Thiết kế bài giảng đại số và giải tích 11 nâng cao (tập 1) phần 2

Bidt phan biet rd cac khai niem quy tac cdng, quy tdc nhan va van dung trong tflng trudng hgp cu the.. • GV neu khai niem quy tac cdng Gid sit mdt cdng viec cd the dugc thuc hien theo ph

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Ndi dung chfnh eua chuong II:

Quy tae ddm: Gidi thieu quy tac cgng va quy tac nhan va nhiing iing dung cua cac quy tdc nay

Hoan vi - ehinh hgp - td hgp : Day la ba quy tac ddm cu thd nhdm dd dem cac phdn tii cua tap hgp hffu han theo cac quy luat thii tu ggi la hoan vi, chinh hgp va td hgp

Nhi thiic Niu-ton : Nhdm tim he sd ciia mdt khai tridn (a + b)"

Phep thit va bidn cd: Day la nhflng khai niem quan trgng cua xac sudt, Trong bai cdn dua ra nhiing quy tdc tfnh xac su&

Xae sudt cua cac bidn cd

II MUG TIEU

1 Kie'n thurc

Nam dugc toan bd kidn thiic co ban trong chuong da neu tren, cu thd :

Hinh thanh nhflng khai niem mdi cd hen quan ddn cac quy tac ddm Tfnh dugc sd cac td hgp, sd cac chinh hgp va sd cac hoan vi cua mdt taj hgp gdm n phdn tit

Phan biet dugc su khac nhau cua chinh hgp va td hop

• xay dung dugc khdng gian mdu, each xac dinh bidn cd va xac suat

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Cdn than, chfnh xac trong lap luan va tfnh toan

Cam nhan duge thuc td eua toan hge, lihdt la ddi vdi xac sudt

III CAU TAO C O A CHUONG

Ndi dung eua chuong gdm hai phdn du kidn duge thuc hien trong 21 tiet, phan phdi cu thd nhu sau :

PM/iA.Tdhgp(8tidt)

§ 1 Hai quy tac ddm co ban 1 tidt

§2 Hoan vi, chinh hgp va td hgp 3 tidt Luyen tap 2 tidt

§3 Nhi thfle Niu-ton 1 tidt Luyen tap 1 tidt

Phdn B Xac sua't (11 tidt)

§4 Bidn ed va xac sudt cua bien cd 2 tiet Luyen.tap 1 tidt

§5 Cac quy tac tfnh xac sudt 2 tiet Luyen tap 2 tidt

§6 Bidn ngdu nhien rdi rac 2 tidt Luyen tap 2 tidt

6n tap va kiem tra chuong 2 2 tidt

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• Hai quy tac ddm co ban : quy tac cdng va quy tac nhan

• Bidt ap dung vao tiing bai toan : khi nao dung quy tdc cdng, khi nao diing quy tdc nhan

2 KT nang

Sau khi hge xong bai nay HS sfl dung quy tac dem thanh thao

Tfnh chfnh xae sd phdn tfl ciia mdi tap hgp ma sdp xdp theo quy luat nao

dd (cgng hay nhan)

3 Thai do

Tu giac, tfch cue trong hge tap

Bidt phan biet rd cac khai niem quy tac cdng, quy tdc nhan va van dung trong tflng trudng hgp cu the

Tu duy eac vdn dd cua toan hge mdt each Idgic va he thdng

II CHUAN Bj cClA GV VA HS

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III PHAN PHOI THOI Ll/ONG

Bai nay chia lam 1 tidt

IV TIEN TRINH DAY HOC

• GV neu bai toan trong SGK

GV dat ra mdt vai cau hdi nhu sau:

?l| Hay vidt mdt sd mat khdu

GV chia ldp thanh 4 td, mdi td viet mdt so mat khdu, sau dd cho mdt ban trinh bay xem cac td cd trung nhau khdng?

• Thuc hien [HIJ trong 3'

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Ggi y tra ldi c^u hdi 3 Khdng du doan dugc

• GV neu khai niem quy tac cdng

Gid sit mdt cdng viec cd the dugc thuc hien theo phuang dn A hodc phuang dn B Cd n cdch thuc hien phuang dnAvdm cdch thuc hien phuang dn B Khi dd cdng viec cd the dugc thitc hien bdi n + m cdch

Quy tac cdng bdi nhieu phucmg an

Gid sic mdt cdng viec cd the dugc thuc hien theo mdt trong k phuang

a« AJ, A2, , A|j Co nj each thuc hien phuang dnA^, n2 cdch

thue hien phuang dn A2, vd n^ cdch thuc hien phuang dn A.^ Khi

do cdng viec cd the dugc thue hien bdi nj + n2 + • • • + n,j cdch

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GV thuc hien vi du 2 Vf du nay chi mang tfnh minh hoạ

