BÀI THI GIẢI TOÁN TRÊN MÁY TÍNH CẦM TAY CẤP TỈNH

MT S THI CP TNH

GII TON TRN MY TNH CASIO T NM 2012 2014

CA MT S TNH THNH

thi gii ton trn MT CASIO tnh AN GIANG

thi gii ton trn MT CASIO tnh BNH THUN

thi gii ton trn MT CASIO tnh CAO BNG

thi gii ton trn MT CASIO tnh CN TH

thi gii ton trn MT CASIO NNG

thi gii ton trn MT CASIO tnh IN BIN

thi gii ton trn MT CASIO tnh HU GIANG

thi gii ton trn MT CASIO tnh GIA LAI

thi gii ton trn MT CASIO tnh KIN GIANG

thi gii ton trn MT CASIO tnh LNG SN

thi gii ton trn MT CASIO tnh LONG AN

thi gii ton trn MT CASIO tnh SC TRNG

thi gii ton trn MT CASIO tnh TY NINH

thi gii ton trn MT CASIO tnh TP.HCM

thi gii ton trn MT CASIO tnh THA THIN HU

thi gii ton trn MT CASIO tnh VNG TU

B INH TAY IMPORT EXPORT JSC Add: 110 112 Hau Giang Street, District 6, HCM City, Vietnam

Tel: 08.3969 9999 Fax: 08. 3960 2478 - 08.3969 6666 Email: [email protected]

Website: www.bitex.com.vn

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AN GIANG

1. thi gii ton trn MT CASIO nm 2012 - 2013 - Mn Ton (THCS): Tr 7

2. thi gii ton trn MT CASIO nm 2012 - 2013 - Mn Ton (THPT): Tr 12

3. thi gii ton trn MT CASIO nm 2012 - 2013 - Mn Ton (GDTX): Tr 19

4. thi gii ton trn MT CASIO nm 2012 - 2013 - Mn L (THPT): Tr 26

5. thi gii ton trn MT CASIO nm 2012 - 2013 - Mn L (GDTX): Tr 41

BNH THUN

1. thi gii ton trn MT CASIO nm 2013 - 2014 - Mn Ton (THCS - Lp 8): Tr 51


3. thi gii ton trn MT CASIO nm 2013 - 2014 - Mn Ton (THPT - Lp 11): Tr 68


CAO BNG

Nm 2012 - 2013




4. thi gii ton trn MT CASIO nm 2012 - 2013 - Mn Ha (THPT): Tr 114

5. thi gii ton trn MT CASIO nm 2012 - 2013 - Mn Sinh (THPT): Tr 126

CAO BNG

Nm 2013 - 2014






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CN TH





NNG



IN BIN






HU GIANG

Nm 2012 - 2013






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Nm 2013 - 2014







Nm 2014 2015



GIA LAI

Nm 2013 - 2014







GIA LAI

Nm Nm 2014 - 2015







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KIN GIANG




LNG SN







LONG AN

1. thi gii ton trn MT CASIO nm 2013 - 2014 - Mn Ton (GDTX Lp 11): Tr 568

SC TRNG






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TY NINH







TP.HCM



THA THIN HU





5. thi gii ton trn MT CASIO nm 2014 - 2015 - Mn Ton (GDTX - Lp 12): Tr 698

VNG TU




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MTBT lp 9 trang 1

S GIO DC V O TO THI CHN HC SINH GII CP TNH

AN GIANG NM HC 2012 2013 Mn thi : GII TON BNG MY TNH CM TAY

Lp : 9

Thi gian lm bi : 150 pht (Khng k thi gian pht )

IM

(bng s)

IM

(bng ch)

CH K

gim kho 1

CH K

gim kho 2

S MT M

do ch kho ghi

Ch : thi gm 2 trang, th sinh lm bi trc tip vo bn thi ny v ghi p s vo kt qu. Cc kt qu tnh ton gn ng, ghi chnh xc ti 5 ch s thp phn sau khi lm trn. Th sinh s dng loi my no th in k hiu loi my vo sau :

Bi 1: (2,0 im)

Tnh

Kt qu:

Bi 2 (2,0 im)

Tnh tng :

Kt qu:

Bi 3: (2,0 im)

Tm c chung ln nht v Bi chung nh nht ca hai s:

Kt qu:

Bi 4 : (2,0 im)

Tnh gi tr ca vi

Kt qu:

Bi 5: (2,0 im)

Tm bn ch s cui cng ca s:

Kt qu:

Bi 6:(2,0 im)

Tm tt c cc cp nghim nguyn

ca phng trnh:

Kt qu:

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MTBT lp 9 trang 2

Bi 7: (2,0 im)

Hnh thang cn ABCD c hai y l AB

v CD, cho

.Tnh gn

ng chu vi v din tch hnh thang bit

.

Kt qu

Bi 8: (2,0 im)

Cho bn ng thng

Bn ng thng trn ct nhau ti bn im A; B; C; D.

a) Tm ta cc im A; B; C; D.

b) Tnh din tch t gic to bi bn ng thng trn.

Kt qu:

Bi 9: (2,0 im)

Cho dy s c cc s hng

a) Tnh ( dng phn s ti gin) b) Tnh gn ng

Kt qu:

Bi 10: (2,0 im) th sinh ghi s lt cch gii v p s vo sau y

Cho hnh vung ABCD c cnh 28cm . V na ng trn ng knh hai gc phn t ng trn tm A v B bn knh AB nm trong hnh vung (hnh v). Tnh hiu din tch hai hnh 1 v 3.

Cch gii:

4cm

A B

DC

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MTBT lp 9 trang 3

S GIO DC V O TO P N THI CHN HC SINH GII CP TNH


Lp : 9

Bi 1: (2,0 im)

2,0 im

Bi 2 (2,0 im)

2,0 im Bi 3: (2,0 im)

1,0 im

1,0 im

Bi 4 : (2,0 im)

2,0 im

Bi 5: (2,0 im)

2,0 im

Bi 6:(2,0 im)

Mi nghim 0,5

Bi 7: (2,0 im)

1,0 im

1,0 im

Bi 8: (2,0 im)

1,0 im

1,0 im Bi 9: (2,0 im)

1,0 im

1,0 im

Bi 10: (2,0 im)

Cho hnh vung ABCD c cnh 28cm . V na ng trn ng knh hai gc phn t ng trn tm A v B bn knh AB nm trong hnh vung (hnh v). Tnh hiu din tch hai hnh 1 v 3

Cch gii:

Cnh hnh vung a=28cm din tch hnh trn bn knh AB

Mt khc

tr hai ng thc ta c

Vy

p s : 139.62824

Phn kt qu 1,0 im phn nu cch gii 1,0 im

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MTBT lp 9 trang 4

TM LC CCH GII THI MTBT LP 9

Bi 1:

Bi 2:

Bi 3

Thc hin bm

Vy c chung ln nht ca a v b l

Bi chung nh nht ca a v b l .

Bi 4

Bm my biu thc x cho gi tr

; gn gi tr va tnh cho bin X

Bm biu thc cho ra kt qu

Bi 5

Dng my bm cc gi tr sau y

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MTBT lp 9 trang 5

Nhn xt rng u c bn ch s cui cng l 0625 nh vy

c bn ch s cui cng l 0625 nn s c bn ch s cui cng l 3125.

Bi 6:

Do x,y u l s nguyn nn

Vy h c 6 nghim nguyn

Bi 7 D thy hnh thang cn c

nn cc tam gic ADM;

ABM;BCM l cc tam gic u c ng cao

. Chu vi hnh thang l

Din tch

Bi 8 bng cch gii h phng trnh ta tm c cc giao im

Ta c t gic l hnh vung c cnh

Din tch t gic

Bi 9:

Lp quy trnh bm my bng cc dng lnh

ly cc gi tr ta c

Bi 10

Cnh hnh vung a=28cm din tch hnh trn bn knh AB

Mt khc

tr hai ng thc ta c

Vy

4cm

N

A B

MD C

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MTBT lp 12 trang 1



Lp : 12 Thi gian lm bi : 150 pht (Khng k thi gian pht )

IM

(bng s)

IM

(bng ch)

CH K

gim kho 1

CH K

gim kho 2

S MT M

do ch kho ghi

*Ch :

thi gm 2 trang, th sinh lm bi trc tip vo bn thi ny v ghi p s vo kt qu.

Cc kt qu tnh ton gn ng, ghi chnh xc ti 5 ch s thp phn sau khi lm trn. Th sinh s dng loi my no th in k hiu loi my vo sau :

Bi 1: (2,0 im)

Cho hm s c th i

qua cc im

a) Hy tnh cc gi tr .

b) Tnh gn ng khong cch gia hai im cc tr ca

hm s vi tm c.

Kt qu:

Bi 2: (2,0im)

Mt cht phng x theo thi gian s b phn hy t nhin. Cng thc tnh khi lng

cht phng x Cacbon cn li sau thi gian t nm l

vi l khi lng cacbon lc ban u. Ngi ta tm trong mt mu c c mt lng Cacbon v xc nh c n mt i 15% lng Cacbon ban u ca n. Hi

mu c c tui khong bao nhiu nm?

