Lai gidi
Ta C O mot nam c6 365 ngay, mot ngay c6 24 gio, mpt gio c6 60 phut va mot phiit C O 60 giay
Vay mot nam c6 24.365.60.60 = 31536000 giay.
V i van toe anh sang la 300 nghin km/s nen trong vong mot nam no di dupe
31536000.300 = 9,4608.10'^ k m . ; ; -i;-:::;
ca 2 . B A I T A P L U Y E N T A P
Bai 1.45: So dan cua mot tinh la A = 1034258 ± 300 (nguoi). Hay tim cac chi> so chac va viet A d u o i dang chuan.
Hu&ng dan gidi
Ta C O : = 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 ô 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 C O sai so t u y f t 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 d u n g cua n c6 6 chir so chac;
c) Gia trj gan diing cua 7t c6 3 ehi> so'chac.
Huang dan gidi
a) V i C O 5 chu- so chac nen so gan diing cua n dupe viet d u o i dang chuan la 3,1416 (hay T I * 3 , 1 4 1 6 ) .
Sai s o t u y e t d o i c i i a so gan dung la = |3,1416 - TI| < 0,000008.
b) V i C O 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) V i c 6 3 e h i i s o c h a c n e n T i * 3 , 1 4 va A^ |3,14 - n| < 0,001593.
O N T A P C H U O N G I
Bai 1-48: Cho O x y , lap menh de keo theo va nienh de tuong d u o n g 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, O y " . "' " ""'' ' Huang dangiat
p ^ Q : " N e u diem M nam tren phan giac cua goc Oxy thi M each deu hai canh Ox, Oy ": dung.
Q => P : " N e u diem M each deu hai canh Ox, Oy thi M nam tren phan giac ciia goc Oxy " : dung.
p ci> Q : " D i e m M nam tren phan giac ciia goc Oxy neu va chi neu (khi va chi khi) diem M each deu hai canh Ox, O y " : diing.
Hay : P ằ Q : "Dieu kien can va dii de diem M nam tren phan giac ciia goc Oxy la M each deu hai canh Ox, O y " : dimg. . / Bai 1.49: Cho djnh li : "Cho so t u 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 d i i " .
d) Hay phat bieu djnh li dao (neu c6) cua dinh li tren roi diing cac thuat ngij
"dieu ki^n can va d i i " phat bieu gpp ca hai dinh li thuan va dao.
Huong dan gidi
a) P : " n la so t u nhien va n'^ chia he't cho 5", Q : " n chia he't cho 5".
b) Voi n la so t u 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 t u nhien, dieu kien can de chia he't cho 5 la n chia he't cho 5.
c) Voi n la s o t u 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 t u 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 s u tap can t i m la i a ; b j v o i a b. ., - .-rv''•
• V i X CO 7 phan t u nen c6 7 each chon phan tir a. Sau k h i chon a t h i X con 6 phan t u , do do v o i moi each chon a, ta c6 6 each chon phan t u b, n h u vay c6 7.6 = 42 cap (a ; b) theo each chon nay.
N h u n g v o i each chon tren t h i v o i hai phan t u bat k"i a, b ta da chon lap lai hai Ian, do la hai cap {a;b) va (b;a), n h u n g chi c6 d u y nhat t a p { a ; b } .
42
Do do, CO —="1^ tap con cua X chua d u n g hai phan t u .
Bai 1.51: Xet t i n h d i i n g sai ciia menh de sau va neu m?nh de p h u d j n h ciia no.
a) 3 x € Q : 4 x ^ - 1 = 0 ; b ) 3 x G Z , x ^ = 3 ; c) V n G N * : 2 " + 3 la m o t so nguyen to ; d) Vx e M, x^ + 4x + 5 > 0 .
e) V X G I R , X ' * - x ^ + 2 x + 2 > 0
' Huang dan gidi
a) Giai p h u o n g t r i n h : 4x^ - 1 = 0 o x = ± i e Q . Vay menh de da cho d u n g . M e n h de p h i i d j n h \/x&Q: Ax^ -
b) Ta CO x^ = 3 <=> x = ±V3 . V i ±N/3 i Z nen menh de da cho sai.
