De HDC thi QG MTCT mon Toan THCS nam 2016

"trang" A'NB' xem hinh ve, Suy ra di~n tlch 6 tam giac trang yang ngoai bang dien tlch luc giac.. V~y dien tich 6 tam giac trang yang ngoai la:..[r]

HtT

c F

Call 3.Tinh gia tri ella U 36•

Bili 4 (10 lliim)

Cau 1.Tai mQtsieu thi, mot cai 16vi s6 n g e 6 gia ga e la 3250000 d5ng Nhan dipngay l~) sieu thi giam gia hai lfut, tanthir nh§t giam l a% so voi gia g5c, I§nthir hai giam 2b % sovoi gia khi etaduoc giam ln thir nhat, Do d gia cua 16 vi s6 g lucna y chi co la 1992900d5ng H oi m 6 i IAnsieu thi giam gia duoc bao nhieu ph~n tram?

Cau 2 Cho da thirc f (x) :;;: (3 x2 +2x - 7t Tinh t6ng cach~sacua da tlnrc chinh xacd~ndonvi Cau 3 Ngiroi ta vi~t 22 ch s6ph~n th~p phan, bAtdiu ill ngay sau d§u phay cua </2015 (gifr nguyen thu nr) duoi dang X1X2X3" X21X22 Tim xn.

Bai 5 (J0 diim) Cha l\lC gilk d~u (g9i 13e§p 1) A BCDEF co qmh AB=a ~ 36cm TiI cae trung di~m eua m6i c~nh dvng mQt h,lcgiae d~u A' B ' C ' D' E 'F ' va hinh sao 6 dnh ding c dinh 1a cae trun di~m A ' , B ' , C', D ' , E ', F ' Ph~n trung tam ella hinh sao 1ftIIf giac d€ dp2

MNP Q RS V6'i IIfC giae nay ta l~ilinn tuong t1,r nhu d6i v ilIfe giac ban dAu A BCDE F v dugc hinh sao m6'i va h,lc giac d~u dp 3 D6i v6'ih, l Cgiac d~u cftp 3, ta l~ilam tuan gtI!nhu tren va duge h.lcgiae d~u cAp 4 va d~nday thi dung l~i A A r

Cae canh hinh sao dUQ'eke x<;>e, con cae hinh thoi trong

hinh du<),cchia thanh 2 phftn: phAnk e X C va ph~n d~tring

(nhu hinh ve) Rieng h)e giae d~u cAp4 dugc d~triing

a) Tinh di~n tich phAn hinh dug c dB tdng

b) Tinh ti s5 ph§ tram gifra di~n tieh ph§.n hinh duQ'cd~

tr~ng~adi~ndch hinh hic giac ban dAu

Bili 1.(10 d t¬ m)

C" 1au Tinl1 1gan ung;. d' graiatrin anlc nlJ.iar va'nh'a n at cua amh~ h' so:.y = x+210 5 .

x- +1 Call 2 Tinh x thea a, b, ebiet: ) a + b Vc - x =2+ ~ "- a - b -V -:- c= - =x.Hay tinhgia tri ellax voi

a= 303; b=313; c= 14

Cau 3 Trang mat phang toa dQ Oxy, tinh khoang each gitra cac gia diBm ella duong th~ng

y =5 x +113- J15 v parab l y =x 2

Bili 2 (10 tli i m )

Call 1.Cho tam giac ABC Cac diem D, E, F thea thfr tt.r thuQe ca canh AB , Be, CA sao cho

AD =!3AB ) BE = l_3BC, CF = !3'C A. Gai A£H , P t~ tum. l a ziaodiBm ella AE voi CD,AEvoiB F ,

c <

BF v6'i CD.Tinh dien tich tamgiacA1N Pbi~t tam gia ABC cod9daiba canh lit2015; 2016;2017 Cau 2 Cho hinh vu o ng c o c nh bang 2if3.