• Thuc hidn [H2j trong 5'

Muc dich Kidm tra xem hge sinh da biet van dung quy tac cdng hay chuạ

Hoqt dpng cda GV

Cdu hoi 1

Cd bao nhieu de taị

Cdu hoi 2

GV đi sd va hdi xem cd

bao nhieu each chgn

• GV neu each phat bieu khac cua quy tac cdng neu trong chu ỵ

Sdphdn tit cua tap hgp hitu hgn Xdugc ki hieu la |x| (hodc n(X)) Quy tdc cdng cd the dugc phdt bieu dudi dgng sau :

Neu Ava Bid hai tap hgp hitu hgn khdng giao nhau thi sdphdn tilt cua A uB bdng sdphdn tit cua A cdng vdi so phdn tif cua B, tUc Id

| A u B | = |A| + |B|

Quy tdc cdng cd thd md rdng cho nhidu hanh đng

- Neu AJ, A2, ; Â Id k tap hOu hgn vd A; n Aj = 0 vdi i ^ j

(vdii,j = 1, , k) thi\AiKj A2^ yJ Ậ\=\AI\ + | A 2 | + " - + |A,^|

- Hai tap hgp A, B bdt ki thi\A u B| = | A | + | B | - | A n B|

Gia sfl tfl nha An đn nha

Binh cd 1 con dudng thi tfl

nha An đn nha Cudng cd

bao nhieu each chgn?

Hoqt dpng cua HS

Ggi y tra Idi cau hdi 1

Cd 6.1 = 6 con dudng

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Hoqt dpng cda GV

Cdu hoi 2

Hdi An cd bao nhieu each

chgn dudng di den nha

Cudng?

Hoqt dpng cda HS

Ggi y tra ldi c^u hdi 2

Cd 4 6 = 24 each di tfl nha An qua nha Binh ddn nha Cudng

• GV neu quy tdc nhan

Gid sit mdt cdng viec ndo do bao gdm hai cdng dogn A vd B Cdng dogn A cd the ldm theo n cdch Vdi mdi cdch thuc hien cdng dogn A thi cdng dogn B cd the ldm theo m cdch Khi do cdng viic cd the thuc Men theo nm cdch

f Thue hien |H3| trong 5'

Muc dich Kidm tra xem hge sinh da biet van dung quy tac nhan hay ehua

Cdu hoi 1

Mdi each dan nhan cd bao

nhieu cdng doan, hay kd ten

cae cdng doan dd

Cdu hoi 2

Cd nhidu nhdt bao nhieu

chide ghd dugc ghi nhan

khac nhau?

Ggi y tra ldi cau hdi 1

Viec lap mdt nhan ghd bao gdm 2 cdng doan Cdng doan thfl nhdt la chgn 1 chft cai trong 24 chft cai Cdng doan thfl hai

la chgn 1 sd trong 25 sd nguyen duong nhd hon 26

Ggi y tra Idi cau hdi 2

Cd nhidu nhdt la 24.25 = 600 chide ghd dugc ghi nhan khac nhau

• GV cho HS md rdng quy tdc nhan cd nhidu hanh ddng

Gid sit mdt cdng viec ndo dd bao gSm k cdng dognA^, A2, , Aj^ Cdng dogn Aj cd the thue hien theo Uj cdch, cdng dogn A2 cd thi thuc hien theo n2 each, , cdng dogn A^ cd the thuc hien theo

ny cdch Khi dd cdng viec cd the thuc hien theo nin2 •••ny cdch

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• Thuc hien vf du 4

Hoqt dpng cda GV

Cdu hoi 1

Mdi each lam mdt bidn sd xe

may cd bao nhieu cdng doan,

hay kd ten cac cdng doan dd:

Ggi y tra Idi c^u hdi 2

Thed quy tac nhan, ta cd tdt ca

26 9 10 10 10 10 = 2340000 (bien sd xe)

• Thuc hien Vl du 5

Hoqt dpng cda GV

Cdu hoi 1

Cd bao nhieu day gdm 6 kf ttt

mdi kf tu hoac la mdt chft cai

(trong bang 26 chft cai) hoac la

mdt chft sd (trong 10 chft sd tfl

0 ddn 9)

Cdu hoi 2

Cd bao nhieu day gdm 6 kf tu

ndi d cau>a) khdng phai la mat

khdu?

Cdu hoi 3

Cd thd lap duge nhidu nhdt bao

nhieu mat khdu?