M e n h de p h i i d j n h Vx G Z, x^ ;^ 3
c) V o i n = 5 t h i 2 " + 3 = 35, so nay chia het cho 5 (khong nguyen to). D o do menh de da cho sai.
M e n h de phii d j n h " B n e N * : 2" + 3 khong phai la m o t so nguyen to"
d) M e n h de d u n g v i x^ + 4x + 5 = (x + 2)^ + 1 > 0, Vx € K.
M e n h de p h u d j n h 3x e K, x^ + 4x + 5 < 0 ' i J i o s o a
e) x"* - x^ + 2x + 2 = (x^ - 1 + (x + 1)^ nen menh de da cho d i i n g M ? n h de p h i i d j n h 3x G K, x^ - x^ + 2x + 2 < 0
Bai 1.52: Xet d j n h I f : " N e u x la so thuc am t h i x + i < - 2 ".
x a) Viet d j n h l i tren d u o i dang k i hieu.
b) D j n h l i tren c6 d j n h If dao k h o n g ? Giai thich.
c) Su dung thuat ngi> "dieu ki^n can" va "dieu kien d i i " de phat bieu djnh If tren.
Huang dan gidi
a) x e M , x < 0 ^ x + l < - 2 . ^ '
b) M e n h de dao la x G K, x + - < - 2 => x < 0 ".^ •
1 „ x ^ + 2 x + 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 d i n h l i tren c6 d i n h l i dao. ^ ; g^i 1.53: C h u n g m i n h bang phan ehung d j n h If sau:
C h u n g m i n h rang v o i 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:> 3ô + 1 = 3(2^ + l) + -\=6k + 4 = 2{3k + 2), v o i 3^ + 2 G N
=> 3ô + 1: chan
Dao : Cho 3n + 1 : chan, ta chung m i n h n : le
D u n g p h u o n g phap phan chung: ) . Gia s u n : chan, tuc \an = 2k(k e N) * j
=> 3ô + 1 = 6A: + 1 => 3?7 + 1 : le: trai v o i gia thie't. Vay 3n + l: le. '^^
T u hai phan thuan va dao ta duoe: ô : le o 3ô + 1: chan Bai 1.54: Cho cac tap hop:
A = { x G Z i l < x < 6 ) ,
B = { X G Q I ( 1 - 3 X ) ( X ' ' - 3 x ^ + 2 ) = 0|, C = (0;l;2;3;4;5;6} * a) Viet cac tap h o p A, B d u o i dang liet ke cac phan t u , tap C d u o i dang chi ro
tfnh dac t r u n g cua phan t u .
b) T i m A n B , A u B , A \ B , C g ^ ^ A A n B . c) C h u n g m i n h rang A n ( B u C ) = A.
* Huang dan gidi a) A - . { - l ; 0; 1; 2; 3; 4; 5 } , B = { - l ; i ; l V I
•I V ,
( l- 3 x) ( x' * - 3 x 2+ 2 ) = 0 o
x = ±l •
x = ±-j2&Q • ,
x = l / 3 •'::>*o C = ( X G N I X < 6 | ' ^ ' - ^ -
b) A n B = | - l ; l } ; A u B = { - l ; 0 ; i ; l ; 2 ; 3 ; 4 ; 5 } , A \ = {0;2;3;4;5), C B u A ( A n B ) = {0;l/3;2;3;4;5)
B u C = i - l ; 0 ; l / 3 ; l ; 2 ; 3 ; 4 ; 5 ; 6 ) , A n ( B u C ) = {-1;0;1;2;3;4;5) = A
fiian loai va pnwimg pnapgiiu tjiii so lu
Bai 1.55: T i m quan he bao ham hay bang nhau giua cac tap hop sau day:
a) A = | X € N | x < 2 } ; B = j x e Q b) A = |x £ 4x^ - 9 = o|; ' B = |x 6 K c) A = j x 6 N | l < x < 4 } ; B = { x e Z
Huang dan gidi
a) T a c o : A = { X G N | x < 2 } = > A = { 0 ; l } . (1) (x^ - x)(x^ - 2) = 0 .
(x^ - x) ( x 2 - 2 ) = 0 X + 4x -I-
x^ - 9 = 0 .