N6i dinh cua hinh vuong voi trung diBmella cac canh (nhu hinh ve)

Tinh dien tich cuabat giac duoc t6 d~m

Bai 3.(10 diim)Cho day s 6: U " =( 1 +.J2)" + ( 1- fir + 1 ,vain 1as6tI,Inhien khac 0

Cau 1 Thill 6s5hang dau cu aday IC- -"' -"" Cau 2 Tim cong thirc t6ng quat tinh U 7l +1 thea U; va U n_ I ' voi nz2

I ~ ~ ' ' \ : ' , : : ' ~ ' ; i Thai gian lambai 9? ph~t \kh~ ng ki th oi g ia ngi a a )

'.~ - ,, _ ~ _ _ _ 1 Ngay thi :."O IJ / 20 1 Chll y : -Thi s inh trinlibay tom tlit e ach giai V a G giey thi do c an b e; co i thi phat;

- N€u eMba i kh on g coy eu cdu rieng thiki tqua lam tron d i n 4chit s8th ~ pp ha

cuoc THI GIAI TOA-NTREN MAyTiNH CAMTAY

NAM201

BO GIA o Due vA BAo TAO.

5+~ S - 4.J15 +4 [13

2

NMp cong thirc tim duoc cua x a tren vao may va Slr dung lenh CALC voi

a==303;b=313;c= 14taauQ'c: x ~1 3, 99 86 3 08 ,

K~tqua: x=1 ,998

C"au 3 Toa dQ~.giao dO(em a n;' I' ghiree m cua. I'1 1;:p uonh g trill' :h ! V-= 5X +, .J13 - Jl5

y = ; c

Giai phirong trinh hoanh dQ giao diem: i =5x+Jl3 - Jl5

<;::::> x' - 5x+ Jl5 -.Jlj=0(*)

5- ~25- 4M + 4Jlj

Dung cong thtrc nghiern tm duoc 2 n hiem: XI = - -' - -

-2

B i 1.(10 cti a m)

Ca u 1 Tinh g~n dung gia cd Ion nh tva n o nh§t cua hamsf>y==x +,20 5 ,

: c+I

-Cau 2 Tinhxtheo a, b,Cbiet: ~a +bVc - x =2+)a - b V c - x ,Hay tinh gia tric~a

xvoi a= 303; b==3 3; C = 1

Cau 3 Tron m~t phang toa dQ Oxy, tinh khoang each gina cac giao di~m cua duong

th~ng y=5x + m -FtS vaparab l y=x 2 ,

Gi ii i

Cau 1 Gia StYhams6coGTLN va GTNN thi phai t 6ntai xthoa man:

x + 20 15 ~ Ph h x + 2 01 5 , h'A ~

Bl'e~ d;"n 01 '}'I A ,2015-J2015 2 +1 20 1 +)2 1 2 +1

Maxy=2 15+)202 5 +1 ;M'1 ny=201S-J2021 ~+1 '

Nhap vao may ta co:Max y~2015,OOOI24; Min y -,2 06947.1O-4

K~tqua: Max y ==2 15,0001 ;Min y =- 0,000 1

, -

Cau 2 Binh phuong hai v~ vabi~n06 itaduoc: bV c - x =2 + 2 ) a - b V c - x ,

Voi dj~u kien y ;:: 2.Binh phuong hai v~ phirong trinh (*) d€giaita ducc:

Tha yY2 vaob v c - x = y => b \t c - x =2 va - 1 ~ c - x= b

8( a -1 ).ra=I

Ta duoc, : x =c- b) ,

N A M 2 1

Mo : Toan Lap: 9 Cap THCS

Fl U ' O NG D A N C IAf HO ~ C DAps o

BO CIAo D\lC VADAO T~O

DETHI CH iNH T R U e

A.B C

S.,IfW' = ~ ~p(p - (l )( p - b )(p - c) ~2514 0,515

K~tqua: S== 251410,5151

K

c

V~y: StlIldNP=SdABe - S tJA C M- StJ.4B,v- StJ B C= -:;StJ

Bi lI i

Cau 1.D ~t S LlAD M = x ~Sd 4BM == 3 ilA D M =3x.