Hoqt ddng cda HS

Vl ed 26 + 10 = 36 each chon nen theo quy tae nhan, ta cd the lap duge

36 day gdm 6 kf tu nhu vay

Ggi y^tra ldi cau hdi 2

Vi mdi kf tu cd 26 each chgn nen theo quy tac nhan, sd day gdm 6 ki tur khdng phai la mgt mat kh'du la 26

Ggi y tra ldi eSu hdi 3

cd36^-26^

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HOATD6NG4

TOMTATBAIHQC

1 - Gia sfl mdt cdng viec co thd dugc thuc hien theo phuong an A hoac phuong

an fi Cd n each thuc hien phuong an A va m each thue hien phuong an B Khi

dd cdng viec cd thd dugc thuc hien bdi n + m each

- Gia sft mdt cdng viec cd the dugc thuc hien theo mdt trong ^phuong anAj, A2, , Ajj Cd Wj each thuc hien phuong anAj, n2 cdch thuc hien

phuong an A2, va n^ each thue hien phuong an Aj^ Khi dd cdng viec ed

the dugc thuc hien bdi Uj + n2 + • • • + nj^ each

2 - Gia sfl mgt cdng viec nao dd bao gdm hai cdng doan A va B Cdng doan A

cd thd lam theo n each Vdi mdi each thuc hien cdng doan A thi cdng doan B

cd the lam theo m each Khi dd cdng viec cd thd thuc hi6n theo nm each

- Gia sfl mdt cdng viec nao dd bao gdm k cdng doan Aj, A2, , Aj^ Cdng

doan AJ cd thd thue hien theo Uj each, cdng doan A2 cd the thuc hien theo

n2 each, , cdng doan Aj^ cd thd thuc hien theo n^ each Khi dd cdng viec

cd the thuc hien theo njn2 Uj^ each

BOAT DONG 3

MOT S 6 CAU HOI TRAC NGHI£M

Cdu 1 Mdt bai tap gdm 2 cau, hai cau nay cd cac each giai khdng lien quan

den nhau Cau 1 cd 3 each giai, cau 2 cd 4 each giai So each giai dd thuc hien eac cau trong bai toan tren la tren la

(a) 3; (b) 4;

(c)5; (d)6

Trd ldi Chgn (c)

Cdu 2 DQ giai mdt bai tap ta cdn phai giai hai bai tap nhd Bai tap 1 cd 3

each giai, bai tap 2 cd 4 each giai Sd cac each giai dd hoan thanh bai tap tren la

(a) 3; (b)4;

(c)5; (d)6

Trd ldi Chgn (d)

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Cdu 3 Mdt Id hang duge chia thanh 4 phdn, mdi phdn duge chia vao 20 hop

khac nhau Ngudi ta chgn 4 hop dd kidm tra chdt lugng

Sd each chgn la

(a) 20.19.18.17; (b) 20 + 19 +18 + 17;

(c) 80.79.78.77 ; (d) 80 + 79 + 78 + 77

Trd ldi Chgn (e)

Cdu 4 Cho cac chft sd: 1, 3, 5, 6,*8 Sd cac sd chan cd 3 chft sd khac nhau cd

duoc tfl cac sd tren la:

(a) 12; (b) 24;

(e) 20; (d) 40

Trd ldi Chgn (b)

Cdu 5 Cho cae chft sd: 1, 3, 5, 6, 8 So cae sd chan cd 4 chft sd khac nhau cd

dugc tfl cac sd tren la:

(a) 4.3.2; (b) 4 + 3+ 2;

(c) 2.4.3.2; (d) 5.4.3.2

Trd ldi Chgn (c)

Cdu 6 Cho cac chft sd: 1, 3, 5, 6, 8 Sd cac sd le cd 4 chfl sd khac nhau cd

dugc tfl cac sd tren la:

(a) 4.3.2; (b)4 + 3+2;

(c) 3.4.3.2 ; (d) 5.4.3.2

Trd ldi Chgn (e)

Cdu 7 Mdt ldp hge cd 4 td, td 1 cd 8 ban, ba td cdn lai cd 9 ban

a) Sd each chgn mgt ban lam 1 ^ trudng la

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c) Sd each chgn 2 ban trong mdt td lam true nhat la

Hudng ddn Sfl dung cac phuong phap ddm sd phdn tfl cfla mdt tap hgp

Thep quy tac cdng, ta cd 5 + 4 = 9 each chgn ao so mi

Bdi 2

Hudng ddn Sfl dung quy tae nhan

Chft sd hang chuc cd the chgn trong cac chft sd 2, 4, 6, 8; do do cd 4 each chgn Chft sd hang don vi cd thd chgn trong eac chft sd 0, 2,4,6, 8; do dd cd 5 each chgn Vay theo quy tac nhan, ta cd 4.5 = 20 sd co hai chft sd ma hai chft sd cfla nd deu chan

Bdi 3

Hudng ddn Sfl dung quy tac nhan va quy tac cdng

a) Theo quy tdc cgng, ta cd 280 + 325 = 605 (each chgn)

b) Theo quy tac nhan, ta cd 280.325 = 91000 (each chgn)

