GI I TON TRN MY TNH BT- C P THCS M T SNGUYN T C CB N KHIGI I BAI
TON TRN MY TNH BT 1. Nu trn thi khng c ch thch g thm , kt qu bi ton
phi ghi y tt c cc ch s c trn mn hnh MTBT2. Tuyt i khng ghi kt qu
trung gian ra giy nhp v sau ghi li kt qu y vo my trong qu trnh tnh
ton , iu y lm kt qu cui cng b sai lch. Gii quyt iu ny bng cch lu kt
qu trung gian trong b nh ca my , v tnh lin tc cho n khi ra kt qu
cui cng 3. Nu thi ch yu cu ghi kt qu bi ton , trong qu trnh gii HS
c th vn dng tt c cc kin thc m gio vin trang b , tt nhin trong c c
cc kin thc thuc chng trnh THPT . Nu thi yu cu trnh by li gii , phn
trnh by ch cn vn tt , ngn gn nhng quan trng l : phi dng kin thc bc
THCS gii quyt bi ton .4. Khi ra yu cu trnh by qui trnh bm phm , HS
s dng my tnh loi no (CASIO FX 500 MS- CASIO FX 570 MS CASIO FX 570
ES CASIO FX 500 VN PLUS ) phi ghi r qui trnh ny p dng cho dng my
tnh no . 5. Kt qu bi ton khng c ghi di dnga. 10n (kt qu l mt s c
nhiu hn 12 ch s )m phi ghi chnh xc s ch s ca kt qu. 1 GV : LM NGUYN
THAO TR NG THPT L C NGHI P GI I TON TRN MY TNH BT- C P THCS PH N I
: NH NG BI T P RN LUY N KN NG TNH TON Bi 1 A = 2 2(1986 1992).(1986
3972 3).19871983.1985.1988.1989 +
B= 3:(0,2 0,1) (34,06 33,81).4 2 426: :2,5.(0,8 1,2) 6,84:(28,57
25,15) 3 21 1+ + 1+ ]D = 1 12 31 13 41 64 55 71 37 91 25 71 13 521
3 11 11 11 11+ + 11+ + 11 11 + + ] ] 1 1 1 1+ + 1+ + 1 1 + + ];E =
1 11 18 21 17 31 16 41 15 51 14 61 13 72 8++ ++ ++ ++ ++ ++ +3 21 3
4 6 7 921 : 3 . 13 4 5 7 8 115 2 8 8 11 123 . 4 :6 5 13 9 12 15 1 _
_ _+ + 1 , , , 1 ] 1 _ _ _+ + 1 , , , ]FBi 2 Tnh M 4 3 2 2 3 44 4 3
3 2 27x y 6x y 5x y 4xy9x y 7x y 5x y + + khi cho x = 1,432v y =
0,321N = 4 3 2 3 3 3 2 3 43 3 2 2 4 42,3x y z 3,2x y z 5x y z3,7x y
7,3y z 10x z ++ khi cho x = 2, 123, y = 23v z = 11125 3 3 2 23 3 2
2 23x y 4x y 3x y 7xP(x, y)x y x y x y 7- + -=+ + +vi x = 1,23456
;y = 3,121235Bi 3:Tnh A = 2 2 2 2 225(10 x) 42(9 x) 14(8 x) 15(7 x)
4(6 x)100 + + + + khi cho x = 8,69 Tnh B = 3 3 33 3 217(3 x) 18(4
x) 19(5 x)9x 5x 7x 1 + + + +khi cho x = 1,987Tnh C = 9 8 7 6 5 4 31
2 3 4 5 6 7+ + + + + ++ + + + + + , D = 7 6 5 48 7 6 519 17 15 1311
12 13 14+ + ++ + +Tnh E = 8x 5x 2x x16x 13x 10x x+ + ++ + +khi cho
x = 55,5552 GV : LM NGUYN THAO TR NG THPT L C NGHI P GI I TON TRN
MY TNH BT- C P THCS TnhF 291945 831910 2631931 322010 1981945 = + +
+ +Bi 4a) Bit sin = 0,3456 ,tnh A = 3 3 23 3 3cos (1 sin ) tg(cos
sin ).cot g + + + b) Cho sin3x = 0,978(x l gc nhn) . Tnh B = 2 3 45
3 23cos x 7sin x 5tg xcotg x 3cos 3x sin 2x ++ c) Chotang x = 2,74
Tnh C = 2 26 2 25sin x 7cos xtg x 2(cos x sin x)++ +d) Cho cos x
=0,569Tnh D =3 4 33 5 2 2tg x sin x 2cos x:sin x 3cos x cos x sin x
_ _ + + , ,e) Tnh E vi ' 0 ' 030 57 , 30 25 2 2 2 2 2 2E [(1 tg )(1
cot g ) (1 sin )(1 cos )] (1 sin )(1 cos ) + + + f) Cho bit
cotang3x = 0,1234 . Tnh gi tr biu thc sau :+ - -=+3 4 22 3sin x cos
x 1 sin xFtg 2x cotg 2xg) Bit cos2= 0,5678 . Tnh H= 2 3 2 33 3 4sin