Kt qu:

Bi 3: (2,0 im)

Tm nghim gn ng (, pht, giy)

ca phng trnh:

Kt qu:

Bi 4: (2,0 im)

Cho dy s c cc s hng nh sau

a) Tm s hng th 10 v s hng th 30 ca dy.

b) Tnh tng 30 s hng u tin ca dy.

Kt qu:

Bi 5: (2,0 im)

Tnh gn ng cc nghim ca phng trnh

Kt qu:

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MTBT lp 12 trang 2

Bi 6: (2,0 im)

Cho tam gic ABC cn ti A ni tip trong ng trn bn knh

. Tm gi tr ln nht ca ng cao k t B ca tam gic ABC.

Kt qu

Bi 7: (2,0 im)

Trn mt phng Oxy cho . Tnh gc B (, pht, giy) v din tch ca tam gic ABC.

Kt qu:

Bi 8: (2,0im)

Gii h phng trnh:

Kt qu:

Bi 9 : (2,0 im)

Tnh th tch khi t din ABCD bit:

.

Kt qu:

Bi 10 (2,0 im)

Mt si dy in VC30 thit din l hnh trn c ng knh k c v bc l chiu di v mt Rulo c dng hnh tr ng knh , chiu di (hnh v). Ngi ta qun cht si dy in vo Rulo kht vi nhau theo tng lp, ht lp trong ra lp ngoi. Gi s s vng dy

mi lp u bng nhau. Tnh gn ng s lp v s vng dy qun trn Rulo.

Cch gii (tm tt):

p s:

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MTBT lp 12 trang 3



Lp : 12 Bi 1: (2,0 im)

Cu a)

1,0 im

Khong cch: 1,0 im

Bi 2: (2,0im)

Kt qu: 1343 nm 2,0 im

Bi 3: (2,0 im)

Mi nghim 0,5

im

Bi 4: (2,0 im)

1,0 im

. 1,0 im

Bi 5: (2,0 im)

Mi nghim 0,5

im

Bi 6: (2,0 im)

. 2,0 im

Bi 7: (2,0 im)

1,0 im

1,0 im

Bi 8: (2,0im)

2,0 im

Bi 9 : (2,0 im)

2,0 im

Bi 10 (2,0 im)

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MTBT lp 12 trang 4

Mt si dy in VC30 thit din l hnh trn c ng knh k c v bc l chiu di v mt Rulo c ng knh , b ngang (hnh v). Ngi ta qun cht si dy in vo Rulo kht vi nhau theo tng lp, ht lp trong ra lp ngoi. Gi s s vng dy mi lp u

bng nhau. Tnh gn ng s lp v s vng dy qun trn Rulo.

Cch gii (tm tt):

Mi lp dy c Lp th 1 c ng knh 17,5cm+0,5cm =18cm R=9cm

Tng chiu di dy qun lp th nht l

Lp th 2 c ng knh

Tng chiu di dy qun lp th 2:

Lp th 3 c tng chiu di

16

Lp th n c tng chiu di

Tng chiu di dy ca n lp l

Thay ta c Nh vy s lp l 10; s vng 9 lp l 16.9=144 vng

Chiu di dy ca 9 lp l

Chiu di dy lp cui cng

S vng dy lp cui cng

C 4 vng lp th 10

Vy tng s vng dy l 148 vng. C 10 lp

Nu ng p s 1,5 im phn nu cch gii 0,5 im

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MTBT lp 12 trang 5

TM LC CCH GII

Bi 1: th hm s c th i qua cc im

ta c h phng trnh

Khong cch gia hai im cc tr

Bi 2: Khi lng Cacbon mt 15% nn lng cacbon cn li l 85% ta c phng trnh

Vy khong 1343 nm

Bi 3:

t phng trnh tr thnh

Vi

Bi 4:

Xt dy

Vy C th gii bi ton trn bng tnh trc tip nh my bng dng lnh

vi gi tr ban u

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MTBT lp 12 trang 6

Bi 5: Gii phng trnh

iu kin dng mode table nhp hm s

d tm khong nghim t -3 n 10 ta nhn thy nghim ri vo khong (-3,-2) (-2,-1) (-1,0)(0,1).

Nhp phng trnh gii bng shift solve chon im d nghim trong cc khong trn ta c bn

nghim

Bi 6: t

p dng nh l Sin ta c

BH l ng cao

Xt hm s

Bi 7: Gii h tm giao im ta c

Bi 8

Bi 9 Din tch y

ng cao AH:

Bi 10: Mi lp dy c

Lp th 1 c ng knh 17,5cm+0,5cm =18cm R=9cm


Lp th 2 c ng knh



16

O

A

CB

H

22

2

2

M

A

B

CD

Hwww.

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MTBT lp 12 trang 7



Thay ta c

Nh vy s lp l 10; s vng 9 lp l 16.9=144 vng



S vng dy lp cui cng

C 4 vng vy tng s vng dy l 148 vng. C 10 lp

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MTBT lp 12 GDTX trang 1



Lp : 12(GDTX) Thi gian lm bi : 150 pht (Khng k thi gian pht )

IM

(bng s)

IM

(bng ch)

CH K

gim kho 1

CH K

gim kho 2

S MT M

do ch kho ghi

*Ch :

thi gm 2 trang, th sinh lm bi trc tip vo bn thi ny v ghi p s vo kt qu.

Cc kt qu tnh ton gn ng, ghi chnh xc ti 5 ch s thp phn sau khi lm trn. Th sinh s dng loi my no th in k hiu loi my vo sau :

Bi 1: (3,0 im)

Cho hm s c th i

qua cc im

a) Hy tnh cc gi tr .

b) Vit phng trnh tip tuyn vi th hm s ti A

vi tm c.

Kt qu:

Bi 2: (2,0im)

Mt cht phng x theo thi gian s b phn hy t nhin. Cng thc tnh khi lng

cht phng x Cacbon cn li sau thi gian t nm l vi l khi lng cacbon lc ban u. Ngi ta tm trong mt mu c c mt lng Cacbon v xc nh c n mt i 15% lng Cacbon ban u ca n. Hi

mu c c tui khong bao nhiu nm?

Kt qu:

Bi 3: (2,0 im)

Tm nghim gn ng (, pht, giy)

ca phng trnh:

Kt qu:

Bi 4: (2,0 im)

Cho dy s c cc s hng nh sau

a) Tm s hng th 10 v s hng th 20 ca dy.

b) Tnh tng 20 s hng u tin ca dy.

Kt qu:

Bi 5: (2,0 im)

Tnh gn ng bn nghim ca phng trnh: Kt qu:

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Bi 6: (2,0 im)

Trn mt phng Oxy cho . Tnh ta ba im A;B;C v din tch tam gic ABC.

Kt qu:

Bi 7: (2,0im)

Gii h phng trnh:

Kt qu:

Bi 8 : (2,0 im)

Tnh th tch khi t din ABCD bit:

Kt qu:

Bi 9: (3,0 im)

Mt si dy in VC30 thit din l hnh trn c ng knh k c v bc l chiu di v mt Rulo c dng hnh tr ng knh , chiu di (hnh v). Ngi ta qun cht si dy in vo Rulo kht vi nhau theo tng lp, ht lp trong ra lp ngoi. Gi s s vng dy

mi lp u bng nhau. Tnh gn ng s lp v s vng dy qun trn Rulo.

Cch gii (tm tt):

p s :

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Lp : 12 GDTX Bi 1: (2,0 im)

2,0 im

1,0 im

Bi 2: (2,0im)

Kt qu: 1343 nm 1,0 im

Bi 3: (2,0 im)

Mi nghim 0,5

im

Bi 4: (2,0 im)

1,0 im

. 1,0 im

Bi 5: (2,0 im)

Mi nghim 0,5

im

Bi 6: (2,0 im)

1,0 im

1,0 im

Bi 7: (2,0im)

2,0 im

Bi 8 : (2,0 im)

2,0 im

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Bi 9 (2,0 im)

Mt si dy in VC30 thit din l hnh trn c ng knh k c v bc l chiu di v mt Rulo c ng knh , b ngang (hnh v). Ngi ta qun cht si dy in vo Rulo kht vi nhau theo tng lp, ht lp trong ra lp ngoi. Gi s s vng dy mi lp u

bng nhau. Tnh gn ng s lp v s vng dy qun trn Rulo.

Cch gii (tm tt):

Mi lp dy c Lp th 1 c ng knh 17,5cm+0,5cm =18cm R=9cm


Lp th 2 c ng knh



16



Thay ta c Nh vy s lp l 10; s vng 9 lp l 16.9=144 vng

Chiu di dy ca 9 lp l Chiu di dy lp cui cng

S vng dy lp cui cng

C 4 vng lp th 10

Vy tng s vng dy l 148 vng. C 10 lp

Nu ng p s 1,5 im phn nu cch gii 1,5 im

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TM LC CCH GII

Bi 1: th hm s c th i qua cc im

ta c h phng trnh

Bi 2: Khi lng Cacbon mt 15% nn lng cacbon cn li l 85 ta c phng trnh

Vy khong 1343 nm

Bi 3:

t

Phng trnh tr thnh

Vi

Bi 4:

Xt dy

Vy

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C th gii bi ton trn bng tnh trc tip nh my.bng dng lnh

vi gi tr ban u

Bi 5 Gii phng trnh

iu kin dng mode table nhp hm s d

tm khong nghim t -3 n 10 ta nhn thy nghim ri vo khong (-3,-2) (-2,-1) (-1,0)(0,1).