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HS cdn hidu dugc each chiing minh dinh If ve sd cac hoan vi

Khai niem chinh hgp, cdng thfle tfnh sd cae ehinh hgp chap k cua n phdn tfl

• HS cdn hidu dugc each chihig minh dinh If vd sd cae chinh hgp chap k cua

n phdn tfl

Khai niem td hgp, sd cac td hgp chap k cua n phdn tfl

• HS cdn hidu dugc each chiing minh dinh li vd sd eac td hgp chap k cua n phdn tfl

HS phan biet dugc khai niem : Hoan vi, td hop va chinh hgp

Tu giac, tfch cue trong hge tap

Bidt phan biet rd cac khai niem co ban va van dung trong tflng trudng hgp, bai toan eu thd

Tu duy cac vdn dd cua toan hge mgt each Idgic, thuc te va he thd'ng

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II CHUAN B! CUA GV VA HS

1 Chuan bj cua GV

Chudn bi cae eau hdi ggi md

Chudn bi phdn mau va mdt sd dd dung khac

2 Chuan bj cua HS

• Cdn dn lai mdt sd kidn thfle da hge vd quy tac cdng va quy tdc nhan

6 n tap lai bai 1

III PHAN PHOI THOI LUONG

Bai nay chia lam 3 tiet:

Tiet 1: TU ddu den hit muc 2

Tiet 2 : Tiep theo den hit muc 3

Tiet 3 : Tiep theo den hit muc 4 vd bdi tap

• GV neu va hudng ddn HS thuc hien vf du 1

GV cho HS didn va chd trdng theo each eua minh, sau Ao liet ke lai

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• Neu dinh nghia

Cho tdp hgp Acdn(n> 1) phdn tic Khi sdp xip n phdn tie ndy theo mdt thU tu, ta dugc mdt hodn vi cdc phdn tA cua tap A (ggi tdt la mdt hodn vi cda A)

• Thue hien [HIJ trong 5'

?l| Mdt tap hgp cd 1 phdn tfl cd bao nhieu hoan vi?

?2| Mdt tap hop cd 2 phdn tfl cd bao nhieu hoan vi?

?3| Mdt tap hgp cd 3 phdn tfl cd bao nhieu hoan vi?

• GVneu dinh ll 1:

Sdcdc hodn vi cua mdt tap hgp cd n phdn tii la

?^ = nt = n(n-l)(n-2) l

• GV hudng ddn HS chiing minh dua vao quy tac nhan

• GV ndu vf du 2, vi du nay chi mang tfnh minh hoa

• ThiJfc hien |H2J trong 5'

Ggi y tra Idi c^u hdi 1

mdi viec lap sd la mdt hoan vi

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HOAT DONG 2

2 Chinh hgp

a) Chinh hgp la gi

• GV neu cau hdi:

Cho mdt tap hgp A gdm n phdn tfl Viec chgn ra k phdn tfl dd sap xep ed thfl tu

?4| Neu k = n, ta duge mdt sdp xdp ggi la gi?

?5| Ndu k < n, ta dugc mdt sap xdp ggi la gi?

• GV neu vf du 3 va hudng ddn HS thuc hien

• GV neu dinh nghia ,

?6

Cho tap hgp A gom n phdn tic vd sd nguyen A vofl < k < n Khi lay

ra k phdn tit cua A vd sdp xep chdng theo mdt thU tti, ta dugc mdt chinh hgp chap k cda n phdn tit cda A (ggi tdt la mot chinh hgp chdp k cua A)

Hai chinh hgp khac nhau la gi?'

?7| Chinh hgp khac hoan vi d didm nao?

• Thuc hien [H3J trong 5'

(a, b), (b, a), (a, c), (c, a), (b, c), (c, b)

Co 6 chinh hgp

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• GV neu nhan xdt:

Hai chinh hgp khdc nhau khi vd chi khi hodc cd it nhdt mdt phdn tic cda chinh hgp ndy md khdng la phdri tic cua chinh hgp kia, hodc cdc phdn tit cua hai chinh hop gidng nhau nhung dugc sdp xip theo thit

GV hudng ddn HS chiing minh dua vao quy tac nhan

• GV neu nhan xet trong SGK

TU dinh nghia ta thdy mdt hodn vi eua tap hgp n phdn tit la mdt ehinh hgp chap n cua tap dd nen A" = P^ = n!

• GV neu vf du 5 cho HS thue hien Co thd thay bdi vf du khac

• GV neu ehu y tfong SGK

Vdi 0 <k< n thi ta cd thi viit cdng thUe (1) dudi dgng

" (n-k)!

