tuyen chon 500 bai tap toan 10 nxb ha noi 2005 2 885

tuyển chọn,chọn lọc các bài toán toán lớp 10 dành cho học sinh khá giỏi.Với 500 bài tập từ dễ đến khó .nhà xuất bản hà nội . sản xuất năm 2005. giúp các em có thể vận dụng vào bài tập nâng cao kĩ năng học toán

B a i 287 (Cfiu hoi trftc nghi^m) ,

Phi^cfng trinh nao sau day c6 hai nghi$m X i , X2 thoa man bit't d^n

thii-c xi < 0 < X2 < 2

a) 3x^ - 5x + 1 = 0

.2

c) Sx"* - 2x - 4 = 0 b) 3x - X + 4 = 0

B a i 288 (Cau hoi trie nghi^m)

Vdi nhffng gia tri nao ciia a thi phi ^c^ng trinh 3x^ - ax + a = 0 c6 hai

nghi^m X i , X2 thoa man - 2 < X i < 2 < X2 hoSc X i < - 2 < X 2 < 2 ?

a) Tim gia t r i n h o n h a t ciia h a m so' y = x^ - 2x + 2

b) Cho phtfoTng t r i n h (x^ - 2x + 2f - 4(x - l)'^ + 2m = 0 (*) Xac d i n h m

de phifoTng t r i n h (*) c6 n g h i $ m

* Hiidng ddn

a) Bien ddi y ( f + hkng so' A > A

b) D a t t = r o i bien doi phiTcrng t r i n h (*) t h a n h phifcrng t r i n h baC

hai in so t (dieu k i e n ciia t : cau a)

Phuong t r i n h da cho c6 nghiem <=> Phuomg t r i n h (theo t ) c6 nghiein

t >

G l A l a) T a c6 y = x^ - 2x + 2 = (x - 1)^ + 1 > 1, Vx £ R

*Hudngd&n

a) B i e n doi y = h ^ n g so' ( ) va liAi y x**, x^ > 0 b) D a t t = - x"* - 4x^ + 5 r o i bien doi phiTcfng t r i n h da cho t h ^ n h phirong t r i n h bac h a i a \o t (dieu k i ? n t < )

Sau d6 giai gio'ng b a i 289

Dap so : b) max(y) = 5 v i y = 5 - (x" + 4x^) < 5 (Dfi'u = xdy ra o x = 0)

R , • c) Phucfng t r i n h da cho o> t = - x"* -4x^ +5 (t < 5)

t^ + 2t + m - 10 - 0 (*) * Tiep tuc giong b ^ i 289, t a c6 : - 25 < m < 11 V m < - 25 <=> m < 11

167

E KI^M TRA CAC KIEN THLfC VE TAM THUTC BAC HAI

B a i 291 (Cau h6i trSc nghiem)

Tam thtfc bglc hai f(x) = (m^ - 3)x^ + 2mx c6 bang xet dau

B a i 292 (Cau hoi trfic nghi^m)

Xac dinh cac gia tri cua m de bat phifoTng trinh x^ - 4x + 2m - 1 < (I

Bat phifcfng t r i n h da cho v6 nghiem <=> - 4x + 2m - 1 > d

nghiem diing vdi moi x E R <=> A' < 0

Dap so': m > — (cau b)

2

2x - 1 >0 1a :

B a i 293 (Cau hoi tr&c nghiem)

T|lp nghiem cua b a t phufoTng t r i n h _

-C a u nao sau day dung ?

a ) m > l b ) m < l c ) l m l > l d ) | m | < l

4[ HUdng ddn f(x) > 0 k h i x e (xi; X2) nen a = l - m < 0 c : > m > l

Dap so': m > 1 (cau a)

B a i 295 (Cau hoi t r i e nghiem)

Tam thiJc f(x) = 2x^ - ax - 3 c6 hai nghiem X i , Xg thoa man dieu ki^n — + — = 5 Tinh a ?

Dap so : a = - 15 (cau c)

B a i 296 (Cau hoi trac nghiem)

Tam thii-c f(x) = x^ - 2mx + 4 c6 gia tri nho nhat bSng 3 Tinh m ? a) m = 1 b ) m = - l c) I m I = 1 d) I m I > 1

* Hudng ddn

• f(x) = (x m)^ + 4 > 4

-V a y m i n f f ( x ) ] = 4 - = 3

Dap so : I m I = 1 (cau c)

B a i 297 (Cau hoi trSc nghiem)

Ham so y = ^2x^ - m x + 2 c6 tgp xac dinh S = R k h i : a) I m l <4 b) I m l <4 c) I m I < 16 d) I m I < 16

* Hudng ddn

H a m so da cho xac d i n h t r e n R o 2x^ >o, V x e R ' < = > A < 0

^ap so : I m I < 4 (Cau a)

169

B a i 298 (C&u hoi trfic nghiem)

Cho tarn thtfc f(x) = - 4x + 3 T|lp nghiem ciia phi^Ong trinh

f(x + 1) = 0 la :

a ) S = | l ; 3 ) b ) S = | - l ; - 3 | c) S = |0; - 2| d) S = {2; 0)

% HUdng d&n '

• g(x) = f(x + 1) = (x + 1)^ - 4(x + 1) + 3

Dap so': S = (2, 01 (cau d)

B a i 299 Cho a, b, c la dp dai ba canh cua mpt tarn giac

ChiiTng minh rSng tarn thrfc f{x) = (b + c)x^ - 2(a + b + c)x + 2(a + b + c )

ludn C O gia tri dUOng

• H a m so (1) xac d i n h o x^ + 4x - m + 2 > 0 (G9i D la tap xac dinh

cua ham so)

Vecta la doan thSng da dinh hirdng, nghla la da chon mot diem mut

lam diem dau, diem mut con lai la diem cuoi

doan th^ng vecta

• Vecta CO diem dau la A, diem cuoi la B, ki hieu : AB

• Vdi hai diem phan biet A va B, ta c6 hai vecta khac nhau :

AB va BA

• Vecta CO diem dau va di§m cuoi trung nhau, chang han: AA , MM ,

goi la "vector - khong", ki hieu 0 , '