=> S t14 CM = 2 St14B M == 6x (1) ( MBM va M CM co chun canh day AM va d uemgcao

MCM g§p d6i duong cao Ll ABM) A

=:>StjADC ==S LIA D M+SMCM = 7x

=>5M BC ==3S tJA DC = 2 1x (2 ).

(1)&(2)=>SM C M == _ ' Sd A1J C== - SiJABC.

Chung minh nrong nr:

S d ABN == '7S tJ ABC; S a BPC = '7StJ A C

Bfli 2 (10 diim)

Cau 1.Cho tam giac AB C. Ca di~m D E , F theo tlnr nr thu9Ccac canh A B , Be C A sao cho

A D = !.3A , BE = !.3Be , CF:::; !.C3' A. Goi M N. P I§nhrot. 1ftziao di~mella AE v6i C D, A

c·

vci BF, BF vat C D Tinh dj~n tich tam gia MN Pbi~t tam giacA B Cc odQdai ba canh lit2015;

2016; 2017

Can 2 Cho hinh vuong c6 canh b~ g 2 VJ.

N6idinh cua hinh vuong v6i trung diem cua cac canh (nhu hinh ve)

Tinh dien tich cua bat giac duoc t o d~m

Nhap vao may va gan X I :::;5- J2S - 4 Mof-4.JU ch A;

2

, 5+~25- 4.Jl5+4J13

Ap dung cong thtl'c tinh khoang each gifra2 di~m A(Xl; y,) va B(X2; Y2)ta co:

A :::;J< x2- x Y +( Y 2 - YI )~=~ (:s - X I) + ( x / - X 1 )1 -=~ (B -AY" +(B~-A : ~ ~2 ,94367785

Kat qua: A B == 24,9437

Cau 1.NMp ham (1+fir+(1- fir+1va o may va sird n lenh CALC voi xhi I, 2,3,

4, 5, 6ta co:

K€t qua: U J= 3; U 2= 7;U 3=15; U J==3S , Us== 83; U 6= 199

Cau 2 Gia sirc6 Un +1 :::: d U j+ e.U s r +f

{

d.7+e.3+ f =15

Theo cau I taco: d.l S+e.7+ f=3

dJ 5+ e t5 +f =83

Giai h~p irong trinh n y bang may tnh ta co d=2; e= I;f= -2

K~tqua: U I/+ I =2 U I/ U " J , 2

Bai 3 (1 0 dii m) Ch d y s6: u, =(1+ fir + (1- J2)"+1,voi n1<1s6nr nhienkhac O

Cau 1.Tinh 6s6 han d~u cua day

Cau 2.Tim c6n thirc t6ng quat tinh U " ~ lthea U" va U _ 1, v i II ~ 2

Call 3.Tinh gia tri ella U} 6 '

~ 1 ducc S US T :; 3~a !

S =:0 - 2<1 4 - { j· = -a· = - :::: 1 3 6 722)4

K~ t qua: S=} )3867

US T = C S Gne n

SinU ST :::sinC S G= sinC SK

C K 2a J5 a.J5 4

= - = - - ; - =

Do US = x thi u r== ~ x;ST ::: ~X

U K =KD = 2 a.f5

( )1 (2 aJS ) ~ 4 1

-Co US x=a J5

10

V I x=- - =

Ma SK =U K - US==D K _ US =3a f5

CS2 =S K 2 +CK 2 ~ CS= a $

2

CK= CD.CG :;:;2a.a = 2 {1 15

DG a f5 5'

KG= aJS

5

CK2 = KD KG ~ KD= CK2 =4 a.J5

f) ~tC G ::: a ;