Ta quy udc

Ol = lvdAl=l

• GV dUa ra cae eau hdi cung cd nhu sau:

Hay chgn dung sai ma em cho la hgp If

?8| Hoan vi n phdn tfl la chinh hgp chap n cua n

(a) Dung; (b) Sai

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?9

?10

?11

Aj;iadungkhik>n

(a) Dung; (b) Sai

A„ la dung khi k < n

(a) Dung; (b) Sai

• GV neu dinh nghia

Gid sit tap Aeon phdn tit (n >1) Mdi tap con gom k phdn tut cua A ditgc ggi la mdt td hgp chdp k cua n phdn tiic dd cho

• Thuc hien |H4| trong 3'

Cd 4 td hgp

b) Sdcdc to hgp

• GV neu cac cau hdi:

?12| Hai td hgp khac nhau la gi?

?13| Td hgp chap k cua n khac chinh hgp chap k cua n la gi?

• GV neu dinh li

Kf hieu Cn la sd cac td hgp chap k cua n phdn tfl (0 < k < n)

Ta cd dinh If sau day

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DINH Lf 3

Sdcdc tdhgp chap k cda mdt tap hoped nphdn tic(l <k <n) Id ^

v_ A!: _n(n-l)(n-2)-(n-k + l)

" k! ki

• GV hudng ddn HS chflng minh dinh li

• GV hudng ddn HS thuc hien vi du 6 va vf du 7 nham cung cd kidn thfle vd tdhgp

1 Cho tap hop A gdm n phdn tfl (n > 1)

Mdi kdt iqua cua su sap xdp thfl tu n phdn tfl cua tap hgp A dugc ggi la mdt

hodn vi cfla n phdn tfl dd

Hai hoan vi cua n phdn tfl chi khdc nhau d thfl tu sap xdp

P„ la sd cac hoan vi cua n phdn tfl Ta cd

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3 Gia sfl tap A ed « phdn tfl (n > 1) Mdi tap eon gdm k phdn tfl eua A dugc ggi

la mdt td hop chdp k cua n phdn td da eho

Cn la so cac td hgp chap k eua n phdn tfl (0 < k<n)

MOT SO CAU HOI T R A C NGHlfiM K H A C H QUAN

Hdy chgn khdng dinh ddng trong cdc khdng dinh sau, tic bdi 1 den bdi 4 Cdu 1 Cd 3 ban nam va 2 ban nfl sap vao 1 hang dgc

a) Sd each sap xdp la :

(a)C^; (b)Ci; (c)5!; (d)A3

Trd ldi Chgn (e)

b) Sd each sap xdp dd hai ban nft dflng hai ddu hang la :

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Trd ldi Chgn (b)

e) Sd each ldy ra 1 ban nam va 1 ban nfl la :

(a) 2; (b)C§; (c)5; (d)3

Trd ldi Chon (c)

f) Sd each ldy ra 2 ban nam va 1 ban nft la :

(a) 2; (b)Ci; (c)5; (d)3

Trd ldi Chgn (c)

g) Sd each ldy ra 1 ban nam va 1 ban nft la :

(a) 2; (b)C|; (c)5; (d)3

Trd ldi Chgn (c)

Mgt ldp hge cd 20 ban nam va 15 ban nft

a) Sd each ldy ra 4 ban nam va 4 ban nft di thi dau thd thao la :

(a)C^o; (h)C\y,

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(c)C^5+C^o; (d)C^5

Trd ldi Chgn (c)

b) Sd each ldy ra 4 ban nam va 4 ban nft va mdt ban phuc vu di thi ddu

the thao la:

Cd^ 5 Hay didn dflng, sai vao d trdng cua nhiing khang dinh sau:

(a) Sd each chgn 4 trong 7 ngudi di du hdi nghi la A7 [ J

(b) Chgn 4 trong 7 ngudi di du hdi nghi la C7 [ J

(c)C^=35 D

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Hucmg ddn Dua vao chinh hgp

Cd A | = 8.7.6 = 336 ket qua ed the

Bdi 7

Hudng ddn Sd doan thang la sd cac td hgp

Sd cae vecto la sd cac chinh hgp

a) vay sd doan thang ma hai ddu mflt la hai didm thudc P chfnh bang

sd td hgp chap 2 cua n phdn tfl, tfle la bang C^ = ,2 _ n ( n - l )

b) Sd vecto cdn tun bang sd chinh hop chap 2 cua n phdn tfl, tfle la bang An = n(n-\)

Bdi 8

Hudng ddn Khdng phan biet chfle vu thi dp dung td hgp

Phan biet chfle vu thi sfl dung chinh hgp

a) Cd C7 = 35 each chgn

b)Cd A7 = 210 each chgn

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Luyen tap (tiet 5, 6)