2) Phifofng, hi^oTng, dp dai ciia vectof

a) Dinh nghla

Hai vecta goi la ciing phuang khi hai vecta nay Ian liicrt nhm tren

hai dUcJng thang song song nhau hoac trung nhau

H$ qua : Hai vecta cung phuang vdi mot vecta thuT ba thi hai vecta

do cung phi/ang

b) Hai vectcf a va b cung phifcfng thi a va b c6 the cung hxidng

hoac ngUdc hi^drng

Chii y : —>

• 0 cung hudrng vdi moi vecta

• Hai vecta cung hudng vdri mot vecta thiJ ba thi hai vecta do cung hirdrng

• Cho san a va mot diem O, ta c6 duy nhat mot di§m A : OA = a

• Mpi vecta 0 deu bang nhau

II) Phep cpng c a c vectd

1) Dinh nghla tong cac vectof

Cho hai vecta a va b

Tii diem A tuy y, ve AB = a va BC = b Vector AC dUdc goi la tong hai vector a va b , ki hi^u : AC = a + b

Chii y :

• Tong a + b khong phu thuoc vao vi tri diem A

• Quy tac ba diem : AC = AB + BC (ba diem A, B, C tuy y) (hinh 1)

• Quy tac dudng cheo hinh binh hanh

ABCD la hinh binh hanh =5 AB + AD = AC (hinh 2)

A D B^

(Hinh 1)

• Neu b la vector doi cua a t h i a la vector doi ciia b n e n :

a va b la hai vector doi nhau

2) H i ? u c i i a h a i vectof

Dinh nghia

Hi#u cua vector a va vectcf b , ki hi^u a - b , la tong cua vectcf a

va vectd doi cua vector b nghia l a : a - b = a + ( - b )

C h i i y :

• Phep t i m hieu a - b goi la phep trir hai vectcr

- > - > - >

• Cho ba diem bat k i : A, B, C, ta c6 : AC = BC - BA

IV) Phep nhan vectd v6l mpt so

1) D i n h n g h i a

Tich ciia vectcr a vdi so thuc k la mot vectcr, k i hieu k a , dxxac xac

a) k a cung hudrng vdi vectcr a neu k > 0 ngugc hirdng vdi vectd a neu k < 0

• G la trong t a m ciia tam giac ABC o GA + GB + GC = Q

• G la trong t a m ciia t a m gidc ABC o OG = OA + OB + OC

T o a n

B a i 1 Cho tam giac ABC, goi A' la diem doi xii'ng \6i B qua A, B' la diem

doi xxjfng vdri C qua B, C la diem doi xii'ng vofi A qua C

Chu'ng minh rang vdti mpt diem O bat ki ta c6 :

OA + OB + OC = OA' + OB' + OC'

* HUdng dan

• OA = OA' + A"A va phu y A'A = AB

• TiTcfng t u doi vdi OB va OC

• AB + BC + CA =

GIAI

Ta CO : OA = OA' + A'A = OA' + AB (1) (vi A"A = AB )

Tuongtu- OB = OB' + B'B = OB' + BC (2)

OC = OC' + C'C = OC' + CA (3)

Cong (1), (2) va (3) ta CO :

OA + 013 + OC = OA' + OB' + OC' + AB + BI: + CA

= OA' + OB' + O C (dpcni)

B a i 2 Cho luc giac A B C D E F Goi P, Q, R, S, T, U Ian li^grt la trung diem

cac canh AB, BC, CD, D E , E F , FA

Chu'ng minh rdng hai tam giac P R T va QSU c6 cung trong tam

3 G K + ( Q K + K P ) + ( sk + K R ) + ( U K + K T ) = (D c:> 3 G K + ( Q P + S R + U T ) = d (3)

Bai 3 Cho ti? giac ABCD Goi M, N Ian lUgft la trung diem cac canh AB,

B a i 4 Mpt gia d9 dUpc gSn vac tvCdng nhvC hinh l a Tam giac A B C vuong

can cf diem C Ngifori ta treo vao diem A mpt v|it nang 5N

Hoi CO nhang Itfc nao tac dpng vao buTc tvTdng tai hai diem B va C ?

GIAI

T a i diem A , liTc keo F hi/dng t h i n g dijfng xuong difcJi c6 ci^cfng dp

5 N , t a c6 the xem F la tdng cua h a i vector Fj va F2 I a n lufcft n&m

t r e n hai dir6ng t h i n g AC va A B

D i thay : | Fj I = | F | va | F2 I = I F | x/2 (do t a m gidc A B C

vuong can t a i C)

Vay : C6 m p t luc 6p vuong goc vdi biJc tirdng t a i diem C v d i ciTdng

do 5 N , v^ m p t life keo biJc tifcrng t a i diem B theo hi/dng B A v d i

cudng dp "5 V2 N (Xem h i n h l b )

B a i 5 Cho n diem tren mSt ph^ng B a n Minh k i hi^u chiing la A i , A2,

-An- B a n Mai ki hipu chiing la B i , B2, , B„

• V i n diem B i , B2, B„ cung Ih n diem A i , A2, , An nhitog ducfc k i

hieu mpt each khac, cho nen t a c6 :

-> -> ^ -> ->

O B j + O B 2 + + 0 B „ = O A j + O A 2 + + 0 A „

Tir (1) va (2) A j B i + A2B2 + + A ^ B , , = 0 (dpcm)

Bai 6 Cho ba diem phan bipt A, B, C

a) Chiing minh rSng neu c6 mpt diem I nao do va mpt so thiic t sao

- > - > - > • cho l A = t.IB + (1 - t ) l C thi vofi mpi diem I ' ta deu c6 :

• /

I-A = t.I-B + ( 1 - t ) l ' C b) Chu-ng to rfing l A = t.IB + ( l - t ) l C la dieu k i ^ n o^n va dii de ba diem A, B, C thSng hang

G I A I a) Theo gia t h i e t : l A = t I B + (1 - t ) I C , t h i vdi m o i die"m I ' , t a c6 :

I F + I ' A = t f-ir + I'B + ( i - t ) i r + r c

= t i ' B + ( i - t ) r c + ir I ' A = t i ' B + ( i - t ) r c

b) Neu t a chon I ' t r u n g vdi A t h i c6 0 = t A B + (1 - t ) A C , do la dieu

k i e n can va du de ba diem A, B, C t h i n g hang

8ai 7 Cho tii giac A B C D a) Hay xac d}nh vi tri ciia diem G sao cho

ciia tii giac ABCD)