Nhp phng trnh gii bng shift solve chon im d nghim trong cc khong trn ta c bn

nghim

Bi 6 Gii h tm giao im ta c

Bi 8

Bi 9

Din tch y

ng cao AH:

Bi 10: Mi lp dy c

Lp th 1 c ng knh 17,5cm+0,5cm =18cm R=9cm


Lp th 2 c ng knh



16



22

2

2

M

A

B

CD

H

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Thay ta c

Nh vy s lp l 10; s vng 9 lp l 16.9=144 vng



S vng dy lp cui cng

C 4 vng vy tng s vng dy l 148 vng. C 10 lp

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MTCT THPT trang 1

CHNH THC

IM CA TON BI THI

GIM KHO

(H, tn v ch k)

S PHCH (Do ch tch Hi ng

chm thi ghi )

Bng s

Bng ch

GK1

GK2

Qui nh:

- Hc sinh trnh by vn tt cch gii, cng thc p dng, kt qu tnh ton vo trng lin

k bi ton. Cc kt qu tnh gn ng, nu khng c ch nh c th, c ngm nh chnh xc ti

4 ch s phn thp phn sau du phy.

- Th sinh in k hiu loi my tnh s dng vo sau:

- thi gm 10 trang, 10 bi.

Bi 1. Mt vt chuyn ng thng nhanh dn u i trn hai on ng lin tip bng nhau

99,75 m ln lt trong 4,98s v 3,47s. Tnh gia tc ca vt.

n v tnh: gia tc (m/s2)

Cch gii Kt qu

S GIO DC V O TO

AN GIANG

K THI HC SINH GII CP TNH

GII TON BNG MY TNH CM TAY

NM HC 2012 2013

Mn : VT L Lp: 12 Trung hc ph thng

Thi gian lm bi: 150 pht

(khng k thi gian pht )

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MTCT THPT trang 2

Bi 2. T mt khi cht trong sut, gii hn bi hai mt song song, ngi ta ct theo mt

chm cu to thnh hai thu knh mng c quang tm tng ng l O1 v O2 . Hai thu knh

ny c t ng trc (nh hnh v), hai quang tm cch nhau O1O2 = 30cm. t vt sng

nh AB vung gc vi trc chnh ti A vi O1A = 10 cm, O2A = 20 cm. Khi , nh ca

AB cho bi hai thu knh c v tr trng nhau. Xc nh tiu c ca cc thu knh.

n v tnh: tiu c (cm).

Cch gii Kt qu

O1 O2

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MTCT THPT trang 3

Bi 3. on mch AB gm hai on mch AM v MB mc ni tip. on AM gm in

tr thun R1 = 20 mc ni tip vi cun thun cm c cm khng 20 3 , on mch

MB gm in tr thun R2 mc ni tip vi t in. t vo A, B in p xoay chiu c gi

tr hiu dng v tn s khng i th in p tc thi hai u on mch AM v MB ln

lt l: 80cos(100 )( )6

AMu t V

v 5

80 2 cos(100 )( )12

MBu t V

. Tnh h s cng sut

ca on mch AB.

n v tnh : lch pha (rad)

Cch gii Kt qu

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MTCT THPT trang 4

Bi 4 . Mt vt dao ng iu ha vi bin 9,88 cm. Khi vt c tc 79,80 cm/s th

ln gia tc l 599,6 cm/s2. Tnh tn s dao ng ca vt.

n v tnh: tn s (Hz )

Cch gii Kt qu

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MTCT THPT trang 5

Bi 5. Mt ngun m t ti O pht sng ng hng trong khng gian. M v N l hai im

nm trn cng mt tia xut pht t O. P l trung im ca MN. Gi LM, LP, LN ln lt l

mc cng m ti M, P v N. Bit LM LP = 2(B). Tnh hiu: LM - LN

n v tnh: mc cng m (B)

Cch gii Kt qu

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MTCT THPT trang 6

Bi 6. Khi mc in tr R1 = 1,88 vo hai cc ca ngun in c sut in ng E v

in tr trong r th hiu in th mch ngoi l U1= 0,7073V.

Khi mc thm in tr R2 = 0,80 ni tip vi in tr R1 th hiu in th mch

ngoi l U1= 0,8442V. Tnh sut ng v in tr trong ca ngun in.

n v tnh: sut in ng ( V );in tr ()

Cch gii Kt qu

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MTCT THPT trang 7

Bi 7. Trong gi thc hnh, hc sinh mc ni tip mt qut in 110V- 100W vi in tr

R ri mc vo ngun in xoay chiu 220V-50Hz. Bit khi hot ng ng cng sut th h

s cng sut ca qut in l 0,85. Tnh in tr R qut hot ng ng cng sut. ( coi

qut in gm cun dy ni tip vi t in)

n v tnh: in tr()

Cch gii Kt qu

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MTCT THPT trang 8

Bi 8. Dao ng ca mt vt c khi lng 200g l tng hp ca hai dao ng iu ha

cng phng cng tn s, c phng trnh li ln lt l x1=5 3 cos(15t + 3

) cm;

x2=A2 cos(15t +2) cm. Bit c nng ca vt l 0,05625 J. Dao ng x1 sm pha hn dao

ng tng hp mt gc 2

. Vit phng trnh li x2.

n v tnh: li (cm)

Cch gii Kt qu

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MTCT THPT trang 9

Bi 9. Dng mt dy ng ng knh d = 0,6 mm c lp sn cch in mng qun quanh

li hnh tr c ng knh D = 2 cm lm thnh ng dy. Khi ni hai u ng dy vi

ngun in c hiu in th U = 3,3 V th cm ng t bn trong ng dy l 15,7.10-4

T.

Tnh chiu di l ca ng dy v cng dng in qua ng dy. Bit in tr sut ca

ng l 1,76.10-8

m. , cc vng dy c qun st nhau.

n v tnh: cng dng in (A) ; chiu di (m)

Cch gii Kt qu

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MTCT THPT trang 10

Bi 10. Cho hai bn kim loi phng c di l = 4,5 cm t nm ngang song song, cch

nhau d = 2 cm. Hiu in th gia hai bn l 920V. Mt electron bay theo phng ngang

vo gia hai bn vi vn tc v0 = 4,8.107 m/s. Bit electron ra khi c in trng. B

qua tc dng ca trng trng.

a. Tnh vn tc ca electron ti im bt u ra khi in trng.

b. Tnh lch ca electron khi phng ban u khi ra khi in trng.

n v tnh: vn tc (107m/s); lch (mm).

Cch gii Kt qu

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1

S GD T AN GIANG

HNG DN CHM

K THI HC SINH GII CP TNH 2012 2013


Mn thi: VT L Lp 12 THPT

- Mi bi ton c chm theo thang im 2,0 .

- Phn cch gii: 1,0 im ; kt qu chnh xc ti 4 ch s thp phn: 1,0 im.

- Nu phn cch gii sai hoc thiu m vn c kt qu ng th khng cho im.

- Nu th sinh lm ng mt phn vn cho im phn ng.

- Nu th sinh khng ghi kt qu trung gian m ghi kt qu cui cng chnh xc th vn

cho im ti a.

- Th sinh gii theo cch khc ng th vn cho im ti a.

- im ca bi thi l tng im ca 10 bi ton.

- T chm c th thng nht iu chnh cho im chi tit tng trong phn cch gii.

Bi 1.

Cch gii im s

- Cng thc qung ng : 2 01

2s at v t

0,25

- t1= 4,98s ; s1 = 99,75m 12,4002a + 4,98v0 = 99,75 (1) 0,25

- t2= 8,45s ; s2 = 199,5m 35,70125a + 8,45v0 = 199,5 (2) 0,50

Gii h : a = 2,0630 m/s2

v v0 = 14,8932 m/s

Kt qu: a= 2,0630 m/s 1,0

Bi 2.

Cch gii Kt qu

- Thu knh ra mng c tiu c f1>0.

- Thu knh ra dy c tiu c f2 = - f1< 0 (1)

0,25

- nh A2B2 ca AB cho bi O2 l nh o nm trong khong AO2

' 22

2

200

20

fd

f

(2)

0,25

- V nh A1B1 A2B2 nn nh A1B1 cho bi O1 cng l nh o

' 11

1

100

10

fd

f

(3)

0,25

Theo bi : ' '1 2 30d d cm (4) 0,25

- Thay (1) , (2) v (3) vo (4)

=> f1 =20,0000 cm v f2 = - 20,0000 cm

Kt qu : f1 =20,0000 cm v f2 = - 20,0000 cm 1,0

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2

Bi 3

Cch gii Kt qu

Cch 1

- Chn iu kin thch hp sao cho dng in c biu thc:

i = I0cos(100t) (A)

- Biu thc in p hai u on mch AM so vi dng in

1tan 3LZ

R => 1 ( )

3rad

=> in p uAM sm pha hn dng in : 1 ( )3

rad

0,25

=> ' 80cos(100 )( )3

AMu t V

- Biu thc in p hai u on mch MB so vi dng in

Do in p hai u on mch MB tr pha so vi in p hai u on mch

AM: 7

12

=> ' 80 2 cos(100 )( )4

MBu t V

0,25

- V uAB = uAM + uMB 0,25

Bm my suy ra : = - 0,0891rad 0,25

=> cos = 0,9960

Kt qu : cos = 0,9960 1,0

Cch 2

on AM: 40 2

240

AM

AM

UI

Z

tanAM = 1

3 '3

LAM

Z

R

+ ' ' '7

12 4AM MB MB

tan MB = 22

1C CZ

Z RR

on MB: 2 2

2 2 240 2 2 40MB

MB C C

UZ R Z R R Z

I

Cos = 22

21

21

)()( CL ZZRR

RR

= 0,9960

0,25

0,25

0,25

0,25

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3

Bi 4.