(1 cos ) cos (1 sin )(1 tg )(1 cot g ) 1 cos+ + + + + + h) bit
tg=tg350.tg360tg370 tg520.tg530, tnh I = 2 3 2 33 3tg (1 cos ) cot
g (1 sin )(sin cos )(1 sin cos ) + + + + + + i) 2 o ' o ' o ' 2 o
'2 o ' 2 o 'sin 33 12 sin56 48.sin33 12 sin 56 48M2sin 33 12 sin 56
48 1+ -=+ +Bi 5 : Gii cc phng trnh sau a) 2x 7x53 53 64 34 45 15 26
2+ + ++ ++ + b) 2x 5x11 41 42 52 53 63 64 7 + ++ ++ +c)
1541135471134 ).282271() 75 , 1323 .(115)116157124 (3110+
xd)494545)30192423770591 ( :181717 ).618 : 14 :3149 ( 14+ + x e) x
301 22 21 12005 61 92006 31 92007 11 92008 91 22009 33215+ =+ ++ ++
++ ++ ++ ++3 GV : LM NGUYN THAO TR NG THPT L C NGHI P GI I TON TRN
MY TNH BT- C P THCS Bi 6:Tnh kt qu ng ca cc tch sau :a) M =
2222255555 2222266666 b) N = 20032003 20042004 c) P = 13032006
13032007d) Q = 3333355555 3333377777e) R= 3333344444 3333366666f)I
= 2603193126032010 g) J= 2632655555 2632699999 .h) 13579873i)
123456782Bi 7 :Tnh a) A = 2 399x 1 1 20 4:5x 5 5 5x 1 x x y xy+ _+
+ + , vix = 215; y7b) B = 2 2 2 22 2 2 22 x x y y x y.x x xy xy xy
y x xy y _ + + + + + ,Khi x =3 2 v y = 2,25c) C = 3 a a 4(a 2) 2 a
5: 1 (a 0,a 16)16 a a 4 a 4 a 4 _ _+ ++ + + + , , khi a = 2007 2008
2009 + +d) D = 2 x x 1 x 2: 1x x 1 x 1 x x 1 _ _+ + + + , ,khi x =
2009 2010 2011 + +e)221 11 + -x4 xE= 1 1 1 11 +x x4 x 2 x _ , _ _ ,
, Vix = 1,15795836.4 GV : LM NGUYN THAO TR NG THPT L C NGHI P GI I
TON TRN MY TNH BT- C P THCS PH N II : T NG SAI PHN H U H N Nguyn tc
chung gii bi tontng sai phn hu hn : nn xt s hng tng qut v tm cch
phn tch chng cho hp l trit tiu hoc n gin . Trong mt s trng hp , ta
c th dng chc nng tnh tng ca dng my CASIO fx 570 ES, tuy nhin thi
gian x l ca my kh lu .Dng 1 :Cc s hng ca tng c dng phn s , trong mu
l cc tch c qui lut v t l hng s Cch gii chung : Xt s hng tng qut v
tm cch phn tch chng hp l trit tiu Tnh cc tng sau a) 1 1 1 1
1.....1.2 2.3 3.4 998.999 999.1000+ + + + +b) 1 1 1 1 1.....2.4 4.6
6.8 996.998 998.1000+ + + + +c) 1 1 1 1 1......3.5 5.7 7.9 997.999
999.1001+ + + + +d) 1 1 1 1 1.............3.5.7 5.7.9 7.9.11
993.995.997 995.997.999+ + + + +e) 1 1 1 1....1.2.3.4.5 2.3.4.5.6
3.4.5.6.7 996.997.998.999.1000+ + + +Dng 2:Tng ca cc tch c qui lut
Tnh cc tng sau :a)1.2 + 2.3 + 3.4 + 4.5 + 5.6 ++999.1000 b) 1.3 +
3.5 + 5.7 + 7.9 +.+ 99.101c)1.2.3+ 2.3.4 + 3.4.5 +..+ 997.998.999d)
2.4.6 + 4.6.8 + 6.8.10 + .. + 96.98.100 e) 1.4 + 4.7 + 7.10 + + 301
.304 f) 2.4.6.8 + 4.6.8.10+ 6.8.10.12 +.+ 100 . 102 .104 . 106 Phng
php:Bin i s hng tng qut uk v dng uk = ak ak-1V d : Tnh tng S =
1.2.3+ 2.3.4 + 3.4.5 +..+ n(n+1)(n+2) Uk = k(k +1)(k+2) cn ak =
1k(k 1)(k 2)(k 3)4+ + +S =u1 + u2 + + un = (a1 a0) + (a2 a1) +
+(an- an-1) = an a0= 1n(n 1)(n 2)(n 3)4+ + +Dng 3 :Tng bnh phng ,
lp phng cc s t nhin lin tip : Tnh cc tng sau a) A = 12 + 22 + 32 ++
n2 Cng thc : A = n(n 1)(2n 1)6+ +b) B = 13 + 23 + 33 + +n3 Cng thc
: B = 2n(n 1)2+ _ ,Trn y l cc dng bi tp kinh in v tng sai phn hu hn
thng gp . Thc t khi i thi c rt nhiu bi tp rt kh i hi phi t duy c