(Diem G nhii the' gpi la trong tam

t h d n g do goi la di^dng thiing 0 - Ic c i i a tam g i a c A B C )

G I A I a) G o i B ' la d i e m do'i xiJng v d i B qua O, t a co B'C 1 B C

« Xac diuh diem F

Tucfng ty, ta c6 : M F = M B + CA o Bli' = CA

V$y F \k dinh thu- tiT cua hinh binh hknh ve tren hai canh C A v^

C B (xem hinh ve)

Ta da c6 : CD = A B ; A E = BI: va B F = CA nen D, E, F k h o n g phu thuoc vao v i t r i cua M

b) So sanh MA + MB + MC va MD + ME + M F

M D + M E + M F = ( M A + AD ) + ( M B + BE ) + ( M C + C¥ )

= ( M A + M B + M C ) + ( A D + BE + CF ) (1) -> ' '

A E = BC chufng m i n h t r e n : '

• Ta l a i CO : < >;

A F = C"B v i CBFA la h i n h binh h a n h

=> A E + A F = B1: + C B = 0 = > A E = - A F =:>Ala t r u n g diem EF

• Tifcfng tir : B la t r u n g diem D F C la t r u n g diem D E nen : ;

2 D A = D E + D F

• 2F'C = F D + F"E

2 E B = ED + E"F

Cong ve theo vf, ta ducfc :

2(D'A + ¥C + E'B) = (D'E + E D ) + ( D F + F D ) + ( F E + E F ) = 0

o D A + F C + EB = ( ) o A b + B E + CF = 0 ( 2 ) ' v

T i r ( l ) va (2) => M D + M E + M F = M A + M B + M C (dpcm)

a i 12 Cho tam giac A B C npi tiep trong difoTng tron (O)

Gpi H la trUc tam tam giac ABC va B' la diem doi xrfng vdfi B qua tam O

A H = B'C va A B ' = HC • ' • ' '

183

= ( A B ' - A B ) + ( AC - AC' ) + ( A D ' - A D )

= ( A B ' + A D ' ) - AC' - ( A B + A D ) + AC

Ma AB' f A D ' = A C ( V i AB'C'D' la hinh binh hanh)

AIB + A D = AC (Vi A B C D la hinh binh hanh)

Nen BB' + C'C + D D ' = AC' - AC' - AC + AC = 0 (dpcm)'

b) H a i tam giac B C D va B ' C D ' c6 c u n g trong tam

Vdi diem G bat k i ta c6 :

G"B + GC' + 0 0 = ( GB' + B'B ) + ( G C + CC') + ( GD' + D'D )

= (GB' + GC + GD' ) + ( B'B + CC' + D'b )

^ ( G B ' + GC + GD' ) - ( BB' + C'C + D D ' )

= GB' + GC + GD' - 0 = GB' + GC + GD' ) ( ) Neu G la trong tam tam giac B C D t h i GIB + G C + GD - 0 , i u c

do til (*) ta cung c6 GB' + GC + GD' = 0 G cung la trong tam

tam giac B'CD' (dpcm)

Vay trong t a m hai tam giac B C D va B'CD' triing nhau

B a i 14 C h o tam giac A B C va di^ofng thSng d T i m d i e m M t r e n dufong

Do d l i vectcf u nho nhat k h i va chi k h i 4 0 M nho nhat hay M la

h i n h chieu vuong goc ciia O tren d

Chu y : Cucli chgn dicin O sao cho v = 0 ,

' ' ' ' • ^ • •• • , •

G la trpng tam tam giac ABC, ta c6 :

V = (OA + OB + OC) + OC = 3OG + OG + GC = 40G + GC Vay dc v = 0 ta chon diem O sao cho GO = —GC

2 0 0 ' = ( OA + 0"b ) + ( A'B + DC ) + ( BO' + CO')

OA + ob = 0 (vi O la trung di§m A D )

Ma BO' + CO' = 0 (vi O' la trung diem BC)

M B

l A va I D cung phu'cfng I , A, D thang hang (dpcm)

B a i 17 C h o tam g i a c A B C , M v a N l a h a i d i e m dxicfc x a c d i n h bcfi

• Ap dung t i n h chat sau : 1 , '

G la t r o n g t a m t a m giac ABC va M la diem tuy y, ta c6 :

M A + M B + MC = 3 M G de chu'ng m i n h DG va DG' cung phuang

G I A I

• Ta CO : 3 D ^ = DA + DB + DC (1) vi G la trong t a m t a m giac ABC

• Tu'ong tuf ta cung c6 :

b) T i m tpa dp t r u n g d i e m I c u a d o a n t h i i n g A B c) T i m tpa dp c i i a d i e m M s a o c h o 2 M A = - 5 M B

• Cong thiic (1) dung trong moi tritang hap nghia Id cong thi'ic (1)

dung khi A, B, C d tren cung mot true (thdng hang) hoac A, B, C

B a i 20 T r e n true x'Ox, eho ba diem A, B, C c6 toa dp Ian Itf^t la a, b, c

Tim tpa dp cua diem 1 sao cho lA + I B + I C = 0

B a i 21 T r e n true x'Ox cho bon diem A, B, C, D tuy y ChiJng minh :

a) AB.CD + AC.DB + AD.BC = 0

b) Goi I, J , K, L Ian l\i(ft la trung diem cac canh AC, BD, AB, CD

Chtfug minh rhng I J va K L c6 chung trung diem

• HUdng ddn

a) Goi toa do cua A, B, la a, b, Sau d6 t i n h AB , CD ,

T i n h A B CD = (theo a, b, )

b) T i m toa do ciia I , J , K, L roi t i m toa do t r u n g diem cua I J vk K L

(xem l a i cau b bai 19)

va la toa do t r u n g diem cua K L

2 Vay I J va K L c6 chung trung diem (dpcm)

B a i 22 T r e n true (O; i ), cho ba diem A(- 4), B ( - 5), C(3)

Tim diem M tren true da cho sao cho MA + MB + MC = 0 Sau

M A 2 M B _ J 3

M B ~ 3 ' M C " 5 Vay

B a i 23 Cho a, b, c, d thu" la tpa dp ciia cac diem A, B, C, D tren true x'Ox

a) Chiirng minh rang khi a + b ^ c + d thi ta ludn tim diidc diem M sao

cho M A M B = M C M D

Ap dung : Tim tpa dp ciia M, neu co, biet A(- 2), B(5), C(3), D(- 1)

b) K h i AB va CD co cung trung diem thi diem M of cau a co xac dinh

OA + OB OC + OD I ^\ 1 • - U K •

- I = O i l (xem lai cau b bai 19)