Cch gii im s

2 2

2 4 2 21

a v

A A ; t x=

2 0,25

Suy ra : A2x

2 v

2x a

2 =0

97,6144 x2 6368,04 x - 359520,16 = 0 0,25

Gii: : x1= 101,5170 v x2= -36,2803(loi) 0,25

10,0756x 0,25

- Tn s 1,60362

f

Hz

Kt qu: f= 1,6036 Hz 1,0

Bi 5.

Cch gii im s

LM(B)- LP(B)= lgM

P

I

I

=2 0,25

=> 2

2

210M P

P M

I R

I R

=> RP= 10RM (1)

0,25

Mt khc: 2RP= RM +RN

=> RN = 2RP - RM = 19RM (2) 0,25

LM(B)- LN(B)= lgM

N

I

I

=lg2

2

N

M

R

R

=lg(192)= 2,5575 (B)

0,25

Kt qu: LM(B)- LN(B)= 2,5575 (B) 1,0

Bi 6.

Cch gii Kt qu

Tnh: 111

0,3762U

I AR

0,25

221 2

0,3150U

I AR R

0,25

p dng : 1 1

2 2

E - I r = U (1) 0,3762 0,7073

E - I r = U (2) 0,3150 0,8442

E r

E r

0,25

0,25

Gii h: (1), (2) 1,5488

2,2369

E V

r

Kt qu: 1,5494

2,2386

E V

r

1,0

Nu t h:1 1 1 1

1 2 2 2 1 2

1,5486

( ) ( ) 2,2361

R E U r U R E V

R R E U r U R R r

Cho im.

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4

Bi 7.

Cch gii im s

Ta c : PQ=UQIcos I = 1,0695A 0,25

=> 102,8518Q

Q

UZ

I

0,25

=> RQ = ZQcos= 87,4240

Mt khc : 205,7036U

ZI

0,25

Ta c:

2 2 2

2 2 2

Q Q LQ CQ Q LQ CQZ Z R R Z Z R Z Z

2

231735,4783 Q QR R R

R2 +174,8480R 31735,4783 = 0 0,25

Gii phng trnh ta c

R = 111,0160 v R = -285,8640 ( loi )

Kt qu: R=111,0160 1,0

Cch khc:

Qut in: cos 1,0695q qP U I I A ( 1)

Ur = Uq.cos = 110.0,85= 93,5000 V (2)

2 2 2( ) 3357,75L C q rU U U U 00 (3)

Mt khc: 2 2 2( ) ( )r R L CU U U U U ( 4)

Thay s: 2 187 36300 0R RU U

=> UR =118,7316 V

RU

RI

= 111,0160 .

0,25

0,25

0,25

0,25

Bi 8.

Cch gii im s

- Tnh: 2 2

2

1 2WW 0,05 5

2m A A m cm

m

0,5

- V dao ng tng hp tr pha so vi x1 mt /2

=> x = 5cos(15t -6

) cm

=> x2 = x x1

0,25

- Bm my ta c : A2= 10,0000cm v 2 = - /2 0,25

Kt qu: x2=10,0000 cos(15t - 2

) cm

1,0

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5

Bi 9.

Cch gii im s

- Cm ng trong ng dy

7 7 14 10 4 10N

B I Il d

0,25

- Thay s ta tnh c

I = 0,7496 (A)

- Chiu di L ca si dy: 2

2

L LR

S d

= >L=

.4

2dR (1)

0,25

Mt khc:dl L L

N ld D D

(2)

0,25

T (1) v (2) suy ra 3d

4D

Rl

(3)

0,25

Vi U

RI

= 4,4023

Thay s ta c: l = 0,6754 m

Kt qu: I= 0,7496 (A)

l = 0,6754 m

0,5

0,5

Bi 10.

Cch gii Kt qu

a. Gia tc ye

q.Ua

m .d ;

x = v0.t; 2

tay

2

y ;

vx = v0; vy = ayt.

=>

2

2 2 2

0 y 0

e 0

e .U.lv v v v

m .d.v

0,25

Thay s tnh c: v = 4,8596.107m/s

0,25

b.

2

2

e 0

e U.ly

2m .d.v

0,25

Thay s tnh c; y = 3,5554 mm

0,25

Kt qu : v = 4,8596.107m/s

y = 3,5554 mm

0,5

0,5

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MTCT GDTX trang 1/10

CHNH THC

IM CA TON BI THI GIM KHO

(H, tn v ch k)

S PHCH (Do ch tch Hi ng

chm thi ghi )

Bng s

Bng ch

GK1

GK2

Qui nh:

- Hc sinh trnh by vn tt cch gii, cng thc p dng, kt qu tnh ton vo trng lin

k bi ton. Cc kt qu tnh gn ng, nu khng c ch nh c th, c ngm nh chnh xc ti

4 ch s phn thp phn sau du phy.

- Th sinh in k hiu loi my tnh s dng vo sau:

- thi gm 10 trang, 10 bi.

Bi 1. Mt vt chuyn ng thng nhanh dn u theo phng Ox c phng trnh

x = 5 + 2,32.t + 4,79t2 ( m, s ). K t thi im ban u, hy xc nh:

a) Thi gian cn thit vt i ht qung ng 3m.

b) Vn tc ca vt sau thi gian 15 giy.

n v tnh: thi gian (s); vn tc(m/s)

Cch gii Kt qu

S GIO DC V O TO

AN GIANG

K THI HC SINH GII CP TNH


NM HC 2012 2013

Mn : VT L Lp: 12 GDTX

Thi gian lm bi: 150 pht

(khng k thi gian pht )

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Bi 2. T mt khi cht trong sut, gii hn bi hai mt song song, ngi ta ct theo mt

chm cu to thnh hai thu knh mng c quang tm tng ng l O1 v O2 . Hai thu knh

ny c t ng trc (nh hnh v), hai quang tm cch nhau O1O2 = 30cm. t vt sng

nh AB vung gc vi trc chnh ti A vi O1A = 10 cm, O2A = 20 cm. Khi , nh ca

AB cho bi hai thu knh c v tr trng nhau. Xc nh

tiu c ca cc thu knh.

n v tnh: Tiu c (cm).

Cch gii Kt qu

O1 O2

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Bi 3. on mch AB gm hai on mch AM v MB mc ni tip. on AM gm in tr

thun R1 = 20 mc ni tip vi cun thun cm c cm khng 20 3 , on mch MB

gm in tr thun R2 mc ni tip vi t in. t vo A, B in p xoay chiu c gi tr

hiu dng v tn s khng i th in p tc thi hai u on mch AM v MB ln lt

l: 80cos(100 )( )6

AMu t V

v 5

80 2 cos(100 )( )12

MBu t V

. Tnh h s cng sut ca

on mch AB.

n v tnh : lch pha (rad)

Cch gii Kt qu

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Bi 4. Mt vt dao ng iu ha vi bin 9,88 cm. Khi vt c tc 79,80 cm/s th

ln gia tc l 599,6 cm/s2. Tnh tn s dao ng.

n v tnh: tn s (Hz )

Cch gii Kt qu

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Bi 5. Mc cng m ti im A pha trc ngun m O mt khong d =AO =4,5 m l

LA=70dB. Coi ngun m nh mt ngun pht sng cu v ly cng m chun l

I0 = 10-12

W/m2.

a) Xc nh cng m ti A.

b) Xc nh mc cng m ti B cch O mt khong 2,5m.

n v tnh: cng m( 10 -3

W/m2) ; mc cng m (dB)

Cch gii Kt qu

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Bi 6. Khi mc in tr R1 = 1,88 vo hai cc ca ngun in c sut in ng E v in

tr trong r th hiu in th mch ngoi l U1= 0,7073V.