chiu su v nhy bn thng qua qu trnh rn luyn v tip thu cc kin thc c bn
v tng hu hn 5 GV : LM NGUYN THAO TR NG THPT L C NGHI P GI I TON TRN
MY TNH BT- C P THCS Bi 1: Tnh cc tng sau : (bi tp luyn thi vng quc
gia nm 2010) A = 2 + 12 + 36 + 80 + 150 + .. + 1343100B = 1 + 9 +
25 + . + 4004001C = 1 + 27 + 125 + . + 1030301D = 12 + 42 + 72 + +
87025E = 22 + 52 + 82 + . + 355216Hng dn A : Tng ca bnh phng v lp
phng cc s t nhin lin tipB :Tng bnh phng ca cc s t nhin l lin tiptnh
bi cng thc 34n n3C :Tng lp phng ca cc s t nhin l lin tiptnh bi cng
thc n2(2n2 1)D : S hng tng qut dng un = (3 n- 2)2, dng cng thcD =
2n(6n 3n 1)2 E: S hng tng qut dng un = (3 n- 1)2, dng cng thc E =
2n(6n 3n 1)2+ Bi 2 : Tnh tng 2 2 2 2 2 21 1 1 1 1 1B= 1 1 ... 11 2
2 3 2009 2010+ + + + + + + + +y l bi tp thng xuyn c rn luyn trong
cc nm va qua thng s cui cng cho n nm hin ti , mt iu may mn l trong
k thi QG ngy 19/3/2010 ca em Nguyn Mnh Cm , bi ny c cho li v ng ti
s .Hng dn : Tnh 21 11k 1 k _+ ,Bi 3:Tnh tng ( thi QG ln 10 )1 1 1
1A= + ...1 3 3 5 5 7 2009 2011+ + ++ + + +Bi 4: thi Casio Tnh Lm ng
nm 2009 Tnh tng (biu din di dng phn s)( ) ( ) ( ) ( ) ( ) ( )2 3 2
3 2 32 2 2 2 2 23 1 3 1 3 1......1 1 1 1 2 2 2 2 2009 2009 2009
2009 111 111+ + + + + + 111+ + + + + + ] ] ]Bi 5:Tnh tng( thi casio
Huyn n dng nm hc 2008-2009)K = 1 1 1.............2 1 1 2 3 2 2 3
2009 2008 2008 2009+ + ++ + +Bi 6:Tnh tng S =2 2 2 2 2 21 1 1 1 1
1............n 1 n 3n 2 n 5n 6 n 7n 12 n 9n 20 n 19n 90+ + + + + ++
+ + + + + + + + + + vi n = 3Bi 7: K hiu na n 1 ]l s tnhin gn nht ca
n. Tnh tngA=1 2 3 ...... 2010 1111+ + + + ] ] ] ]Bi 8::K hiu[ ] xl
phnnguyn ca x (s nguyn ln nht khng vt qu x) . Tnh tngB =1.2.3.4
2.3.4.5 3.4.5.6 ..... 2007.2008.2009.2010 1111+ + + + ] ] ] ]Bi
9::Cho tng Cn = 1.2.3 + 2.3.4 + 3.4.5 ++ n( n+ 1)(n + 2)a) Tnh
C2010b)Chng minh rng 4Cn + 1 l mt s chnh phng 6 GV : LM NGUYN THAO
TR NG THPT L C NGHI P GI I TON TRN MY TNH BT- C P THCS Bi 10 :Cho
hm s f(x) = 220102 x x + ..Tnh S = f(1) + f(2) + .+ f(2010)Bi 11 :
Cho dy s 1k k 1a 19,157a a 0, 25 (2 k 1000) ' + Tnh tng S = 1 2 2 3
999 10001 1 1.......a a a a a a+ + ++ + +( thi ca Thanh ha) Bi 13 :
Tnh tng S = 1.3 + 2.4 + 3.5 + 4.6 + .+ 9998.10000Bi 14:Cho n 1nn 13
SS1 3.S+ vi n N, n 2 Tnh tng S = S1 + S2 + S3 + ..+ S2010 + S2011Bi
15: Cho Sn = 1 -2 +3 4 + 5 6 + .+ (1)n.n Tnh tng S = S2007 + S2008
+ S2009 + S2010 Bi 16 : Cho hnh vung th nht cnh a , ni trung im cc
cnh hnh vung th nht ta c hnh vung h hai , ni trung imhnh vung th
hai ta c hnh vung th 3 .tip tc nh th cho n hnh vung th n a) Lp cng
thc tnh tng Tn = S1 + S2 + S3 + .+Snvi Sn l din tch hnh vung th n
b) Tnh tng S ca 50 hnh vung u tin vi a = 1182010Bi 17 : Cho S = 1
+11 +111 + 1111 +.+ ncs 11111111111......11 4 4 2 4 4 3a) Tnh tng S
theo nb) Tnh S khi n = 5 , n = 9 , n = 12 Bi 18: Chof(x) = 21x 3x 2
+ +Tnh S = f(1) + f(2) + .+f(2010)Bi 19 : Tnh tng 1 + 3 + 9 + 27 +
.+ 14348907Kin thc b sung cho bi 17 v 18 v 19 Hng ng thc an bn = (a