2 2

hay a + b = c + d Vay diem M khong xac dinh

B a i 24 Cho A, B la hai diem tren true (O; i ) va I la trung diem ciia doan

AB Chufng minh rang vdi moi diem M ta luon co :

Cliii.yen d e 3

HE TRUC T 0 6 t>P OECfiC VUONG GOC

K i e n thi?c coT ban

I) He true toa dp vuong g o c

Trong m a t phSng, cho true x'Ox c6 veeto don v i i , true y'Oy c6

vecto don v i j sao eho i 1 j

D i n h n g h i a

He gom h a i true noi t r e n goi la he true toa do Deeac vuong goc, k i

hieu mp(Oxy)

• True x'Ox goi la true hoanh do • True y'Oy goi la true tung do

• Diem O goi l a goe toa do

II) T p a dp cua vectci

D i n h l i

Trong mp(Oxy), chp vecto u tuy y K h i do, eo duy nhat m o t cap so

thuc (x; y) sao cho : u = x i + y j

D i n h n g h I a

- * - > - > ->

Neu u = X i + y j t h i cap so' x va y goi la toa dp eiia vectcr' u

do'i vdi mp(Oxy), k i hieu : u = (x; y)

Trong mp(Oxy), eho diem M tiiy y K h i do, toa do ciia vecto O M goi

la toa do ciia diem M , k l hieu M(x; y)

— > • - > — >

Tom t a t : O M = X i + y j o M(x; y ) 2) D i n h l i

T p a d p d = - 2 7 o c = 0 i* - 2 j d = (0; - 2)

B a i 2 6 V i e t vector u diidi d a n g u = x i + y j k h i biet toa dp c i i a u :

u = (2; - 3) ; u = (- 1; 4) ; u = (2; 0) ; J = (0; - 1) ; u = (0; 0)

B a i 28 Cho ba diem A = (- 1; 1), B = (1; 3), C = (- 2; 0) trong mp(Oxy)l

a) ChuTng minh ba diem A, B, C thSng hang

b) Tim ti so ma diem A chia doan thSng B C , diem B chia doan thdng

AC va diem C chia doan thang AB

GIAI a) ChiJCng minh A, B, C thdng hang

Ta CO : AB = (XB - XA; y s - yA) = (2; 2) = 2( T + j ) (1)

AC = (xc - XA; y c - yA) = (- 1; - 1) = - ( T + j ) (2)

So sanh (1) va (2) ta dirge : AB = - 2 AC =i> AB va AC cung phirong

=> A, B, C t h i n g h^ng

b) » A chia dog.n thdng BC theo ti so k = ?

Theo ket qua cau a, ta c6 : AB = - 2 AC Vay diem A chia doan thSng BC theo t i k =

• B chia doan thdng AC theo ti so' k' = ?

Vay B chia doan t h ^ n g AC theo t i so' k' =

• C chia doan t h i n g AB theo ti so k " = ?

B a i 29 Cho tam giac ABC vori A = (xi; yi), B = (X2; y2) va C = (X3; ya) trong

mp(Oxy) Tim tpa dp trpng tam G cua tam giac A B C

B a i 30 Cho ba diem A = (4; 6), B = (5; 1), C = (1; - 3) trong mp(Oxy)

a) Tinh chu vi cua tam giac ABC » b) Tim tpa dp tam dvCdng tron ngoai tiep tam giac A B C va ban kinh

dxfdng tron do

• DiTcfng thang (AB) qua M o M, A, B thSng hang

o AB va AM cung phuorng, ma AB = (6; - 3) va AM = (4; y - 1) nen

• ABCD la hinh binh hanh o AD = BC

• Goi (x; y) la toa do ciia D, tinh toa r j ciia AD va BC roi cho

• Tim toa do AB va DC roi so sanh hai vector nay '

• Tam I ciia hinh binh hanh ABCD la trung diem ciia AC (hoac ciia BD)

D a p so : I

B a i 3 5 T r o n g m p ( O x y ) , c h o A ( - 2; 4), B ( - 4; 3), C ( 2 ; 1), D ( l ; 3) a) A , B , C C O t h a n g h a n g k h o n g ? T a i sao ?

b) C h i i ' n g m i n h A B C D l a mpt h i n h t h a n g c a n , d a y B C

• HUdng ddn

a) Xet phirong cua AB va BC ?

b) Churng minh AD // BC va AB = DC, AD ^ BC

Tir (1) va (2) ==> ABCD la hinh thang can c6 ddy BC, AD

Bai 36 Trong mp(Oxy), chb ba diem A(- 2, - 5), B(2; 3), C(0; 4)

a) Chufng minh A, B, C khong 6 tren cung mpt dxicing thSng

b) Tim diem D tren true hoanh Ox sac cho ABCD la hinh thang c6

ABCD la hinh thang day AB va CD o AB va DC cung phuang

Tinh AC^ BC^ AB^ r6i so sdnh AC^ vdi AB^ + BC^

Dap so : D(- 2; 0) va ABCD la hinh thang vuong tai B y^ C

=> ABC la tam giac vuong can tai A

a) Tinh AB, BC, AC ta dugc ket qua

b) Diing cong thufc tinh dien tich tam giac vuong

= 10 (dvdt)

Bai 38 Trong mp(Oxy), cho tam giac ABC, biet A(- 1; 5), B(- 3; 1) va

C(l; 4) Chiang minh ABC la tam giac vuong

Tinh chu vi ciia tam giac ABC ? ^ {.;;(}

• Hiidng dan

Tinh AB, AC, BC roi diing dinh l i Pitago dao "''-^

Dap so': Tam giac ABC vuong tai A va c6 chu vi 2p = 3 N/S + 5 Bai 39 Trong mp(Oxy), cho A(- 2; 1), B(6; 0)

a) Tim diem M tren true tung sao cho tam giac AMB vuong tai M

b) Chon M (cau a) c6 tung dp difotng

Tim diem C sao cho AMBC la hinh chu" nhgit \

* Hiidng d d n

Tinh A M ^ B M ^ A B ^ Diing dinh l i Pitago

b) Vi A M B = 90° nen A M B C la hinh chuT nhat

o A M B C la hinh binh hanh o M A = B"C

A M B = 90° nen A M B C la h i n h chiJ nhat

ci A M B C la h i n h binh hanh o M A = BC

B a i 40 Trong mp(Oxy), cho bon diem A(- 3; 0), B ( - 6; 2), C ( - 2; 8), D ( l ; 6)