CO=2a

DG=aJS

~ 1

t

J

Cau 2

Cau2 T6ng ca h~56 cua da thirc J ( x ) Iiigia tn cua dathirc tai x=I GQitAng cac h~

s6 cua da thirc 180A, ta co :A=f () ;:;(3+2- 7)64 =26-1.f)~yrang :

264= ( 232 t =42949672962. D?t 42949 =X, 6729 =Y,taco:

Bili 4 (10 d i m)

sieu thi giam gifthai l~n, l§ thir nh~tgiam la% so voi gia g8c, IAn tlnr hai giam 2b% so

1992900 d6ng Hoi m6i l5.n sieu th]giam gia d cc bao n ieu phan tram?

dan vi

G in;

Cau 1 Lin thir nhc1t giam: l a % = x% (1 sx < 20 )

L~n thir hai giam: 2 b%=y% (20~y < 30) veri x,Y N.

Sau l~n giam thir nhAt s6 ti n m ot cai 16 i song lit: 3250000- 3 2 50000 x% (d6ng)

(3250000 - 3 2 500 0 x % ) - 3 5 000- 3 5 000 0x % )y%= 199 900

<:::> (100- x ) 100- y ) ==6 32 (* ).

Giai plnro g trinh (*)bang each Slrdung lenh CALC tren MTCT taduoc x =16) dod

y= 2 7.

636 0

Cau 3 Tinh b~ng may thay U 1 kh6ng trim man hin (U 1 1l =7761799), con U3 tran man

hinh Do d tatim each bien d i va tinh U36theo UI S

U1 -1 =(l-J2),s+(l+J2)ls

Binh phuon hai v~ va bien doi ta co:

(Uwl) 2 =( 1 _.fi ) 36 +( 1 + fi ) 3 6+ 2

(U1 -1)2-1 =(1-.J2 /6 +(1+ J2)"6+I= U36

MaU'R == 776179 :::::> Uwl == 7 6179 => U31i == 77617982-1

Ta di t nh chinh xac M == 77617982

Th?tv?y:d~ta=7761;b=7 8

M.::; (a.l 03 +bi ::;a2 10 +2 b 03+b2

Tinh a2 , 2ab, b2 bAng may tinh r6i su ra a2.106 ; 2ab.103 ; b2 • Sau d6 cong tai b~ngtayta

d uo c:

a2•i0 6 ::;

2ab.03=

b2 =

Cong lai ta co:

U3 == M- = 6 245 0 1 2803

G iiii

a) Chia iuc giac thanh 6 tam giac d~u c o c nh lita bang 3duong cheo di qua2 dinh d6i

, A ' d' '(' 6 a [J 3 a)Jj Chi I h' h 24

xung qua tam, ttl' 0taco » = ' -4 - = " 2 ' ia ucgra A C DE F t an, tam giac

c F

Bai 5 (10 di i m ) Ch ) \ IC giac d~u (goi lildp I AB CD EF cocanh AB = a = 3 cm T cac

trung diem c a m 6 i canh dun I11 Qt 1 \ IC gia d u A' B 'CD ' E ' F' v hinh sao 6 c n cimg

codin hic c trung diem A ' , 8',C, D ' , E ' P ' Phan trun t a mcua hinh sa 1<1luc gia d€u

cAp 2 i' vfNP Q RS Voi l \I C giac nay ta lai lam nrong tv n u d6i vci I \ I C giac ban d§u