I MUC TIEU

1 Kien thurc

HS dn tap lai

Quy tdc cgng va quy tae nhan

Khai niem, cdng thfle tfnh so cac td hgp, ehinh hgp hoan vi

HS phan biet dugc khai niem : Hoan vi, td hgp va chinh hgp

TU giac, tich cue trong hge tap

Bidt phan biet rd cac khai niem co ban va van dung trong tflng trudng hgp, bai toan eu the

TU duy eac vdn dd cua toan hge mdt each Idgic, thuc td va he thdng

II CHUAN Bj C O A GV VA HS

Chudn bi cac cau hdi ggi md

Chudn bi phdn mau va mdt sd dd dung khac

2 Chuan bi cua HS

Cdn dn lai mgt so kidn thfle da hge d bai 1 va bai 2

Bai nay chia lam 2 tidt:

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Hoqt dpng cda GV

Cdu hoi 1

Gia sfl ed mdt eau trac

nghiem, hdi cd mdy phuong

an?

Cdu hoi 2

Bai thi cd 2 cau thi cd bao

nhieu phuomg an?

Cdu hoi 3

Bai thi cd 10 cau thi cd bao

nhieu phuong an?

Cd 4^° = 1048576 phuong an tra ldi

HOAT DONG 2 Bdi 10

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Cd 9.10.10.10.10.2 = 180000 vay

>' cd 10

sd nhu

Mdi phuong an tren cd bao

nhieu each di?

GV chia HS lam 4 td, mdi td lam cau Dua vao quy tac nhan

Ggi y tra ldi cAu hdi 3

Cdng 4 phuong an tren lai

mdt

Trang 26

HOAT DONG 4 Bdi 12

Ggi y tra ldi cau hdi 2

Cd 2^ = 8 trang thai trong dd cd 1 trang thai thdng mach Cd 7 ttang thai khdng thdng mach

Goi y tra ldi c&u hdi 3

Cdu hoil

Viec chgn ra 4 ngudi cd didm

cao nhit la td hgp hay chinh

hgp?

Ggi y tra Idi c^u hoi 1

La td hgp vi khdng cdn thfl tu

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Chgn 3 ngudi sap thfl tu vih.it,

nhi, ba la td hgp hay chinh

Hoqt dpng cda GV

Cdu hoi 1

Viec chgn ra 4 ngudi xep cac

giai nhdt, nhi, ba, tu la to hgp

Cd bao nhieu kdt qua cd thd,

ndu bidt rdng ngudi gift ve so

47 dugc giai nhdt?

Cdu hoi 4

Cd bao nhieu kdt qua cd the,

ndu bidt rang ngudi gift ve sd

Hoqt ddng cda HS

Ggi y tra Idi c^u hdi 1

La chinh hgp cd thfl tu

Ggi y tra ldi cau hdi 2

Cd AJOO = 94109400 kdt qua ed the Ggi y tra ldi cau hdi 3

Alg = 941094 kdt qua cd the

Ggi y tra Idi cau hdi 4

4 Alg = 3 764 376 kdt qua cd thd

Trang 28

47 tning mdt trong bdn giai?

Cd bao nhieu each chgn 5 em

theo yeu cdu bai toan?

Hoqt dpng cda HS

Sd each chgn 5 em trong 10 em laCfg •

Ggi y tea ldi cau hdi 2

So each chgn 5 em toan nam laCg

Ggi y tra ldi c^u hdi 3

Sd each chgn cd ft nhdt mdt nft la Cfg

- ci = 196

Chfl y : Cd thd giai theo each khac

Cd bao nhieu each chgn 5 em

theo yeu edu bai toan?

Hoqt dpng cda HS

Sd each chgn 5 em toan nam laCy

Sd each chgn 4 nam va 1 nft laC7C3

vay dap sd bai toan la C^ + C^C^ = 126

Trang 29

§3 Nhi thurc Niu-ttfn

(tie't 7)

I MUC TIEU

1 Kien thurc

HS nam duge:

Cdng thfle nhi thfle Niu-ton

He sd cfla khai tridn nhi thfle Niu-ton qua tam giac Pa-xcan

2 KT nang

Tin dugc he sd eua da thfle <khi khai tridn (a+ b)"

• Dien duge hang sau eua nhi thfle Niu-ton khi bidt hang d ngay trude dd

3 Thai do

Tu giac, tfch cue trong hoe tap

Sang tao trong tu duy

Tu duy cac vdn dd cua toan hge mdt each Idgic va he thd'ng

Chudn bi cac eau hdi ggi md

Chudn bi phlun mau, va mgt sd dd dung khac

Cdn dn lai mdt sd kidn thfle da hge vd hang dang thfle

• On tap lai bai 2

III PHAN PHOI THdl Ll/pNG

Trang 30

• GV neu cac cau hdi sau:

?l1 Neu cac hang dang thfle (a + b)^ va (a + b)^?