Chufng minh ABCD la hinh chff nh^t

• Hiidng dan

• Chufng m i n h AB va DC bang nhau

T i n h A B ^ B C ^ AC^ roi dung dinh l i Pitago dao

B a i 41 Trong mp(Oxy), cho A(- 4; 3), B ( - 1; 7) Co tim dtfdc diem C de

OABC la hinh vuong khong ? Giai thich ? Neu c6, hay cho biet toa

dp ciia C ? (O la goc toa dp)

• Hxidng dan

• T i n h OA, OB, AB roi chufng m i n h tam giac OAB vuong can t a i A

nen t 6 n t a i C sao cho OABC la h i n h vuong

• TCr AB = OC ta se t i m diTofc toa do C

G I A I

« Ta C O : OA = 5, AB =: 5, OB VSO = 5%/2

OA = AB [OA^ + AB^ = OB^

Vay ton t a i diem C de OABC la h i n h vuong

B a i 42 Trong mp(Oxy), cho bon diem M(l; 2), N(- 3; 5)* P(0; 9), Q(4; 6)

Chu-ng minh MNPQ la hinh vuong

b) V e p h a n giac trong c u a B , p h a n giac i i ^ y c ^ ^ D t a i I

=> I l a tarn dacfng t r o n noi tiep t a m giac A B C

* Hudng dan

• M(x; 0) {|| • T i n h MA^ v a M B ^ roi cho M A = M B

B a i 49 Trong mp(Oxy), cho diem M(2; 2) DrfcJng thiing (d) qua M va cAt tia

Ox va tia Oy Ian Ividt tai A va B

Xac dinh vi tri ciia (d) sao cho khoang each tiJf 0(0; 0) den (d) dai

nhat Trong trxfdng hdp nay, tim tga dp ciia A va B ?

* IlUdng dan

Ve OH 1 (d), H e (d), chufng minh 0I^^< h^ng so

K h i OH dai nhat t h i tani giac OAB vuong can t a i O n e n :

Vay niax(OH) = 2 ^2 o H = M , luc do OM 1 (d)

Vay khoang each tCr 0(0; 0) den (d) dai nhat o (d) 1 OM

K h i (d) 1 OM t h i tam giac OAB vuong can t a i 0(0; 0)

(Vi O M vira la dtfdng cao vifa la duorng phan giac ve tii 0(0; 0) ciia

tam giac OAB)

Vay OA = OB = OM V2 = 4

Do do : A(4; 0) va B(0; 4)

B a i 50 Trong mp(Oxy), cho ba diem A(a; b), M(- 1; 3) va N(3; 5)

Chtfng minh rfing dieu ki^n can va dxi de tam giac AMN vuong tai

B a i 51 Trong mp(Oxy), cho bon diem A(- 2; 5), B(6; - 1), C ( - 1; - 2), D(5; 6)

Chu"ng minh rfing tx? giac ABCD npi tiep difofng tron c6 tam la trung diem AB ABCD la hinh gi ?

• HUdng ddn

• Cach 1 : Chufng m i n h tam giac ABC vuong tai C va tam giac A D B vuong tai D

(b^ng each t i n h A C ^ B C ^ A B ^ roi dung dinh If Pitago dao) ! '

• C a c h 2: T i m toa do t r u n g diem I ciia A B roi t i n h I A, I B , IC, I D Dap so : A B C D la hinh ehuT nhat

B a i 52 Trong mp(Oxy), cho bon diem A(4; - 1), B ( - 2; 1), C(4; - 5), D(2; - 7)

a) Tim diem I d tren true tung va each deu hai diem A va B

b) VoTi diem I tim dvtdc of cau a, chiJng minh rSng I cung each deu C

va D Ket luan gi ve txi" giac ABDC ?

* Hudng ddn

a) Gpi 1(0; y)

I each deu A va B nen l A = lb

T i n h lA^ va I B ^ theo y roi cho lA^ = I B ^ Dap so : 1(0; - 3)

!• b) K h i CO toa do I , t i n h IC va I D ta c6 ket qua

So sanh bon doan l A , I B , IC, I D ArtS;:; f

Tu" giac ABDC ngi tiep ducfng tron t a m I , ban k i n h R = 2^5

B a i 53 Trong mp(Oxy), cho ba diem A(- 4; 6), B ( - 2; - 1), C(3; 4)

Tim diem M tren true hoanh sao cho | MA + MB + MC I ngfin nhfi't

207

2) T i so' Irfgfng giac ctia mpt goc

Trong he true toa do Oxy, ve niira dirdng t r 6 n don \n k i n h

R = 1), nufa ducrng t r o n nay cdt true x'Ox t a i A va A', c^t t i a Oy t a i

C vdi A ' ( - 1; 0), A ( l ; 0), C(0; 1)

a) D i n h n g h i a

T r e n nufa difdng t r o n don v i (.ban k i n h R = 1) lay diem M(x; y)

Dat a = A O M (0*^ < a < 180°), ta eo :

• Tung dp y eiia M goi la sin ciia goc a, k i hieu sina va ta e6 sina = y

• Hoanh do x cua M goi la cosin cua goc a, k i hieu cosa va t a c6

y

b) T i so Ivtifng giac ciia mOt s6' g6c cAn nh<3r

c) Dau ciia cac ti so Itfdng giac

• sina > 0 v6i moi goc a e [0°; 180°]