Khi mc thm in tr R2 = 0,80 ni tip vi in tr R1 th hiu in th mch

ngoi l U1= 0,8442V. Tnh sut ng v in tr trong ca ngun in.

n v tnh: sut in ng ( V );in tr ()

Cch gii Kt qu

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Bi 7. Mt on mch xoay chiu gm cun dy c t cm L, in tr thun r v t in

c in dung C mc ni tip. Cho dng in xoay chu i=2 2 cos100 t (A) chy qua th

in p hiu dng hai u on mch l 12 2 V, hai u cun dy l 20V v gia hai bn t

l 28V. Tnh : L, r v C

n v tnh: r() ; L(mH) ; C(F)

Cch gii Kt qu

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Bi 8. Dao ng ca cht im c khi lng 200g l tng hp ca hai dao ng iu ha

cng phng cng tn s, c phng trnh li ln lt l x1=5 3 cos(15t + 3

) cm; x2=A2

cos(15t +2) cm. Bit c nng ca cht im l 0,05625 J. Dao ng x1 sm pha hn dao

ng tng hp mt gc 2

. Vit phng trnh li x2 .

n v tnh: li (cm)

Cch gii Kt qu

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Bi 9. Hai con lc c di khc nhau 22cm, dao ng cng mt ni, trong cng mt

khong thi gian, ngi ta thy con lc th nht thc hin c 30 dao ng, con lc th hai

thc hin c 36 dao ng. Tnh chiu di ca mi con lc.

n v tnh: chiu di (cm)

Cch gii Kt qu

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Bi 10. Mt vt dao ng iu ho, khi vt c li x1= 3cm th vn tc 1 20 3 /v cm s ,

khi vt c li 2 4 2x cm th vn tc 2 20 2 /v cm s .Tnh bin dao ng ca vt.

n v tnh: bin (cm)

Cch gii Kt qu

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thi MTCT lp 8 Nm 2014 Trang 1

S GIO DC V O TO THI GII TON TRN MY TNH CM TAY

BNH THUN Mn Ton Lp 8 Thi gian: 150 pht (khng k thi gian pht )

THI CHNH THC Ngy thi: 19/01/2014

IM BI THI GIM KHO I GIM KHO II PHCH Bng s Bng ch

Lu : - thi ny c 4 trang, gm 10 bi ton, mi bi 5 im.

- Th sinh lm bi trc tip vo bn thi ny, trnh by li gii tm tt (nu c yu cu)

v ghi kt qu tnh ton vo cc trng lin k bn di bi ton. Cc kt qu gn ng ly 5 ch

s thp phn.

Bi 1: a) Tnh gi tr ng ca A=190120144

b) Tnh gi tr biu thc: 1 1 1

1 2 4 5 ... 2014 20153 6 2016

B

Kt qu

Bi 2: a) Khi vit phn s 97

197 sang dng s thp phn v hn tun hon th chu k gm

bao nhiu ch s?

b) Tm d ca php chia 19012014

cho 2014.

Kt qu

Bi 3: Cho a thc f(x) c bc 6 v h s cao nht l 1. Bit f(5) = 2029; f(6) = 2044;

f(7) = 2063; f(8) = 2086; f(9) = 2113; f(10) = 2144. Tnh f(19) v f(123).

Kt qu

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Bi 4: Hai vi nc cng chy vo mt b nc cn (khng c nc) th y b sau 1

gi 56 pht 25 giy. Nu vi th nht chy trong 24 pht 36 giy v vi th hai chy

trong 32 pht 42 giy th c 24,1% b. Hi vi th hai chy mt mnh th trong bao

lu s y b? (Tnh chnh xc ti giy)

Kt qu

Bi 5: Cho dy s u1 = 1; u2 = 3; un+1 = 2un - un-1 + 2, (n l s nguyn dng).

a) Tnh u20; u40.

b) Tnh u2014.

Li gii tm tt cu b Kt qu

Bi 6: Tm s t nhin nh nht bt u bng ch s 1 sao cho khi chuyn s 1 xung

v tr cui cng th s ny tng ln 3 ln.

Kt qu

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Bi 7: Cho 2 2 21 1 1

Cx y z

a) Tnh C khi x = 2,3(45); y = 123,45; 1,67

8,901z

b) Tnh C bit x, y, z khc 0 tha mn x + y + z = xyz v 1 1 1

1,63205x y z


Bi 8: Cho tm ba hnh ch nht c cc kch thc 1901mm v 2014 mm.

a) Tnh din tch ca tm ba.

b) Gp tm ba theo ng cho. Tnh din tch phn ba chng ln nhau.

Kt qu

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Bi 9: Cho tam gic ABC vung ti A, c BC = 2AB = 19,54 cm

a) Tnh din tch tam gic ABC.

b) Dng ra ngoi tam gic ABC hnh vung BCDE v cc tam gic u ABF, ACG.

Tnh din tch t gic DEFG

Kt qu

Bi 10: Mi ng cho ca mt ng gic li ct ra khi n mt tam gic c din tch

bng a.

a) Tnh din tch ca ng gic theo a.

b) Tnh din tch ca ng gic khi a=19,75cm.

Li gii tm tt cu a Kt qu

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P N S GIO DC V O TO THI GII TON TRN MY TNH CM TAY


P N CHNH THC Ngy thi: 19/01/2014





s thp phn.

Bi 1: a) Tnh gi tr ng ca A=190120144

c) Tnh gi tr biu thc: 1 1 1

1 2 4 5 ... 2014 20153 6 2016

B

Kt qu

A=130650928868009553402019750416 (2,5)

B = - 669,6372607 (2,5)



bao nhiu ch s?


cho 2014.

Kt qu

a) 98 ch s (2,5)

b) D 1711 (2,5)

Bi 3: Cho a thc f(x) c bc 6 v h s cao nht l 1. Bit f(5) = 2029; f(6) = 2044;

f(7) = 2063; f(8) = 2086; f(9) = 2113; f(10) = 2144. Tnh f(19) v f(123).

Kt qu

f(19)=2164763 (2,5) f(123)=2372504250691 (2,5)

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trong 32 pht 42 giy th c 24,1% b. Hi vi th hai chy mt mnh th trong bao

lu s y b? (Tnh chnh xc ti giy)

Kt qu

4 gi 32 pht 49 giy (5)

Bi 5: Cho dy s u1 = 1; u2 = 3; un+1 = 2un - un-1 + 2, (n l s nguyn dng).

a) Tnh u20; u40.

b) Tnh u2014.


b) Ta c un = n2 - n +1 , n l s nguyn dng. (0,5)

Tht vy:

Khi n=1, u1=1 (0,5)

Khi n=k, uk=k2 - k +1(0,5)

Khi n=k+1, uk+1 = 2uk- uk-1 + 2 = k2+k+1= (k+1)

2 - (k+1) +1 (0,5)

a) u20 = 381 (1)

u40 = 1561 (1)

b) u2014 = 4054183

(1)

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Bi 6: Tm s t nhin nh nht bt u bng ch s 1 sao cho khi chuyn s 1 xung

v tr cui cng th s ny tng ln 3 ln.

Kt qu

S cn tm l 142857 (5)

Bi 7: Cho 2 2 2

1 1 1C

x y z

c) Tnh C khi x = 2,3(45); y = 123,45; 1,67

8,901z

d) Tnh C bit x, y, z khc 0 tha mn x + y + z = xyz v 1 1 1

1,63205x y z


2

2 2 2

2

2

1 1 1

1 1 1 1 1 1 1 1 12 1

2. 1

2 0,663587202 1

ax y z

x y z x y z xy yz zx

x y za C

xyz

C a

a) 28,59010724 (2)

Bi 8: Cho tm ba hnh ch nht kch thc 1901mm v 2014 mm.

c) Tnh din tch ca tm ba.

d) Gp tm ba theo ng cho. Tnh din tch phn ba chng ln nhau.

Kt qu

a) 3828614mm2 (2) b) 1809913,641mm

2 (3)

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Bi 9: Cho tam gic ABC vung ti A, c BC = 2AB = 19,54 cm

c) Tnh din tch tam gic ABC.

d) Dng ra ngoi tam gic ABC hnh vung BCDE v cc tam gic u ABF, ACG.

Tnh din tch t gic DEFG

Kt qu

a) 82,66463626 cm2 (2) b) 718,8642769 cm

2 (3)


bng a.




Chng minh c: CE//AB (1)

Tng t ta c BD//AE; AD//BC; AC//DE.

Gi Q l giao im ca CE v BD; t SBQC=x









s thp phn.

Bi 1: a) Tnh ng gi tr ca A = 201320144.

b) 1 1 1

1 2 4 5 ... 2014 20153 6 2016

B

Kt qu



bao nhiu ch s?


cho 2013.

Kt qu

Bi 3: Cho a thc f(x) c bc 6, bit f(5) = -1919; f(6) = -1900; f(7) = -1881;

f(8)=-1862; f(9) = 83645; f(10) = 469056; f(11) = 1 554 835. Tnh f(12) v f(49).

Kt qu

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trong 32 pht 42 giy th c 24,1% b. Hi vi th nht chy mt mnh th trong bao

lu s y b? (Tnh chnh xc ti giy).

Kt qu

Bi 5: Cho dy s u1 = 1; u2 = 3; un+1 = 2un - un-1 + 2, n l s nguyn dng,

a) Tnh u30; u50.

b) Tnh k theo n uk = un+1 . un


Bi 6: Tm s nh nht v s ln nht c dng 1 2 3 4a b c d chia ht cho 2014.

Kt qu

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Bi 7: Cho n n

n n

x xm

x x

a) Tnh m bit 20,14 20,15x , n=3.

b) Khi m =1901 v n = 2014, tnh 2 2

2 2

n n

n n

x xD

x x


Bi 8: Cho hnh ch nht ABCD vi ; 3AB a BC a .

a) Tnh phn din tch ca hnh trn ngoi tip hnh ch nht ABCD ngoi hnh ch nht khi a = 56,4cm.

b) Dng trong hnh ch nht hai na ng trn ng knh AD v CD. Tnh din tch phn chung ca hai na ng trn ny khi a = 64,5cm.

Kt qu

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Bi 9: Thc hin cuc vn ng ton dn xy dng nng thn mi, thn X b tng

on ng (nh hnh v):

Cho kch thc nh sau: AB = 219, 63m; AB = 218,54m; BC = 213,72m;

BC = 210,67m. B rng ca ng l 2,78m v dy ca lp b tng l 15,5cm.

ch ngot B v B, din tch phn (I) thm vo ng bng din tch phn (II) bt i.