b)(an-1+ an-2b + an-3 b2 +.. + a2bn -3 + abn-2 + bn -1 )Bi 20:Tnh
tng M = 3 5 7 201....1.4 4.9 9.16 10000.10201+ + + +HD : 2 22 2 2
22k 1 (k 1) kk (k 1) k (k 1)+ + + +Bi 21: Tnh tng 1 4 9 9801....1.3
3.5 5.7 197.199+ + + +Bi 22 : Tm s t nhin n bit :a) 1 1 1 1
2999.....1.2 2.3 3.4 n(n 1) 3000+ + + + =+b) 1 1 1 1.....2.4 4.6
6.8 2n(2n 2)+ + + ++= 5022009c) 2 2 2 2 2 2 2 21 1 1 1 1 1 1 11 1 1
.... 1 38,4752 3 3 4 4 5 (n 1) n+ + + + + + + + + + + + =-Bi 23:
Cho Un+1 = Un + d (nN*) tha : 2 54 3U U 29U U 5+ = - =Tnh tng S =
U1 +U2 +U3 +..+ U100 Bi 24: Cho f(1) = 1 , f(m + n) = f(m) + f(n)
+mn (m,n nguyn dng). Tnh f(10) ; f(2010)HD : f(n) = f(n 1 +1) = f(n
-1) + f(1) + n -1 = f(n 1) + n 7 GV : LM NGUYN THAO TR NG THPT L C
NGHI P GI I TON TRN MY TNH BT- C P THCS PH N III :a th c v cc bi
ton va th c I KIN THC CN VN DNG TRONG CC BI TON A THC :nh lBezout :
D trong php chia a thc f(x) cho nh thcx a l f(a) H qu : Nu f(a) =
0, a thcf(x) chia ht cho nh thc x a D trong php chia a thc f(x) cho
(ax + b)l f ba - Nu a thcP(x) = anxn + an-1xn-1 +.+a1x + a0( n N)c
n nghim x1 , x2 xntha thc P(x) phn tch c thnh nhn t : P(x) = a(x
x1)(x x2) .(x xn-1)(x xn) IICC DNG TON V A THC : Bi 1 : Tm m a thc
f(x) =4x4 5x3 + m2 x2 mx 80 chia ht cho x 2 Gii :t g(x) = 4x4 5x3
80 ta c f(x) = g(x) +mx2 mx f(x) (x 2 ) f(2) = 0hayg(2) +4m2 2 x =
0 Ta c g(2) = 56 f(2) = 0khi 4m2 2m = 56 4m2 2x 56 = 0 Gii phng
trnhn m , ta c m1 = 4 v m2 = 3,5 Ngha l hai a thc f1(x) = 4x4 5x3 +
16 x2 8x 80 v f2(x) = 4x4 5x3 + 12,25 x2 +3,5 x 80 u chia ht cho x
2 Bi tp tng t : Cho a thc f(x) = x5 3x4 +5 x3 m2x2 + mx+ 861. Tm m
f(x) (x + 3)KQ: 1 21m 5; m 53= =-Bi 2 : Tm a v b sao cho hai a thc
f(x) = 4x3 3x2 + 2x + 2a + 3bv g(x) = 5x4 4x3 + 3x2 2x 3a+ 2bcng
chia ht cho (x 3) Gii:f(x) , v g(x) cng chia ht cho (x 3) khi v ch
khif(3) = g(3) = 0 t A(x) = 4x3 3x2 + 2xv B(x) = 5x4 4x3 + 3x2 2xTa
c f(x) = A(x) + 2a + 3b g(x)=B(x) 3a +2b f(3) = A(3) + 2a + 3b = 87
+2a + 3b f(3) = 0 2a + 3b = 87 g(3) = B(3) 3a + 2b= 3183a +2b g(3)
= 0 3a +2b= 318 Ta c h phng trnh : 2a 3b 873a 2b 318+ =- - + =-Vo
MODE EQN gi chng trnh gii h phng trnh bc nht hai n ta c nghim ( x =
60 ; y = 69) hay a = 60 , b = 69 .Bi tp tng t : Tm m v n hai a
thcP(x) v Q (x) cng chia ht cho (x +4 ) P(x)= 4x4 3x3 + 2x2 x +2m
3n Q(x) = 5x5 7x4 + 9x3 11x2 + 13x 3m + 2n(m = 4128,8 ; n =
2335,2)Bi 3 :Phn tch a thc sau thnh nhn t : 105x2 + 514x 304 Nu
khng c s h tr ca MTBT th vic phn tch a thc trn thnh nhn t l 1 bi
ton kh Gii:nMODE MODE 2 UNhpa = 105 , b = 514 , c = 304Tm c nghim
ca a thc trn : 1 28 38x, x15 7= =-8 GV : LM NGUYN THAO TR NG THPT L
C NGHI P GI I TON TRN MY TNH BT- C P THCS Vy a thc 105x2 + 514x 304
c phn tch thnh 8 38 8 38105 x x 15.7 x x (15x 8)(7x 38)15 7 15 7 -
+ = - + = - + Bi tp tng t : Phn tch a thc sau thnh nhn t a) 65x2 +
4122x +61093 b) 299 x2 2004x + 3337 c) 156x3 413 x2 504 x+ 1265Bi 4
:Cho a thc x5 + ax4 + bx3 + cx2 + dx + e . Bit f(0) = 1 , f(1) = 2