Vdi 0 < a < 90° ta c6 0 < cosa < 1

Vdi 90° < a < 180° ta c6 - 1 < cosa < 0

• tga va cotga (neu khac 0) deu cung dau vdi cosa

II) Cac he thi/c glQa cac ti so lUting glac

c) sin^a + cos^a = 1

2) C a c thufc k h a c

Dinh li :

a) Neu cosa ^ 0, ta c6 : 1 + tg^a = —

b) Neu sina ?^ 0, ta c6 : 1 + cotg a =

cos a

1 sin^ a

3) L i e n h ^ giu^a ti so' lifofng giac ciia h a i goc bik n h a u

a) dSO" - a) va a Id hai goc bu nhau

b) (90 - a) va a la hai goc pht/inhau

Dinh li ,

Hai goc phu nhau t h i sin ciia goc nay bkng cosin cua goc k i a va ngucfc lai, nghia la :

sin(90 - a) = cosa cos(90° - a) = sina Do do

tg(90° - d ) = cotga cotg(90° - a) = tga

l o a n

B a i 55 Biet cosa = - , tinh P = Ssin^a + 4cosV

2

• HUcfng dan

• Biet cosa, dung he thiJc : cos^a + sin^a = 1 de c6 sina

• Biet sina, v i n diing he thuTc do de biet cosa

Ta CO : cos^p + sin^p = 1 (dinh l i )

2V2

c) Ta CO tgx = 2J2 > 0 nen sinx > 0 va cosx > 0

1 ung hai cong thiJc : cos^ x

Jai 57 Chiang minh h^ng d^ng thu"c : (sinx + cosx)^ = 1 + 2sinx.cosx

B a i 58 Chu'ng minh hSng d^ng thu^c : (sinx - cosx)^ = 1 - 2sinx.cosx

* HUcfng ddn

Tuong tir bai 57

GIAI

2 2 2

Ta CO : (sinx - cosx) = (sinx) + (cosx) - 2sinx.cosx

- sin^ x + cos^ X - 2blxix.cosx

Ap dung cong thuTc a^ + b^ = (a + b)^ - 2ab, t a c6 :

sin'^x + cos'^x = (sin^ x + cos^ x j - 2 sin^x.cos^x

A = cosy + siny.tgy = cosy + siny

cosy

= cosy + sin"' y cos" y + sin" y 1

Bai 62 Dotn gian bieu thtfc : B = + cosb.^1 - cosb

* HUdng dan

S\i dung cong ihufc \/a.v^ = \/ab (a, b > 0) va chii y 1 - cos% = sin"!

GIAI

B = x/l + cosb.^l - cosb = 7(1 + cosb).(l - cosb)

= sjl - cos" b = \/sin^ b = ! sinb I

Vi sinb > 0 nen j sinb! = sinb, do do : B = sinb

• Ta CO C = sina v/l + tg^a - si sma cos^ a sina cosa

• Vi 90° < a < 180° nen cosa < 0 => | cosa I = „ sina

Su dung dinh h've ti so' lucfng giac ciia hai g6c phu nhau

Cong thiJc cos^a + sin"a = 1

Tom lai cosh2° + cos^78° + cos^l° + cos'^89° = 2 Bai 65 Tinh : T = sin^3» + sin^l5° + sin^TS" + sin's?"

* HUdng dan

Tirang t\l bai 64

Ta CO T = (sin^3° + sin^87°) + (sin^l5° + sin^75°) , GIAI / Ta CO sin^3° + sin^87° = sin^s" + cos^3° = 1

• Tuong tiT sin"15° + sin-75° = sin^l5° + cos^l5° = 1 Vay T = sin^3° + sin^l5° + sin^75° + sin^87° = 2

B a i 6 6 Ddn g i a n b i e u thuTc : A = s i n O O " - x).cos(180** - x)

• Hii6ng ddn

Suf dung d i n h l i ve t i so' liJOng giac c u a h a i goc phu n h a u , h a i goc bii nhau

G I A I ^

• sin(90° - x) = cosx v i (90° - x) v a x l a h a i goc phu n h a u

, cos(180° - x) = - cosx v i (180° - x) v a x l a h a i goc bii n h a u

Do do : A = sin(90° - x).cos(180° - x ) = cosx( - cosx) = - c o s \

• sin(180° - x) = sinx v i (180° - x) v a x l a h a i goc bii n h a u

Do do : B cos(90° - x).sin(180° - x) = s i n x s i n x = sin^x

B a i 68 B i e t r&ng sinl5** = — — T i n h c a c t i so liitfng g i a c c u a g o c Is"

B a i 69 Chiang m i n h r&ng : cotg^a - cos^a = cotg^a cos^a

2 cos a , 2 2 , , cos a ^— = cotg a cos a (dpcm) sin a

B a i 71 Cho 90" < a < 180**, chii-ng minh rllng :

,Jcos^a(l - tga) + sin^a(l - cotga) = sina - cosa (*)

T = cos'^ a 1 - sina

cosa y

+ sin a 1 - cosa

sina

= v/cos^a - 2 sin a cos a + sin^ a

= ^(cosa - sina)^ = i cosa - sina I

Vi 90" < a < 180° nen cosa < 0 va sina > 0, do do cosa - sina < 0

=> I cosa - sina I = sina - cosa

Vay T = sina - cosa (dpcm)

B a i 72 Chiing minh neu 0° < a < 90" thi :

^sin^a(l + cotga) + cos^a(l + tga) = sina + cosa

^f- Hii&ng dan

Giai gio'ng bai 71 va chii y rfing 0° < a < 90° nen cosa > 0, sina > 0

=> I sina + cosa I - sina + cosa

B a i 73 Chi?ng minh dSng ihHc ^^"^ — ^ = cosx + sinx

cosx - sinx

^^• Hiidng dan

Bien ddi ve t r a i t h a n h ve phai bang each chu y :

1 = cos"x + sin'^x va a^ - b" = (a - b)(a + b)

G I A I

, 2cos-x

Ta CO Y 2cos^x - ^cos^ X + sin^ x j

cos X - sin X cos x - sin x

„2 „ : , 2 cos'"^ X - sin^x (cosx - sinx)(cosx + s i n x )

Giai giong bai 73

B a i 75 Biet tga = 2, tinh gia tri bieu thufc T =

^^ HUdng dan

2sina - 3cosa 3sina + 2cosa

, „, • , s i n a Chia tijf va mau cua T cho cos a va chu y = tga

cosa

G I A I

2 sin a 3 cos a T„ on T = cosa " cosa _ 2tga - 3 _ 2(2) - 5 ^ 1

3 sin a 2 cos a 3tga + 2 3(2) + 2 8 cosa cosa

B a i 76 Tinh gia tri bieu thiiTc T = -j=— , biet cotga = - 41

V3cosa + 4sina

• HUdng ddn

, , cosa , Chia t u va mau cua T cho sina va dung cong thUc = cotga

G I A I

• V i 0° < a < 90° nen cosa > 0, sina > 0 r^ E > 0

• Ta CO E^ = (sina + cosa)'^ = sin^a + cos^a + 2sina.cosa

= 1 + 4V2 9 + 4>/2 1 + 2V2 Vay E = i ^ : ^ - (vi E > 0)