Tnh th tch si b tng on ng , bit si chim 57,86% th tch b tng.

Kt qu

Bi 10: Mi ng cho ca mt ng gic li ct ra khi n mt tam gic c din

tch bng a.

a) Tnh din tch ca ng gic theo a. b) Tnh din tch ca ng gic khi a = 19,75cm.


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P N GII TM TT S GIO DC V O TO THI GII TON TRN MY TNH CM TAY


P N CHNH THC Ngy thi: 19/01/2014





s thp phn.

Bi 1: a) Tnh ng gi tr ca A = 201320144.

b) 1 1 1

1 2 4 5 ... 2014 20153 6 2016

B

Kt qu

A=164266458830585844199592870416 (2,5)

B=14,30268772 (2,5)



bao nhiu ch s?


cho 2013.

Kt qu

a) 141 ch s (2,5)

b) D : 1468 (2,5)

Bi 3: Cho a thc f(x) c bc 6, bit f(5) = -1919; f(6) = -1900; f(7) = -1881;

f(8)=-1862; f(9) = 83645; f(10) = 469056; f(11) = 1 554 835. Tnh f(12) v f(49).

Kt qu

f(12) = 3998294 (2,5) f(49)=155 349 099 285 (2,5)

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trong 32 pht 42 giy th c 24,1% b. Hi vi th nht chy mt mnh th trong bao

lu s y b? (Tnh chnh xc ti giy).

Kt qu

3 gi 23 pht 4,23 giy (5)

Bi 5: Cho dy s u1 = 1; u2 = 3; un+1 = 2un - un-1 + 2, n l s nguyn dng,

a) Tnh u30; u50.

b) Tnh k theo n uk = un+1. un


b) Ta c un = n2 - n +1 , n l s nguyn dng. (0,5)

Tht vy:

Khi n=1, u1=1

Khi n=k, uk=k2 - k +1 (0,5)

Khi n=k+1, uk+1 = 2uk- uk-1 + 2 = k2+k+1= (k+1)

2 - (k+1) +1(0,5)

Do un.un+1= n4 + n

2 +1= (n

2 +1)

2 - (n

2 +1) +1= un

2+1 (1)

Vy k=n2 +1 (0,5)

a) u30 = 871 (1)

u50 = 2451 (1)

Bi 6: Tm s nh nht v s ln nht c dng 1 2 3 4a b c d chia ht cho 2014.

Kt qu

Min = 10263344 (2,5)

Max = 1925384 (2,5)

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Bi 7: Cho n n

n n

x xm

x x

c) Tnh m bit 20,14 20,15x , n=3.

d) Khi m =1901 v n = 2014, tnh 2 2

2 2

n n

n n

x xD

x x

Li gii tm tt cu b (2) Kt qu

4028

4028

4028

1901; 2014

11901 (0,5)

1

1900 1902 (0,5)

(0,5)

(0,5)

n n

n n

m n

x xm

x x

x

x

x

x

Suy ra D

a) 0,999866135 (3)

Bi 8: Cho hnh ch nht ABCD vi ; 3AB a BC a .

c) Tnh phn din tch ca hnh trn ngoi tip hnh ch nht ABCD ngoi hnh ch nht khi a = 56,4 cm.

d) Dng trong hnh ch nht hai na ng trn ng knh AD v CD. Tnh din tch phn chung ca hai na ng trn ny khi a = 64,5 cm.

Kt qu

a) S=4483,696231 (2)

b) 921,4361647 (3)

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Bi 9: Thc hin cuc vn ng ton dn xy dng nng thn mi, thn X b tng

on ng (nh hnh v):

Cho kch thc nh sau: AB = 219, 63 m; AB = 218,54 m; BC = 213,72 m;

BC = 210,67 m. B rng ca ng l 2,78 m v dy ca lp b tng l 15,5cm.

ch ngot B v B, din tch phn (I) thm vo ng bng din tch phn (II) bt i.

Tnh th tch si b tng on ng bit si chim 57,86% th tch b tng.

Kt qu

107,5261862 (m2)


bng a.




Chng minh c: CE//AB (1)

Tng t ta c BD//AE; AD//BC; AC//DE.

Gi Q l giao im ca CE v BD; t SBQC=x


-------------------HT----------------------

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thi lp 11 THPT Trang 1/4


BNH THUN Mn Ton Lp 11 Trung hc ph thng

Thi gian: 150 pht (khng k thi gian pht )



Lu :

- thi ny c 04 trang. Gm 10 bi ton, mi bi 5 im.

- Th sinh lm bi trc tip vo bn thi ny. Trnh by li gii tm tt (nu c

yu cu) v ghi kt qu tnh ton vo cc trng lin k bi ton. Cc kt qu gn ng

ly 4 ch s thp phn.

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

Bi 1. Cho dy s ( )nu c xc nh bi:

1

2

1 2

122 3 , , 3n n n

uuu u u n n

.

a (3,0 im). Tnh chnh xc 7 8 9 10 29...S u u u u u .

b (2,0 im). Tnh chnh xc 4 4 46 12 16u u u .

Bi 2. Trong mt phng Oxy, cho ng trn 2 2( ) : ( 1) ( 2) 9C x y c tm l I.

a (3,0 im). Tnh gn ng di cnh ca hnh vung ni tip trong ng trn (C);

b (2,0 im). Vit gn ng phng trnh ng thng qua im (4; 5)M v ct (C) ti hai

im A v B sao cho 0110AIB .

Bi 3. Tnh gn ng tng tt c cc s hng ca mt cp s nhn, bit rng s hng u l 1

256, s

hng th hai l 1

512 v s hng cui l

1

1048576.

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Bi 4.

a (3,0 im). Cho 2013.sin 2 2014.sin 2 .sin 27 7

t t d t

, t .

Tm gn ng d v (rad).

Li gii tm tt Kt qu

b (2,0 im). Gii phng trnh: 4 3 221 43 45 54 8 0x x x x .

(Biu din nghim di dng cn thc)

Bi 5.

a (3,0 im). Tnh chnh xc tng ca tt c cc s nguyn dng nh hn 106 v chia ht cho 3.

b (2,0 im). Tm s nguyn dng n nh nht sao cho 3n l mt s c ba ch s u v ba ch

s cui u l ch s 1.

Li gii tm tt Kt qu

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Bi 6. Cho hm s 3 2( ) 3 1f x x x c th (C).

a (3,0 im). Tm gn ng ta giao im ca (C) v ng thng 2y ;

b (2,0 im). Vit gn ng phng trnh tip tuyn d vi (C), bit d to vi trc tung mt gc 50

0

Bi 7. Mt bi thi trc nghim gm 50 cu hi, mi cu hi c 4 p n, trong ch c mt p n

ng. Mi cu tr li ng c 0,2 im, tr li sai b tr 0,05 im. Mt hc sinh tr li ht tt c cc

cu hi v mi cu chn mt p n bt k.

a (3,0 im). Tnh gn ng xc sut hc sinh c 2,5 im;

Li gii tm tt Kt qu

b (2,0 im). Tnh gn ng xc sut hc sinh b im m.

Bi 8. Tnh gn ng tng tt cc nghim thuc on [10;99] ca phng trnh:

4 sin cos3. 1 0

1 cos 2 cos

x x

x x

.

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Bi 9. Tm tt c cc nghim gn ng ca h phng trnh: 2 2

2

2 ( 1)( 1) 6 8 31

x y x y x xxy x x

.

Bi 10. Cho hnh chp S.ABC c 4SA BC , 5SB AC , 6SC AB .

a (3,0 im). Tnh gn ng (, pht, giy) s o ca gc ASB ;

b (2,0 im). Gi M, N ln lt l trung im ca SA v BC. Tnh gn ng (, pht, giy) s

o ca gc gia hai ng thng AN v CM.

-----HT-----

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HNG DN CHM KHI 11 Nm 2014

Bi Ni dung im

1 a. Ta c 29 6 6648829 ( 21) 6648850S S S .

b. 4 4 4 136 12 16 1,318696562.10 13186965616656u u u

3,0

2,0

2 a. Ta c 2 3 2 4,2426a R

b. 2,5953 5,3811 0x y hoc 0,3853 3,4587 0x y

3,0

2,0

3 13 0,0078S 5,0

4

a. Vi 0t , ta c 2

.sin 2013.sin 2014.sin 15,60635181 ( )7 7

d B

Vi 1

4t , ta c

2.cos 2013.cos 2014.cos 68,40711627 ( )

7 7d C

Suy ra tan 0,2243004594 ( )B

DC

, 70,16474737sin

Bd

D .

b. 1 57 7 61

,7 6

1,0

1,0

1,0

2,0

5

a. 166666833333S

b. 3n c tn cng l 11 th n c tn cng l s 1. Th trn my cc s 11, 21, 31,

, 91 ta c s 71 khi ly tha bc ba c tn cng l 11.

Tip tc nh trn, 3n c tn cng l 111 th n phi c tn cng l s 471.

Suy ra n c dng 471n m . Kim tra trc tip trn my theo qui trnh X = X+1:

(X.1000+471)3 ta c m = 223.

