
V5
2 2
b) DKXD: (x^ - \f -Ix^^Oc^ (x^ - V2x
-\)(x^
+ V2x -1) ^ 0
V2±V6
x^ - V2x -1 ^ 0
+
N/2X
-1
7^
0
X
^•
X
?t

•
2
-V2±N/6
Suy ra tap xac
dinh
cua ham so la
j5^j^JV2-V6
V2+V6
-V2->/6
-V2 + 76