B a i 78 Biet 90" < x < I8O" va sinx.cosx = ' -•

Tinh gia tri bieu thii"c T = cosx - sinx

• Hudng dan

• 90° < X < 180° => cosx < 0 va sinx > 0 ^ T < 0

T i n h T^ = => I T I = ma T < 0 nen T =

Dap so : T = - V2

B a i 79 Biet tga + cotga = m

Tinh theo m gia tri bieu thuTc : T = tg^'a + cotg^'a

Ta'co T = (tga + cotgaXtg^a + cotg^a - tga.cotga)

= (tga + cotga)[(tga + cotga)^ - 3tga.cotga| - m(in

cos^a + 4sin^a + 2(sin^a + cos^aj

6sin^a + Sees a 6sin a Scos^a

2""" + 2

COS a cos a tg2a - 4 - 4 1

Giai giong bai 81 (sau k h i bien d6i T theo sin^x, cos^x : chia tilr va

m l u ciia T cho sin^x) , '

Can nh&: tgacotga = 1 ,

-B a i 84 t i n h gia trj bieu thtfc : T = >ytgl5''.tg30".tg45".tg60°.tg75*'

* Hudng ddn

Giai giong bai 84 vk chii, y tg45'' = 1

Bap so : T = 1

C l i u y e n d o 5

TiCH VO Ha<3fNQ CUfl Mfil VECTO

K i e n thdrc C O b a n 1) G o c c u a h a i vector

Dinh nghia

—> -> ->

Cho h a i vecto a va b khac 0

- > - - » - > - >

Tix diim O tiiy y , ve OA = a va OB = b

So do ciia goc AOB gpi la so do ciia goc giiira hai vector a va b ,

kf hi^u : a , b

B

a , b

0 khong phu thupc v i t r i di§m 0

Tich v6 hUoing c u a hai vectd a va b , ki hi^u a b , la mot so

thifc dUdc xac djnh bdl cong thufc : a b = | a | | b | cos( a b )

b) He qua 1 a b = 0 < = > a J b

-> - ^

He qua"2 Neu hai vectP A B va CD cung nam t r e n mot true t h i :

A B CD = AB.CD ( A B , CD Ian lifot l a dp d a i dai so ciia A B va

CD t r e n true do)

H $ q u a 3

B i n h phaong v6 hudng ciia mot vecto bang b i n h phaong do dai ciia

vecto do, nghia l a a^ = l ^ l

A

3) C o n g thufc h i n h c h i e u a) Dinh nghia

A' Cho vecto a = A B va diidng t h ^ n g (d) Goi A ' va B ' l a h i n h chieu vuong goc ciia A va B t r e n (d)

Vecto a' = A ' B ' goi l a h i n h chieu ciia vecto a t r e n (d)

b) D i n h h

~> >

Tich v6 hifdng ciia hai vecto a va b bang tich v6 hiTdrng ciia vecto

a va h i n h chieu b' ciia vecto b t r e n diicrng t h i n g chufa vecto a

a b = a b' *

4) C a c t i n h c h a t cof b a n c i i a t i c h v 6 hifdrng Dinh li

" > - - > — > _ ,

Vdi moi vecto a , b , c va moi so thiic k, t a c6 :

- > - » - > - >

a) T i n h giao hoan : a b = b a b) T i n h phan pho'i do'i v(Ji ph^p cong vecto : " '

T o a n

Bai 85 Cho tarn giac ABC vuong tai A, AB = a, BC = 2a^ Drfa vao dinh

nghia cua tich v6 hvCdng hay tinh AB.AC, AC.CB, AB.CB

GIAI C Tinh AB AC

Tirong t\i, ta c6; AB C^ = BA BC = BA.BC.cos B = a.2a ^ =-a^

Cdn nhd : cos30° = ^ , cos45" = ^ , cos60° = ^

Bai 86 Ttf dang thtJc a = |a| ^ c6 the suy ra cao d^ng thrfc sau day hay

va b ma chi diing khi a va b cung phucfng

Bai 88 Cho bon diem A, B, C, D bat ki ChuTng minh rdng :

DA.BC + DB.CA + DC.AB =0 (*) Tijf do suy ra each chu-ng minh dinh li "Ba difofng cao ciia mpt tam giac thi dong quy tai mpt diem"

GIAI

DA.BC = DA ( D C - DB) = D " A D C - DA.DB (1) Taco : DB.CA = DB.(DA - D C ) = DB.DA - D B D C (2)

DC.AB = DC.(DB - DA) = DC.DB - DC.DA (3) Cong (1), (2) va (3) ve theo ve ta dugc :

D A B ^ + DB.CA + DC.AB = 0 K") (dpcm)

• Trong tam giac ABC, hai diTdng cao ve tCf A va B cat nhau tai H

Theo he thufc (^^') ta c6 : HA.BC + HB.CA + HC AB = 0 (Chon H = D) (2)

ciia tam giac ABC

Vay trong mot tam giac, ba dudng cao dong quy tai mot diem (diem nay goi la trUc tam ciia tam giac)

B a i 89 Cho tam giac ABC vdri ba dufdng trung tuyen AD, B E , C F

Chi?ng minh rang : B C A D + C A B E + A B C F =0

B a i 90 Cho hai diem A, B co dinh va mpt so duTdng k khong doi

Tim quy tich nhffng diem M sao cho MA MB = k

o M I = ^jk + lA^ = khong doi

, Vay quy tich ciia M la dudng tron tam I , ban kinh yjk + lA^ (k > 0)

B a i 91 Cho hai diem M, N nSm tren difoing tron dufofng kinh AB = 2R

Goi I la giao diem hai difofng thdng AM va BN

a) Chu-ngminh: A M A I = A B A I , B N B I = B A B I

b) Tinh AM AI + B N BI theo R

G I A I

a) • M d tren du&ng tron dirdng kinh AB nen BM 1 AM

hinh chieu ciia AB len dUcJng thang (AM)

Do do : AB AI = AM AI • Tucfng ty BN la hinh chieu ciia BA len dudng thing BI

a i 92 Trong h? toa dp Oxy, cho cac diem : A ( l ; 1), B(2; 4), CdO; - 2)