Vy 223471n

3,0

0,5

0,5

0,5

0,5

6

a. Ta giao im: (2,8794; 2), (0,6527; 2), ( 0,5321; 2)A B C .

b. 0,8391 6,7344y x , 0,8391 0,9438y x , 0,8391 3,3835y x

0,8391 0,9383y x

3,0

2,0

7

a. Gi x l s cu tr li ng. Khi 50 x l s cu tr li sai.

hc sinh t 2,5 im th 0,2. (50 ).0,05 2,5 0,25 5 20x x x x

Gi A l bin c em hc sinh tr li ng 20 cu.

Ai l bin c em hc sinh tr li ng cu th i. Ta c cc bin c Ai c lp v

1 3( ) , ( )

4 4i iP A P A .

p s:

20 30

20 3

50

1 3( ) 7,654701431.10 0,0077

4 4P A C

.

b.

509

50

0

1 3( ) 0,1637

4 4

k k

k

k

P B C

1,0

1,0

1,0

2,0

8 3060,7428S 5,0

9 (1,2483; 0,5528) , (0,1659; 6,8614) , (0,5995; 2,0687) , ( 2,0137; 2,5171) 5,0

10 a. Ta c

2 2 2 2 2 24 5 6 1cos

2 . 2.4.5 8

SA SB ABASB

SA SB

. Suy ra 082 49'9"ASB

b. 041 48'60" .

3,0

2,0

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BNH THUN Mn Ton Lp 12 Trung hc ph thng

Thi gian: 150 pht (khng k thi gian pht )



Lu :

- thi ny c 4 trang. Gm 10 bi ton, mi bi 5 im.

- Th sinh lm bi trc tip vo bn thi ny. Trnh by li gii tm tt (nu c

yu cu) v ghi kt qu tnh ton vo cc trng lin k bi ton. Cc kt qu gn ng

ly 4 ch s thp phn.

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

Bi 1. Cho hm s 3 2( ) 3 1f x x x c th (C).

a (3,0 im). Tm gn ng ta giao im ca (C) v ng thng 2y ;

b (2,0 im). Vit gn ng phng trnh tip tuyn d vi (C), bit d to vi trc tung mt gc 500.

Bi 2.

a. (3,0 im). Cho 2013.sin 2 2014.sin 2 .sin 27 7

t t d t

, t .

Tm gn ng d v (rad).

Li gii tm tt Kt qu

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b (2,0 im). Gii phng trnh: 4 3 221 43 45 54 8 0x x x x .

(Biu din nghim di dng cn thc)

Bi 3. Trong mt phng Oxy, cho ng trn 2 2( ) : ( 1) ( 2) 9C x y c tm l I.

a (3,0 im). Tnh gn ng di cnh ca hnh vung ni tip trong ng trn (C);

b (2,0 im). Vit gn ng phng trnh ng thng qua im (4; 5)M v ct (C) ti hai

im A v B sao cho 0110AIB .

Bi 4. Cho hm s 2( ) .cos 1f x x x .

a (3,0 im). Tnh gn ng "2014

g

, bit rng 1

( )( )

g xf x

.

b (2,0 im). Tm gn ng cc im cc i thuc khong 5

0;2

ca hm s ( )f x .

Bi 5. Mt bi thi trc nghim gm 50 cu hi, mi cu hi c 4 p n, trong ch c mt p n

ng. Mi cu tr li ng c 0,2 im, tr li sai b tr 0,05 im. Mt hc sinh tr li ht tt c cc

cu hi v mi cu chn mt p n bt k.

a (3,0 im). Tnh gn ng xc sut hc sinh c 3 im;

Li gii tm tt Kt qu

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b (2,0 im). Tnh gn ng xc sut hc sinh b im m.

Bi 6. Cho dy s ( )nu c xc nh bi:

1

2

1 2

122 3 , , 3n n n

uuu u u n n

.

a (3,0 im). Tnh chnh xc 8 9 10 30...S u u u u .

b (2,0 im). Tnh chnh xc 4 4 45 10 15u u u .

Bi 7. Tnh gn ng tng tt cc nghim thuc on [10;100] ca phng trnh:

4 sin cos3. 1 0

1 cos 2 cos

x x

x x

.

Bi 8. Tm tt c cc nghim gn ng ca h phng trnh: 2 2

2 2

2 2

2013 2014 2015

2 . 4052169 3 . 4056196 2016

x y

x y

Bi 9. Cho hnh chp S.ABC c SA SB SC , y ABC c 6AB , 7BC v 8AC . Gc gia hai

mt phng (SAB) v (ABC) bng 700.

a (3,0 im). Tnh gn ng th tch ca khi chp S.ABC;

b (2,0 im). Tnh gn ng th tch ca khi cu sinh bi mt cu ngoi tip hnh chp S.ABC.

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Bi 10.

a (3,0 im). Tm gn ng gi tr ca m ( 1m ) 2

1

sin 1 0

m

xe dx m .

b (2,0 im). Tm s nguyn dng n nh nht sao cho 3n l mt s c ba ch s u v ba ch

s cui u l ch s 1.

Li gii tm tt Kt qu

-----HT-----

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HNG DN CHM KHI 12 Nm 2014

Bi Ni dung im

1

a. Ta giao im: (2,8794; 2), (0,6527; 2), ( 0,5321; 2)A B C .

b. 0,8391 6,7344y x , 0,8391 0,9438y x , 0,8391 3,3835y x

0,8391 0,9383y x

3,0

2,0

2

a. Vi 0t , ta c 2

.sin 2013.sin 2014.sin 15,61989827 ( )7 7

d B

Vi 1

4t , ta c

2.cos 2013.cos 2014.cos 68,40402438 ( )

7 7d C

Suy ra tan 0,2245 ( )B

DC

, 70,1647sin

Bd

D .

b. 1 57 7 61

,7 6

1,0

1,0

1,0

2,0

3 a. Ta c 2 3 2 4,2426a R

b. 2,5953 5,3811 0x y hoc 0,3853 3,4587 0x y

3,0

2,0

4 a. " 1,999970675 2,0000

2014g

b. 0,6533 ( )x A , 3,2923 ( )x B , 6,3616 ( )x C .

3,0

2,0

5

a. Gi x l s cu tr li ng. Khi 50 x l s cu tr li sai.

hc sinh t 3 im th 0,2. (50 ).0,05 3 0,25 5,5 22x x x x

Gi A l bin c em hc sinh tr li ng 22 cu.

Ai l bin c em hc sinh tr li ng cu th i. Ta c cc bin c Ai c lp v

1 3( ) , ( )

4 4i iP A P A .

p s: 3( ) 1,601633056.10 0,0016P A .

b. ( ) 0,1637P B

1,0

1,0

1,0

2,0

6 a. 30 7 2836779 ( 8) 2836787S S S

b. 4 4 4 135 10 15 4,777078156.10 47770781561778u u u

3,0

2,0

7 3160,1666S 5,0

8 0,9379; 0,9659x y 5,0

9 a. . 52,8889S ABCV

b. 522,0995V

3,0

2,0

10

a. 1,2136m

b. 3n c tn cng l 11 th n c tn cng l s 1. Th trn my cc s 11, 21, 31,

, 91 ta c s 71 khi ly tha bc ba c tn cng l 11.

Tip tc nh trn, 3n c tn cng l 111 th n phi c tn cng l s 471.

Suy ra n c dng 471n m . Kim tra trc tip trn my theo qui trnh

X = X+1: (X.1000+471)3 ta c m = 223.

Vy 223471n

3,0

0,5

0,5

0,5

0,5

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1

S GIO DC V O TO

CAO BNG K THI GII TON TRN MY TNH CM TAY

BC TRUNG HC NM HC 2012 - 2013

MN :Ton - LP 9 THCS Thi gian lm bi : 120 pht (khng k thi gian giao )

Ngy thi : 12 thng 12 nm 2012

Ch : - thi ny c : 08 trang

- Hc sinh lm bi trc tip vo bn thi ny.

IM

(ca ton bi thi) CC GIM KHO

(H tn v ch k) S PHCH

(Do Ch tch Hi ng ghi)

Bng s Bng ch Gim kho s 1:

............................................

Gim kho s 2:

............................................

BI LM

Trong cc cu di y, em hy gii n cng thc p dng (nu c), khng

vit quy trnh bm my. Tnh ton ri ghi kt qu vo hnh ch nht vi 4 ch s

thp phn (nu c).

Cu 1: (3 im). Tnh gi tr ca cc biu thc:

a) 2 x 3x x 3 x 3

Ax 2 x 3 x 1 3 x

Vi x = 14 - 6 5

b) Cho sin x = 3

5, tnh B =

2 2

2

2 os 5sinx + 3tan

5 tan 6

c x x

x

c) Tnh C = 3 3 3

0,(2012) 0,0(2012) 0,00(2012)

Li gii IM

.

.

.

.

.

.

.

CHNH THC

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2

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

KQ:

A ..

B ..

C ..

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3

Cu 2: (2 im) Tm a, b, c, d N bit: 1 2003

71 273

21

1

1

a

b

cd

Li gii IM

.

.

.

.

.

.

.

.

.

Cu 3: ( 2 im)

Tm s d trong php chia ( 3x5 - 7x

2+ 5x + 8) : (3x - 2) ri ghi kt qu di

dng phn s, hn s v s thp phn.