, f(2) = 3 , f(3) = 2 ; f(4) = 1 .Tnh f(100) Gii :R rng nu ta th
0,1,2,3,4, ch xc nh h s t do , vic cn li l gii h phng trnh bc nht 4
n m my CASIO khng th gii quyt c . Gii bng tay th rt vt v . Bi ton
ny c th gii quyt nh sau : Xt a thc ph k(x) = x2 4x + 1 Ta c : k(0)
= 1 ; k(1) = 2 ; k(2) = 3 ; k(3) = 2 ; k(4) = 1 t g(x) = f(x) k(x)
Ta c : g(0) = f(0) k(0) = 0 g(1) = f(1) k(1) = 0 g(2) = f(2) k(2) =
0 g(3) = f(3) k(3) = 0 g(4) = f(4) k(4) = 0T suy ra 0,1,2,3,4 l
nghim ca g(x) Mt khc g(x) l a thc bc 5 (Cng bc vi f(x) v k(x) l bc
2 m g(x) = f(x) k(x)) v c h s cao nht l l 1 T suy rag(x) phn tch c
thnh nhn t : g(x) = (x 0)(x 1)(x 2)(x 3)(x 4) mg(x) = f(x) k(x)
f(x) = g(x) + k(x) Vy f(x) = x .(x 1)(x 2)(x 3)(x 4)+x2 4x + 1
f(100) = 9034512001 Vn y l lm sao tm c a thc ph k(x) ? Ta gi s k(x)
= ax2 + bx + c v cho gn cho k(x) nhn cc gi tr k(1) = 1k(2) = 3 ,
k(3) = 2 (nhn 3 trong 5 gi tr ca f(x) cho) ta c h phng trnh :a b c
24a 2b c 39a 3b c 2+ + =- + + =-+ + =-
nhp cc h s vo my tm c nghim a = 1 , b = 4 , c = 1 k(x) = x2 4x +
1 . Th tip thy k(0) = 1 v k(4) = 1Vy k(x) = x2 4x + 1 l a thc ph cn
tm . Tt nhin khi th k(0) 1 hoc k(4) 1th buc phi tm cch gii khc .Bi
tp tng t :a) Cho a thc P(x) = x5 +ax4 +bx3 +cx2 +dx + e . Bit P(1)
= 1 ; P(2) = 4 ; P(3) = 9 ; P(4) = 16 ; P(5) = 25 . Tnh cc gi tr ca
P(6) ; P(7) , P(8) , P(9) b) Cho a thc Q(x) = x4 + mx3 + nx2 + px +
q v bit Q(1) = 5 , Q(2) = 7 , Q(3) = 9 Q(4) =11 Tnh cc gi tr Q(10)
, Q(11) Q(12) , Q(13)c) Cho a thcf(x) = x5 + ax4 + bx3 + cx2 + dx +
e Bit f(1) = 1 ; f(2) = 1 ; f(3) = 1 ; f(4) = 5; f(5) = 11 . Hy tnh
f(15) f(16) f(18,25)d) Cho a thc f(x) = 2x5 +ax4 +bx3 +cx2 +dx + e
. Bit f(1) = 1 f(2) = 3 f(3) = 7 f(4) 13 f(5) = 21Tnh f(34,567)Bi
5:9 GV : LM NGUYN THAO TR NG THPT L C NGHI P GI I TON TRN MY TNH
BT- C P THCS Cho P(x) = x5 + ax4 + bx3 + cx2 + dx + 132005 Bit P(1)
= 8 , P(2) = 11 , P(3) = 14 , P(4) = 17 Tnh P(15)Gii : Xt a thc ph
Q(x) = 3x + 5 Ta c Q(1) = 8 ; Q(2) = 11 ; Q(3) = 14; Q(4) = 17 t
k(x) = P(x) Q(x) Ta c k(1) = k(2) = k(3) = k(4) = 0 hayk(x) c 4
nghim l 1 , 2 , 3 , 4 .Li bnh :Ti y , lm nh bi 5 th trt lt bi
vk(x)phi l a thc bc 5 m ta mi ch tm c c 4 nghim !!. Bi ton ny qu
hay !a thc k(x) phi c h s cao nhtl h s cao nht ca f(x) nn k(x) c
phn tch thnh nhn t nh sauk(x) = (x + J) (x 1)(x 2) (x 3) (x 4) . Vn
cn li l tm s J nh th no?Tip tc : Vk(x) = P(x) Q(x) P(x) = k(x) +
Q(x) Hay P(x) = (x + J) (x 1)(x 2) (x 3) (x 4) + 3x + 5 H s t do ca
P(x) l J.(1)(2).(3).(4) + 5 = 132005 hay 24J = 132000 J = 132000:24
= 5500 Vy P(x) = (x + 5500)(x 1) (x 2) (x 3) (x 4) + 3x + 5 P(15) =
132492410Bi tp tng t :Cho a thc f(x) = 2x5 + ax4 + bx3 + cx2 + dx+
115197 Bit f(1) = 1 , f(2) = 1, f(3) = 3 , f(4) = 5. Tnh f(12) (KQ
: 38206101)Bi 6:Chof(2x 3) = x3 + 3x2 4x + 5 a) Xc nh f(x) b) Tnh
f(2,33) Gii: a) t t = 2x 3 t 3x2+=
f(t) = 3 2t 3 t 3 t 33 4 53 3 3 + + + + - + f(x) 3 2x 3 x 3 x 33
4 52 2 2 + + + = + - + b)f(2,33) Qui trnh n phm :3 2( 2.33 3)