AB CDEF va d oc hinh sao rnoi va luc giac d~u d p 3 D6i voi luc giac d~u C§p3, ta l?i

lam nrong nrnhir trenva duoc luc giac c1 ~1Idp 4 vad~ndaythi dirnglai A ' B

Cac canh hinh sao duoc ke X C , concac hlnh thoi trong

hinh duoc chia than 2 p an: p an ke X C va phan d~

trang (nhu hinh ve), Rieng tuc giac d~u dp 4 diroc d€

tr~n

a)Tinh dien tich phan hinh duoc d~trang

b) Tinh ti s6 p ~n tram giua dien tich ph§n hinh d ocd€

trang va dien tich hinh luc gia ban d~u

Ket qua: 184 46 744 07 37 0 9551 6 16

Cau 3 -Bfim tren may </2015 =12,63063011 b~m (i~ ~12,6 d~ kiern rra chir s6eu ico

bi lamtron k 6 g, duoc 0,030 3 I0 6 n n 12015 =1 ,6 06301 66

~ 2015 = 12,6306 0103 + x' +3.12,63063010.x (12,63063 10 +x)

~ 201 =1 ,6306 0103 +x' +3.12,63063010.x.4'2015

~ X 3 +3.1 ,63063010.</20 5x +12,630 3 1 3 - 201 = 0

K~t hQ'Ptinh tren may va tren gi§y taduoc

1 ,6306301 3 - 2015 =(1 ,63 +6 01.1O-7f -2 15

= 12,6 3 +6 0e.1 0-1 +3.1 ,6 6301.10-7 , (1 ,63 +6 01.lO-7 ) - 2015

=2014,6 8 47 +2 0 6 0 8 01.10-1 +3.12,63.6301,1 -7.1 ,630630 - 2 1

=2014,698447 +2 0166088901.10-2 +30154 8.393 55189.10-7 - 2015

=2014,6 8447 +0,00000000 250166088901 +0,3015498393855189- 2015

=-0,0000031603 431501109

V~y x3+3.12,63063010.</2015x - 0,000 0316 364315011099 =0

Khai bao va d ng chirc nang SHIF SOLVE ta d oc

x ~ 6,6033734 10-9 b~m -6,60337 IO-Y duoc 4.798 145.10-1

V~y ~201 ~ 12,630 3 1066 3 7 4798 1 =:> x2] =4

K~t qua: X 2 = 4

X - l O 'U = II

1 8 1 ~ 1 6 ! I ! 6 [I 'TOl I I0 0 0 a 0 o !0 0 0 o ;

2 XY I O) - I , _ fL_ 7D= i 0 5 9 1 8 0 8 o ;0 0 0 o I

A = (X lOs +y)2 =X2.IO'O +2XY.IO) +y

Tinh tren may k t h pv i gily ta c6:

K~t qua: SII'= 1157,44cm1 va34,38%.

b) V~ y ~ :::34,3 %

S AB C OEf

dien tlch luc giac dp 1 A8CDEF.

Tu01lg n r VO', , I 'l each tinhtr e~,n t a co: , !tIN =b =-(/; c = - b

Di'Aen n,ch6t am giac., tran! g eua'I 1,lCgiac., ca~ 3p 1'a: 413'c-21J3 - ' (3)

SI = ~ 3 1 J3 = 3 (/2.j3 S, = ~ .3 b 1 J3 = ~ 3 1 -.[3 = 3a 1 J3

S3= ~ 3c 1J3 = ~ 3(l2fi :::;3 1J3. S4= 3d \ [3 = 3u 1$ = 3a!fi

4 2 4 2· 4 ' 2 1 ' 2 2 8 2 2 '

. .

C/U IY : T 6 c h m t hi di n C ' v ao h t r on g da n g iii i ili c hin di § m c hi t iE t Ca e eac h

g ia i kh dc IlE u dun g, giii m khiio el m C tl- V elO k h ng t ho ng m int if i ch o a d m

,

Di~rn

K H qua

Ba! S (J 0 dilm)

Bai 3.(10 (f ii m)

Bai 2. (1 0m i m )

Ba i 1 (10 tl ii m )

cuoc THI GIAI ToANTREN MAyTiNH CAMTAY

Men: Toan Lap: 9 Cap THCS

DE Till C HiNH T HUC

BO. GIAo Du.e V A DAO T A.O

,.

' ,. ,