Li gii IM

..

.

.

KQ:

r ..

KQ:

a = ..

b = ..

c = ..

d = ..

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4

Cu 4: (2 im)

Cho Q(x) = x4 + mx

3 + nx

2 + px + q .

Bit Q(1) = 5, Q(2) = 7, Q(3) = 9, Q(4) = 11 .

Tnh cc gi tr Q(12), Q(14)

Li gii IM

..

Cu 5: (3 im)

Cho a thc : F(x) = x5 + 2x

4 3x

3 + 4x

2 5x + m 2012.

a) Tm gi tr ca m phng trnh F(x) = 0 c mt nghim l x = 0,8068.

b) Vi gi tr m tm c, so snh m - 2012 vi cc nghim ca phng trnh

2 3 2 5 2 2 5 2 2y y y y

Li gii IM

KQ

.

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5

..

Cu 6: (2 im)

Ba xe ch hng xut pht cng mt lc t ba a im A, B v C v kho hng

D (A, B, C, D thng hng; B nm gia A v C; D nm cng pha vi C so vi im

A). Trong AB = 25km, BC = 10km. Vn tc ca ba xe ln lt l v1 = 58km/h,

v2 = 35km/h, v3 = 20km/h. Bit rng, sau mt thi gian chuyn ng xe xut pht t

B v tr chnh gia xe A v xe C trn on ng CD, tnh thi gian xe B i t

lc xut pht n v tr trn? Khi xe A i c qung ng l bao nhiu?

Li gii IM

........................

................

........................

................

KQ: .

................

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6

.................

Cu 7: (2 im)

Cho ABC vung ti A, v trung tuyn AM v ng cao AH. Bit AC = 7 5 cm,

AM = 11 cm.

a) Tnh gc B, C (lm trn n pht)

b) Tnh di cnh AB, AH.

Li gii IM

............

................

.............

............

................

.............

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7

............

................

........

.

........

............

Cu 8: (4 im)

Cho 4 im A, B, C, D nm trn ng trn tm O bn knh bng 1cm sao cho

AB l ng knh. OC vung gc vi AB v CD i qua trung im ca OB, gi E l

trung im ca OA. Tnh :

a) Din tch tam gic CDE

b) Tnh gc CED (lm trn n pht)

Li gii IM

............................

............

............

............

KQ:.

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8

................

.........

................

........

--------------------------------- Ht ----------------------------------

KQ.

..

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1

HNG DN CHM

Mn: Ton - Lp 9 - THCS nm hc: 2012-2013

(Hng dn chm gm: 04 trang)

Cu Cch gii im

1 a) 2 x 3x x 3 x 3A

x 2 x 3 x 1 3 x

, vi x = 14 - 6 5

A 4,8662

1 im

b) Hc sinh bm my trc tip hoc tnh: Cho sin x = 3

5

B = 2 2

2

2 os 5sinx + 3tan

5 tan 6

c x x

x

=

2

2

2

2

2

3sin2 1 sin 5s inx +

1 sin

5sin6

1 sin

xx

x

x

x

B 2,0102

0,5 im

0,5 im

c) Tnh C = 3 3 3

0,(2012) 0,0(2012) 0,00(2012)

Bm my trc tip hoc bin i ri tnh:

C 1654,9041

1 im

2 Ta c: 2003 1

71273

21

11

291

12

Vy : a = 1; b = 29; c = 1; d = 2

1 im

1 im

3 t P(x) = 3x5 - 7x

2 + 5x + 8, ta c:

(3x5 - 7x

2 + 5x + 8) = (3x - 2).Q(x) + r

S d ca P(x) khi chia cho 3x - 2 l:

r = P(2

3

) = 698

81

= 508 8,6173

81

1 im

1 im

CHNH THC

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2

4 Vi:

Q(1) = 5

Q(2) = 7

Q(3) = 9

Q(4) = 11

Ta c h phng trnh:

1 5 4

8 4 2 16 7 8 4 2 9

27 9 3 81 9 27 9 3 72

64 16 4 256 11 64 16 4 245

m n p q m n p q

m n p m n p

m n p q m n p q

m n p q m n p q

7 3 13 10

26 8 2 76 35

63 15 3 249 48

27

m n p m

m n p q n

m n p p

q

Q(x) = 4 3 24 10 35 48 27x x x x

Q(12) = 7947; Q(14) = 17191

0,5 im

0,5 im

0,5 im

0,5 im

5 a)

b)

V x = 0,8068 l mt nghim ca phng trnh F(x) = 0, nn ta c

m = - x5 - 2x

4 + 3x

3 - 4x

2 + 5x + 2012

=> m 2013,8165

Gii phng trnh 2 3 2 5 2 2 5 2 2y y y y

KX: y 5

2

Nhn c hai v vi 2 , ta c:

2 4 6 2 5 2 4 2 2 5 4

2 5 3 2 5 1 4

2 5 3 3 2 5

y y y y

y y

y y

Suy ra:

2 5 3 0

2 5 9

7

y

y

y

Kt hp vi iu kin xc nh ta c 5

7

2

y

Vy m - 2012 2013,8165 - 2012 = 1,8165 < y

0,5 im

0,5 im

0,5 im

0,5 im

0,5 im

0,5 im

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3

M

C

A B

H

F

E OA B

C

D

H

6

A B C DA' B ' C '

Gi x (gi) l thi gian cn tm

Sau thi gian x (gi):

khong cch gia xe A v xe B:

AB = BB + AB - AA = 35x + 25 - 58x = 25 - 23x

khong cch gia xe B v xe C:

BC = CC + BC - BB = 20x + 10 - 35x = 10 - 15x

V xe B nm chnh gia xe A v xe C nn ta c:

AB = BC

Hay 25 - 23x = 10 - 15x

x = 15

8

= 1,875 (gi) = 1 gi 52 pht 30 giy.

Qung ng xe xut pht t A i c l:

SA = AA = 58.1,875 = 108,75 (km)

0,5 im

0,5 im

0,5 im

0,5 im

7 a)

b)

Ta c : BC = 2 AM = 2 x 11 = 22 (cm)

sinB = 7 5

22

AC

BC

=> B 45021

C 44039

AB = 2

2

(2.11) 7. 5 = 239 15,4596 (cm)

AH = .AC AB

BC

= 7 5. 239

10,9992

22

(cm)

0,5 im

0,5 im

0,5 im

0,5 im

8

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4

a) Ta c : 0

0,5tan 0,5 26 34'

1

OFOCF OCF

OC

K DH AB 026 34'FDH OCF ( so le trong)

M : ODC OCF (do OCD cn ti O)

053 8'ODH ODF FDH

Nn : DH = OD. 0os = 1.cos53 8' 0,6c ODH (cm)

Suy ra : 1 1

. . F .0,6.1 0,32 2

EDFS DH E (cm2)

1 1

. . .1.1 0,52 2

CEFS CO EF (cm2)

Vy : 0,3 0,5 0,8CDE EDF CEFS S S (cm2)

0,5 im

0,5 im

0,5 im

0,5 im

b) Ta c : FH = DH. tan FDH = DH. tanOCF 0,6. 0,5 = 0,3 (cm)

Suy ra : EH = EF + FH 1 + 0,3 = 1,3 (cm)

Nn : 0,6

tan1,3

DHDEH

EH

024 47'DEH

M : 1

tan 20,5

CEO

CEO 063 26'

Vy : CED = DEH CEO 24047+ 63026 = 88013

0,5 im

0,5 im

0,5 im

0,5 im

* Lu chung:

- Th sinh gii cch khc nhng lp lun cht ch, tnh ton chnh xc th vn cho

im ti a bi ;

- im ton bi l tng im cc bi thnh phn, vn gi li 2 s hng thp phn.

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1

BI THI GII TON TRN MY TNH CM TAY CP TNH Lp 12-THPT nm hc : 2012- 2013

Thi gian lm bi: 120' khng k thi gian giao

( thi c 6 trang)

Quy nh: Trong cc cu di y, em hy gii n cng thc p dng (nu cn),

khng vit quy trnh bm my. Tnh ton ri ghi kt qu gn ng vo hnh ch

nht vi 4 ch s thp phn (nu c).

BI LM

1. Cu 1 (2,0 im): Tnh gn ng gi tr ca x: 32

3

2

4

3

5

21

25

2

4

12

x

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Gim kho s 1: .................................................

Gim kho s 2: .................................................

im ..........................

Phch ........................

chnh thc

x ..........................

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2

2. Cu 2 (2,0 im) :

Cho x, y l hai s dng tha mn 2 2

2,317

1,654

x

y

x y

. Tnh gn ng cc gi tr x,y.

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3. Cu 3 ( 2,0 im):

Gii phng trnh: x4 - 32x

3 + 254x

2 + 32x - 63 = 0

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x ............................

y ............................

x1 ........................... x2 ........................... x3 ........................... x4 ...........................

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3

4. Cu 4 (2,0 im):

Gii phng trnh: 2log ( 4x + 15.2

x + 27) + 2 2log

1

4 . 2 3x = 0

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5. Cu 5 (3,0 im):

Cho T = cos2 a + cos

2 2a + cos

2 3a + cos

2 4a + cos

2 5a + cos

2 6a

a, Rt gn biu thc T vi 0180 ( )a k k

b, Tnh gi tr gn ng ca T khi a = 5028

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