2shift STO A alpha A x 3alpha A x 4alpha A 5 + + - + =KQ :
34,57410463Bi 7Cho a thc P(x) = 9 7 5 31 1 13 82 32x x x x x630 21
30 63 35- + - +a) Tnh f(4) , f(3) , f(2) , f(1) ,f(0) , f(1) , f(2)
,f(3) , f(4) b) Chng minh rng vi mi x Z th P(x) nhn gi tr nguyn
.Gii : a) Cu a tht ra l gi gii cu b .D dng tnh c f(4) = f(3) = f(2)
= f(1) = f(0) = f(1) = f(2) = f(3) = f(4) = 0b) Suy ra4 ,3 , 2 ,1 ,
0 , 1 , 2, 3 , 4 l 9 nghim ca ca P(x) P(x) c phn tch thnh nhn t nh
sau : P(x) = 1630 (x 4)(x 3) (x 2 (x 1)x (x + 1) (x + 2) (x +3)(x +
4 )Vi x Z th (x 4)(x 3) (x 2 (x 1)x (x + 1) (x + 2) (x +3)(x + 4 )
l 9 s nguyn lin tip 10 GV : LM NGUYN THAO TR NG THPT L C NGHI P GI
I TON TRN MY TNH BT- C P THCS Tong c t nhn 1 s chia ht cho 2 , 1 s
chia ht cho 5 1 s chia ht cho 7 v 1 s chia ht cho 9 t A = (x 4)(x
3) (x 2 (x 1)x (x + 1) (x + 2) (x +3)(x + 4 ) VCLN(2,5) = 1 A 10
CLN(7,9) = 1 A 63 CLN(10 ,63) = 1 A 630 1A630 l mt s nguyn hay P(x)
lun nhn gi tr nguyn vi mi x Z(Chuyn v a thc trong phn trn c ti vit
vo nm 2006 trong thi gian qua b sung rt nhiu dng bi tp v a thc rt
hay v kh cao hn rt nhiu c update trong phn tip theo ) Bi tp v a thc
cn c phn lin quan : Phn trnh nghim nguyn Bi vit v phng trnh nghim
nguyn do thy Phm Vn Qung(Cu gio vin THPT Lc Nghip c hc sinh t gii
Casio quc gia cp THPT )trnh by nh sau :Phng trnh nghiem nguyenI .
NGHI M NGUYN C AA TH C Cho a thc f(x) = anxn + an-1xn-1 +.+a1x + a0
(n N) Gi trx0 l nghim ca a thc f(x) f(x0) = 0 Nhn xt :a) (1)nan +
(1)n-1an-1 + .+ (1)a1 + a0 = 0th phng trnh f(x) = 0 c nghim nguyn x
= 1 b) an + an-1 +..+a1 + a0 = 0 th phng trnh f(x) = 0 c nghim
nguyn l x = 1 nh l 1: a thc f(x) c nghim nguyn l x0 th x0 l c s ca
h s t do a0 nh l 2 :Nu f(1) 0 v f(1) 0 v x0 l nghim nguyn ca phng
trnh f(x) = 0 th 0 0f(1) f( 1) va 1 x 1 x +l nhng s nguyn Hai nh l
trn l iu kin cn x0 l nghim nguyn ca phng trnh f(x) = 0 T ta c cc bc
tm nghim nguyn ca phng trnh f(x) = 0 nh sau Trng hp 1 : Nu f(1) 0v
f(1) 0 Bc 1 : Tm cc c s khct 1 (x1; x2; x3 .)ca h s t do a0 Bc 2 :
Kim tra cc xi no tha i if(1) f( 1) va 1 x 1 x + l nhng s nguyn chn
Bc 3 : Tnh cc f(xi) va chn khng nh nghim , nu khng c gi tr xi no
thaf(xi) = 0 th phng trnh v nghim (nguyn) Trng hp 2: Nu f(1) = 0
(hoc f(1) = 0 ) th phn tch f(x) = (x 1).g(x)[f(x) = (x + 1) . h(x)]
v quay v trng hp 1 tm nghim nguyn ca gx) [ hay h(x) ] Li bnh Bi ton
tm nghim nguyn l mt bi ton p ca i s thng ng dng trong vic phn tch a
thc thnh nhn t .II BI T P MINH H A V d 1:Tm nghim nguyn ca phng
trnh2x4 + 3x3 24x2 13x + 12 = 0 Gii :t f(x) = 2x4 + 3x3 24x2 13x +
12Ta c : f(1) = 0 x = 1 l nghim ca phng trnh f(x) = 0 Dnglc Horner
chia a thc f(x) cho x +1 2 3 24 13 121 2 1 25 12 0f(x) = (x +
1)(2x3 + x2 25x + 12) . t g(x) = 2x3 + x2 25x + 12 Ta c g(1) 0 v
g(1) 0 , cc c s khct 1 ca 12 l xi =t 2;t 3;t 4;t 6 ;t 12 11 GV : LM
NGUYN THAO TR NG THPT L C NGHI P GI I TON TRN MY TNH BT- C P THCS
Ch c xi = 2 , 3 4 tha mn iu kin i ig(1) g( 1) va 1 x 1 x + nguyn .