?2| Tfnhcae he sd cua (a + b)" va ed nhan xet gi vd he sd

• GV neu va hudng ddn HS giai cap vi du 1 va vf du 2 trong SGK

GV hudng ddn HS thuc hien [HI

Muc dich Kidm tra xem hge sinh da bidt van dung cdng thfle nhi thfle

Niu-ton dd khai tridn da thfle dang (ax - 6 ) " hay chua

Hoqt dpng cda GV

Cdu hoil

Trong khai tridn Niu-ton d

day a, b bang bao nhieu?

Hoqt dpng cda HS

Ggi y tra ldi cSu hdi 1

a = 3x, b = - 4

Trang 31

Ggi y tra Idi c&u hdi 2

sd hang chfla x^ \aCl(3xf(-4f Vay

• Neu dinh nghia:

Trong cdng thdc nhi thdc Niu-tan dmtic 1, cho n=0, l, vd xip

cdc he sd thdnh ddng,ta nhdn dugc tam giac sau ddy, ggi la tam

GV neu quy luat va cho mdt vai HS didn tidp eac ddng sau cua bang

GV dua ra quy luat

-Dinh duge ghi sdi Tiip theo la hdng thic nhdt ghi hai sol

- Niu biit hdng thUn (n >1) thi hdng thii n + 1 tiep theo dugc thiit lap bdng cdch cdng hai sd lien tiip cda hang thU n roi viet kit qud xudng hdng dudi d vi tri giita hai sd ndy Sau do viit sd I

d ddu vd cudi hdng

Trang 32

• Ihuc hien [H2] trong 5'

Muc dich Kidm tra xem hge sinh da bidt thidt lap hang thfl n + 1 tfl hang thfl n ciia tam giac Pa-xcan hay ehua

Ggi y tra ldi cau hdi 2 Hang thfl tam la 1, 8, 28, 56, 70, 56,

28, 8,1

' H0ATD6NG3

TOM TAT B A I HOC

1 (a+b)" =c;a"+c!,a"-ib + + C|;a"-'^b'^+ +CS"^ab"-^+c;;b"

Vdia = b = l , t a e d 2 " = c ; + c ; , + + c;;

Vdi a = l ; b = - l , tacd

o=c;-c!,+.:.+(-l)'^ci;+.•.+(-l)"c^ :

2 Tam giac Pa-xcan dugc lap tiieo quy luat sau :

- Dinh dugc ghi sd 1 Tidp theo la hang thfl nhdt ghi hai sd 1

- Ndu bidt hang thfl n (n > 1) thi hang thfl n + 1 tidp theo dugc thidt lap bang

each cdng hai sd lien tidp cua hang thfl n rdi vidt kdt qua xud'ng hang dudi d

vi trf gifta hai sd nay Sau dd viet sd 1 d ddu va eudi hang

HOAT DONG 4

MOT s d C A U H O I T R A C NGHlfiM K H A C H QUAN

Hdy dien dung sai vdo 6 trdng sau

Cdu 1 Trong khai tridn (a + b)*

Trang 33

Hdy chgn khdng dinh dung trong cdc cdu sau

Cdu 3 Cho phuong trinh lugng giac : -2sirui: = 2

Trong khai trien (a + 2b)® he sd ldm nhdt la

(a) 16; (b) 32;

(c)64; (d)112

Trd ldi (c)

Cdu 4 Cho phuong trinh lugng giac : - 2siiu = 1

Trong khai trien (a + 2b)* he sd cfla don thfle chfla b' la

(a) 16; (b) 32;

(c)64; (d)112

Trd ldi (b)

Trang 34

HOAT DONG 3

HUC»VJG D A N B A I T A E SGK

Bdi 17

Hudng ddn Sfl dung true tidp cdng thfle nhi thfle Niu-ton

Ddpsd Sd hang chfla x'°'y^^ trong khai trien (2x - 3y)^°° la

Hudng ddn Sfl dung true tidp cdng thfle nhi thfle Niu-ton

Sd hang chfla x^ trong khai trien (2 - x)^^ \aC\g(-xf2^^

Cdng thfle nhi thfle Niu-ton

He sd cua khai tridn nhi thfle Niu-ton qua tam giac Pa-xcan

2 KT nang

Tin duge he sd cua da thfle khi khai tridn (a+ b)"