X G Q B =
( x 2 - x ) ( x 2 - 2 ) = 0 ô x^ - x = 0 x^ - 2 = 0
x = 0 V X = 1 X = ±V2 ô O Q 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 • ằ 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 = { x 6 Z | 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 d j n h tat ca cac tap con khac rong X, Y ciia A biet ran;
X u Y = A va ( A n B ) c X ;
b) Hay xac d j n h tat ca cac tap P biet rang (A n B) c P c (A u B).
Huang dan gidi a) T a c o A n B = {4 ; 6 } c X .
D o 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 A u B = ( 0 ; 2 ; 4 ; 6 ; 5 1 , do do cac tap P thoa m a n dieu kic' t; ' ( A n B ) c P c ( A u B ) l a :
{4 ; 6}, {4 ; 6 ; 0}, {4 ; 6 ; 2}, {4 ; 6 ; 5}, {4 ; 6 ; 0 ; 2}, {4 ; 6 ; 2 ; 5}, { 4 ; 6 ; 5 ; 0 } v a { 4 ; 6 ; 0 ; 2 ; 5 } .
Bai 1.57: Cho ba tap hop:
A = { x e R | - 3 < x < l } ; B = { x 6 R | - l < x < 5 | ; C = {x e S ||x| > 2' .
a) Xac djnh cac tap hop sau day va viet ket qua d u o i dang khoang, doan hay nua khoang: A o B, A u B, (B \) n C.
b) C h u n g m i n h rang: C s ( A u B ) = ( C s A ) n ( C s B ) . Huang dan gidi a) T a c o : A n B = [ - 3 ; l ) n [ - l ; 5 ] = [ - l ; ] ) ;
A u B = [ - 3 ; l ) u [ - l ; 5 ] = [ - 3 ; 5 ] ; B \ = [ - 1 ; 5 ] \ [ - 3 ; 1) = [ l ; 5 . c = { X e IK. | x | > 2 } = ( - ô ; - 2 ] u [ 2 ; + x ) ; (B \nC = [2 ; 5
b ) T a c 6 : C^ ( A u B) = C R [ - 3 ; 5] = ( - x ; - 3 ) u { 5 ; + x ) . (1)
CgA = C K [ - 3 ; l ) = ( ^ ; - 3 ) u [ l ; + x); C , B = C , , [ - 1 ; 5 ] = ; - l ) u ( 5 ; + x ) ; ( C K A ) n ( C j , B ) = ( ^ ; - 3 ) u ( 5 ; + x). (2)
(1) v a ( 2 ) c h o : C^ ( A u B) = ( C B { A ) n ( C K B ) . Bai 1.58: Cho hai tap hop A, B bat k i .
C h u n g m i n h rang: A u B = A r i B < = > A = B.
Huattg dan gidi
• Thuan:/4 u B = /4 n B, ta chung m i n h :/4 = B Vx, X e A X € A u B (vi A c A u B) => X e A n B ( v i A u B = A o B ) x € B (vi A n B c B)
N h u t h e : Vx,x e A => X e B, nen A c B (a) , >
V x , x e B= > x G A u B ( v i B c A u B ) = > x € A n B ( v i A u B = A n B ) => x e A (vi A n B c A)
N h u t h e : V x , x e B = > x e A , n e n B c A (b) .i>
Tir (a) va (b) cho/I = B
• Dao: Cho/4 = B, ta Chung m i n h : A u B = A n B ' Ta CO A u B = A u A ( v i B = /\ = /\)
A n B = A n A ( v i B = /l) = /4 (d) T u (c) va (d) cho A u B = A n B .
T u hai phan thuan va dao ta dugc: A u B = A n B < = > A = B
%hu(mq2 H A M S O B A C N H A T V A B A C H A I
§l.DAICUaNGVEHAMS6 m - ^ -.^.i
A . T O M T A T L Y T H U Y E T * ^ 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 ( x 2 )
• H^m so y = f(x) nghich bien (giam) tren K neu v-
Vxi,X2 G K : X I < X 2 =>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 . C A C P A N G T O A N V A P H l / O N G P H A P G I A I .
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