Dng my tnh kim tra ta c g(3) = 0 v g(4) = 0 . Vy phng trnh f(x) = 0
c 3 nghim nguyn l 1 ; 3 ; 4 V d2 :Phn tch a thc f(x) = x5 2x 4 9x3
+ 8x2 22x + 24 thnh nhn t bng cch tm nghim nguynGii Nhn thy f(1) =
0 x = 1 l mt nghim nguyn ca phong trnh f(x) = 0 .Dng s Horner chia
a thc f(x) cho x 1 1 2 9 8 22 241 1 1 10 2 24 0f(x) = (x 1)(x4 x3
10x2 2x 24 ) . th(x) = x4 x3 10x2 2x 24 Ta c h(1) = 36v h(1) = 30
Cc c s khct 1 ca 24 lt 2 ;t 3 ,t 4 ,t 6 ,t8 ,t 12 ;t 24 Cc xi tha i
ih(1) h( 1) va 1 x 1 x + l 2 , 2 , 3 , 4 Dng my kim tra ch cx = 3 v
x = 4l tha h(3) = 0 v h(4) = 0 Dng s Horner chia h(x) cho x + 3 1 1
10 2 243 1 4 2 8 0h(x) = (x + 3) (x3 4x2+2x 8 )Tip tc chia x3
4x2+2x 8 cho x 41 4 2 84 1 0 2 0x3 4x2+2x 8 = (x 4) (x2 + 2) a thc
x2 + 2 khng phn tch c thnh nhn t Vy f(x)= x5 2x 4 9x3 + 8x2 22x +
24 =(x1)(x + 3)(x 4)(x2 + 2) Bi tp :Tm nghim nguyn ca cc phng trnh
sau :a) 2x3 12x2 + 22x 12 = 0b) x5 10x3 20x2 15x 4 = 0 c) x7 3x6 +
5x5 7x4 + 7x3 5x2 + 3x 1 = 0 12 GV : LM NGUYN THAO TR NG THPT L C
NGHI P GI I TON TRN MY TNH BT- C P THCS CC BI T P V A TH C BSUNG
TRONG TH I GIAN G NY (Trch t cc thi cc cp v mt s bi tp do tc gi son
tho v bin tp) Bi 1:a) Cho tam thcf(x) = ax2 + bx + c, tm cc h s
caf(x) bit f(0) = 6; f(5) = 141; f(3) = 93b) Cho tam thc g(x) = ax2
+ bx + c , tm cc h s ca g(x) bit g(3) =59,9 ; g(2) = 32,6 ; g(4)=
75,4Bi 2: a) Cho a thcP(x) = 2x4 +mx3 + nx + pTnh P(2,5) bit P(0) =
17 , P(2) = 37 , P(5) = 1708 b) Cho a thcQ(x) = 5x4 + ax3 + bx2 +cx
+ d bit P(0) = 3 ; P(2) = 67 ; P(2) = 155 ; P(3) = 345 Tnh Q(3,14)
Bi 3:a)Cho a thcf(x) = x5 + ax4 + bx3 + cx2 + dx + e Bit f(1) = 1 ;
f(2) = 1 ; f(3) = 1 ; f(4) = 5; f(5) = 11 . Hy tnh f(15) , f(16) f(
18,25) b) Cho a thc P(x) = x5 +ax4 +bx3 +cx2 +dx + e . Bit P(1) = 1
; P(2) = 4 ; P(3) = 9 ; P(4) = 16 ; P(5) = 25 . Tnh cc gi tr ca
P(6) ; P(7) , P(8) , P(9) c) Cho a thc Q(x) = x4 + mx3 + nx2 + px +
q v bit Q(1) = 5 , Q(2) = 7 , Q(3) = 9 Q(4) =11 Tnh cc gi tr Q(10)
, Q(11) Q(12) , Q(13)d) Cho a thc f(x) = 2x5 +ax4 +bx3 +cx2 +dx + e
. Bit f(1) = 1 f(2) = 3 f(3) = 7 f(4) 13 f(5) = 21Tnh
f(34,567)e)Cho a thc x2 + ax4 + bx3 + cx2 + dx + e Bitf(1) = 0 ,
f(2) = 0 , f(3) = 2 , f(4) = 6 , f(5) = 12 . Tnh f(15)f) Cho a thc
f(x) = x5 + x4 + ax3 + bx2 + cx + d. Bit f(1) =1 , f(2) = 32 , f(3)
= 243 , f(4) = 1024 .Tm f(x) ,tnh f(10) , f(15) , f(20) Bi 4 : Tm m
v n hai a thcP(x) v Q (x) cng chia ht cho (x 3) P(x)= 4x4 3x3 + 2x2
x +2m 3n Q(x) = 5x5 7x4 + 9x3 11x2 + 13x 3m + 2n Bi 5 : Cho a thcv
P(x) = x4 + ax3 + bx2 + cx + dc P(1) = 7 , P(2) = 28 ; P(3) = 63
.TnhP=P(100) P( 96)8+ Bi 6: Cho a thc f(x) = x5 + ax4 + bx3 + cx2 +
dx + eBit f(1) = 2 ; f(2) = 16 ; f(3) = 54 ; f(4) = 128 Tnh gi tr
biu thc M = f (50) f ( 45)665- -Bi 7: Cho a thc P(x) = x3 + ax2 +
bx -1 a) xc nh gi tr hu t a v b x = 7 57 5+l nghim ca a thcb) Vi a
, b tm c , tm cc nghim cn li ca a thc Bi 8: Cho P(x) xc nh vi mi gi
tr nguyn ca x v P(x2 +1) = x4 + 5x2 + 3 a) Tm P(x)b) Tnh P(2008) ,
P(2009) , P(2010)Bi 9: Phn tch a thc sau thnh nhn t : a) x5 + 2x4