Trang 35

• Didn duge hang sau cua nhi thfle Niu-ton khi bidt hang d ngay trudc dd

3 Thai do

Tu-giac, tfch cue trong hge tap

Tu duy cac vdn dd eua toan hge mdt each Idgic va he thdng

1 Chuan bjcua GV

Chuan bi cae eau hdi ggi md

• Chudn bi ph^n mau, va mgt sd dd dung khac

Cdn dn lai mdt sd kien thfle da hgc trong bai 3

III P H A N P H O I THdl LUONG

IV TI§'N TRINH DAY HOC

Hoqt dpng cda GV

Cdu hoi 1

Hay ap dung true tidp cdng thfl

nhi thfle Niu-ton de khai tridn

Ggi y tra ldi cdu hdi 2

Trang 36

Hoqt dpng cda GV Hoqt dpng cda HS

1 + 30x + 405x^+3240x^ + • • • (sfl dung may tfnh)

Trang 37

Khai niem phep thfl

Khdng gian mdu, sd phdn tfl eua khdng gian mdu

Bidn cd va cac tinh chdt cua chung

• Bien ed khdng thd va bidn ed chac chan

2 KT nang

Bidt xae dinh duge khdng gian mdu

Xae dinh duge bidn ed ddi, bidn cd hgp, bidn cd giao, bien cd xung khac eua mgt bidn cd

3 Thai do

Tu giac, tfch cue trong hgc tap

Tu duy cac vaii de cfla toan hgc, thuc td mdt each logic va he thdng

Trang 38

Chudn bi cac eau hdi ggi md

Chudn bi phah mau, va mdt sd dd dung khac

Cdn dn lai mgt sd kidn thfle da hge vd td hgp

On tap lai bai 1, 2, 3

Tietl: Tin ddu den hit dinh nghia cua muc 2

Tiet 2 : Tiep theo den hit vd bdi tap

d) Phep thvc ngdu nhien vd khong gian mdu

• GV neu cac eau hdi sau:

?l| Khi gieo mdt eon sflc sac cd mdy kdt qua ed the xay ra?

?2| Tfl cac sd 1, 2, 3, 4 cd the lap duge bao nhieu sd cd ba chft sd khac nhau?

Trang 39

• GV vao bai:

Mdi khi gieo mdt eon sue sdc, gieo mdt ddng xu, lap cac sd ta dugc mdt phep thfl

• Neu khai niem phep thfl:

Phep thii ngdu nhien (ggi tdt la phep thdc) Id mdt thi nghiem hay mgt hdnh dgng md:

-Kit qud cda nd khdng dodn trudc dugc;

— Cd thi xdc dinh dugc tap hgp tdt cd cdc kit qud cd the xdy ra cua phep thic dd

Phep thd thudng dmc ki hieu bdi chiJcT

Tap hgp tdt cd cdc kit qud cd thi xdy ra cda phep thit dugc ggi Id khong gian mdu cda phep thdc vd dugc ki hieu bdi chit Q (dgc Id d-me-ga)

• GV neu va eho HS thuc hien vi du 1 va vf du 2

• Thuc hien |H1| trong 3'

Miic dich Kiem tra xem hgc sinh cd bidt each md ta khdng gian mdu cua mdi

phep thfl hay ehua

Hoqt ddng cda GV

Cdu hoi 1

Mdi ldn gieo cd mdy kdt qua

eua mdi ddng xu

Khdng gian mfiu la Q = {SSS, SSN, SNS, SNN, NSS, NSN, NNS, NNN}

b) Bien cd

• GV neu vi du 3

• GV neu cac cau hdi:

?3| Khi gieo mdt con sue sdc, tim cac kha nang cac mat xudt hien la sd chan?

Trang 40

?4| Khi gieo hai ddng tidn, tim eac kha nang cac mat xudt hien la ddng kha nang?

Sau dd GV khai quat lai bang khai niem:

Biin cd A lien quan den phep thd T la biin cd md viec xdy ra hay

khdng xdy ra cda A tuy thudc vdo kit qud cua T

Mdi kit qud cua phep thd T ldm cho A xdy ra, dugc ggi la mgt kit

qud thudn lgi cho A

Tap hop cdc kit qud thuan lgi cho A dugc ki hieu IdQp^ Khi do

ngudi ta ndi Men cd A dugc mo td boi tap Q^

• Thue hien |H2| trong 3'

Muc dich Cung ed khai niem "Tap hgp md ta bien ed A" hay tap hgp eac kdt qua

thuan loi eho A

GV dua ra khai niem bidn cd khdng thd va bidn ed chdc chan

Tap 0 dugc ggi la Men cd khong the (ggi tdt la Men cd khong)

Cdn tap £2 dugc ggi Id bien cdchdc chdn

Neu vi du vd bidn cd khdng thd

Neu vf du ve bidn ed chdc chdn

?5

?6

• GVneu quy udc

Khi ndi cho cdc biin cd A, B, md khdng ndi gi them thi ta hieu chung cung lien quan din mdt phep thd

Ta ndi rdng bien cd A xay ra trong mot phep thd ndo dd khi vd chi khi

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