Chiifng minh rSng tam giac ABC vuong tai A

Tinh tich v6 hufdng B A BC va tinh cosB Txidng tU tinh cosC

GIAI

Ta C O :

AC = ( x c - X A ; y c - yA) = (10 - 1; - 2 - 1) = (9; - 3)

AB = ( X B - X A ; ys - yA) = (2 - 1; 4 - 1) = (1; 3) Suy ra AB AC = 9.1 + (- 3).3 = 0

Vay AB 1 AC hay tam giac ABC vuong tai A

Ta cd : BA = (- 1; - 3) va BC = (8; - 6) nen :

BA BC = (- 1).8 + (- 3)(~ 6) = - 8 + 18 = 10 Tinh cosB ,

Ta cd : BA BC = | BA I I BI:; | cosB ;

BA.BC 10 cosB =

B a i 93 Trong mat phSng toa dp Oxy, cho hai diem A ( - 1; 1), B(7; - 5)

Tim diem M tren true tung sao cho tam giac MAB vuong tai M

B a i 94 Trong mp(Oxy), cho hai diem M(l; 4) va N ( - 5; - 4)

Tim diem I tren true hoanh sao cho I of tren di^oTng tron dUcfng

• Dien tich (ABCD) = AB.BC = 15

B a i 96 Trong mp(Oxy), cho I ( - 3; 1), J(5; 0), H(6; 8), K ( - 2; 9)

Chx^ng minh I J H K la hinh vuong Tim tpa dp tam cua hinh vuong

I J H K

* Hudng d&n

Giai giong b a i 95

B a i 97 Trong mp(Oxy), cho tam giac A B C : A(2; 2), B ( - 5; 1), C ( - 6; - 2)

Tim tpa dp trtfc tam H, trpng tam G ciia tam giac A B C

B a i 98 Trong mp(Oxy), cho tam giac ABC : A(4; 4), B ( - 3; 3), C ( - 4; 0)

Chxifng minh rang trijfc tam H, trpng tam G, tam difofng tron ngoai

tifi'p I ciia tam giac thi d tren mpt dxidng th^ng (dtfdng th^ng nay

goi la dUcfng th^ng Euler cua tam giac ABC) va G chia doan

theo ti so bSng bao nhieu ?

\ay • G H va G I cCing phirang => G, H , I t h i n g hang (dpcm)

• G chia doan H I theo t i so' k = - 2

B a i a Trong mp(Oxy), cho sau diem A(2; 0), B(l;^/3 ), C ( - 1; %/3 ), D ( - 2; 0),

E ( - 1;- VS ) , F ( 1 ; Chiing minh rSng A B C D E F la luc giac deu

-• HUd\g ddn

Pinh do dai sau canh : A B , BC, CD, DE, EF, FA va t i n h dp dai OA,

HQ, OC, OD, OE, OF ta se difOc ket qua

Dap so ;

/BCDEF la luc giac deu c6 canh AB = BC = = 2 noi tiep d'dng t r o n t a m 0(0; 0), ban k i n h R = 2

Bai 100 C h o tam giac A B C ( B C = a, C A = c)

b) SiJf dung ket qua cau a d6ng thcri siJf dung dinh nghia :

Bai 101 Cho tam giac ABC (BC = a, CA = c)

Tinh dO dai trung tuyen AM (M la trung diem BC) theo a, b,

* HUdng d&n

1 Dung cong thiJc AM = - (AB + AC ) roi tinh AM va d^ng thcfi

2 dung ket qua bai 100

if- HUdng d&n

Tinh MP theo MA va M D , tinh BC theo M B , MC roi tinh

MP (Chu y : AC 1 BD)

GlAl

Ta CO • 2 M P = M A + M D vi P la trung diem A D

B'C = M C - M B Vay 2 M P B'C = ( M A + M D )( M C - M B )

Bai 103 Cho AA' la day cung ciia difcfng tron tam O va M la mpt diem tuy

y tren doan thfing AA'. » r , _ 5 ^

Chufng minh : 2 MA MO = Mi "MA - MA')

1) Dinh li cosin trong tarn giac

Vdi moi tarn giac ABC, ta c6 : ,;

a) a^ = b'^ + c^ - 2bc.cosA (1) b) b^ = a % c^ - 2ac.cosB (2) c) c^ = a^ + b^ - 2ab.cosC (3)

qua Tam gidc ABC vuong tai A nen cosA = 0, ta c6 dinh l i Pitago :

Tam giac ABC vuong tai A o a^ = b^ + c^

A su>'.*

2) Dinh li sin trong tam giac

Vdi moi tam giac ABC, ta c6 :

ha, hb, he Ian lucft la do dai dudng cao ke t\i cac dinh A, B, C va S la

di^n tich tam giac ABC A a) S = ^ a.ha = ^ b.hb = \c

2i ^

b) S = -a.b.sinC= -b.c.sinA= -c.a.sinB

2 2 2 B c) S = 3 b C -' , trong do R la bdn kinh di/dng tron ngoai tiep tam giac

4R

d) S = p.r, trong do p la niifa chu vi va r la ban kinh diTdng tr6n noi

tiep tam giac ABC

e) S = ^p(p - a)(p - b)(p - c) trong do p la mjfa chu vi va a, b, c ia ba canh ciia tam giac ABC (Cong thiJc Heron)

Cho hai diem co dinh A va B

a) Tap hop nhij-ng diem M sao cho MA^ + MB^ - (trong do k 1^ mot

so duong cho trifdc) la difdng t r o n c6 t a m la t r u n g diem I ciia A B va

ban k i n h R = -J2k^ - AB^ vdi 2k^ > AB^

2 ^

b) Tap hop nhufng diem M sao cho MA^ - MB^ = k (trong do k la mot

so cho trirdc) la duomg thSng vuong gdc vdi AB t a i H , H la diem

B a i 104 Tam giac ABC c 6 b = 7, c = 5, cosA = -

Tinh ha va ban kinh di/cfng tron ngoai tie'p R

Ap dung dinh l i sin :

2.14

4V2

7V2

a sin A = 2R R =

B a i 106 Tam giac ABC c6 AB = 8, AC = 9, B C = 10 Mpt diem M ndm tren

canh BC sao cho BM = 7

Tinl\p dai doan thdng AM