2x3 7x2 8x 4 b) 2x5 + 11x4 104x3 299x2 + 1614x 720c) x4 +5x3 19x2
125x -150Bi 10: Khai trin a thc f(x) = (1 + 2x +3x2)15 ta c a thc
a0 + a1x + a2x2 +..+a30x30 Tnh chnh xc gi tr biu thc E = a0 -2a1 +
4a2 8a3+.536870912 a29 + 1073741824a30 Bi 11: Cho a thc 7 5 31 1 7
6f (x) x x x x210 15 30 35 + . Chng minh f(x) nhn gi tr nguyn vi mi
x Z Bi 12: Cho a thc P(x) = x6 + ax5 + bx4 + cx3 + dx2 + ex + fc gi
tr l 3 ; 0 ; 3 ; 12 ; 27 ; 48khi x nhn cc gi tr 1 ;2 ; 3 ; 4 ; 5
;6. Xc nh cc h s ca a thcv tnh P(x) khi x nhn cc gi tr t 11 n 20 13
GV : LM NGUYN THAO TR NG THPT L C NGHI P GI I TON TRN MY TNH BT- C
P THCS Bi 13 : Cho a thc P(x)=ax3 +bx2 +cx+d bit
P(1)=27;P(2)=125;P(3)=343 v P(4)=735.a)Tnh
P(-1);P(6);P(15);P(2006). ( Ly kt qu chnh xc)b)Tm s d ca php chia
P(x) cho 3x 5.Bi 14: Cho a thc g(x)=8x3 -18x2 +x+6a) Tm cc nghim ca
a thc g(x)b) Tm cc h s a, b, c ca a thc bc ba ca f(x)=x3 +ax2
+bx+c, bit rng khi chia a thc g(x) cho a thcf(x) th c a thc d l
r(x)=8x2 +4x+5c) Tnh chnh xc f(2010)Bi 15 : Cho a thc f(x) = ax3 +
bx2 + cx + d Bit f(1) = 27 , f(2) = 125 , f(3) = 343, f(4) = 735 a)
Xc nh cc h s a, b , c , d b) Tnhf(1) ; f(6) ; f(15) v f(2010)c) Tm
s d trong php chia a thc f(x) cho3x 5 ; 5x + 2 ; 7x 1Bi 16:Cho hai
a thcf(x) = x3 + ax2 + bx 5 vg(x) = x2 + 2ax b Tm a v b bit f(3) =
g(2) v f(2) = g(3) Bi 17 : Phn tch a thc thnh nhn ta) 2x5 + 11x4
104x3 299x2 + 1614x 720b) 24x4 + 74x3 + 35 x2 73x 6014 GV : LM
NGUYN THAO TR NG THPT L C NGHI P GI I TON TRN MY TNH BT- C P THCS
PH N IV : HTH CI X NG I . CC PHP BIN I :1- Th no l hthc i xng ? Biu
thc A = F(x,y) c gi l h thc i xng nu F(x,y) = F(y,x) . C ngha l nu
ta thay i vi tr ca x v y th gi tr ca biu thc khng i . VD : Ccbiu
thc A = x2 + y2 4xy , B = x3 + y3 l cc h thc i xng 2- Mt s php bin
i :Cc php bin i trong vic tnh ton gi tr cc h thc i xng da trn c s l
cc hng ng thc ng nh .Nu t S = x1 + x2, P = x1x2, x1 v x2 l nghim ca
phng trnh bc hai t2 St + P = 0 (Nu cn tm x1 v x2 )(iu kin : S2 4P
0) Ta c mt s php bin i thng dng nh sau :a)1 21 2 1 2x x 1 1 Sx x x
x P++ = =b) ( )22 2 21 2 1 2 1 2x x x x 2x x S 2P + = + - = -c) ( )
( )2 221 2 1 2 1 2x x x x 4x x S 4P - = + - = - x1 x2 = 2S 4P -d)2
2 21 2 1 22 1 1 2x x x x S 2Px x x x P+ -+ = =e) ( ) ( )2 2 21 2 1
2 1 2x x x x x x S S 4P - = + - = -f). ( ) ( )3 3 2 2 21 2 1 2 1 1
2 2x x x x x x x x S(S 3P) + = + - + = -g) ( ) ( ) ( ) ( )2 2 2 26
6 3 3 3 3 3 3 2 31 2 1 2 1 2 1 2x x x x x x 2x x S S 3P 2P + = + =
+ - = - - Ngoi ra cn mt s php bin i khc suy ra t cc php i trn,VD
php bin i (g) c suy ra t(b) v (f )B BI TPBi 1 : Cho x1 + x2 =
5,221x1.x2 = 3,52 . Hy tnh chnh xc n 0,0001 a) 3 31 2x x + b) 6 61
2x x +Gii :(trn my Casio fx 570 MS) Ci t chng trnh tnh ton vi 4 ch
s thp phn :MODE MODE MODE MODE Fix1 4 5,221SHIFT STO A ( )
3,52SHIFT STO Ba) Tnh 3 31 2x x +,n tip ALPHA A(ALPHA2x 3ALPHA B
SHIFT STO CKt qu : 197,4522b) Tnh 6 61 2x x +: n tipALPHA C2x
2ALPHA B3x Kt qu :39074,5874Bi 2 : Cho a + b = 5,809v ab =
8,136364.(a > b)Hy tnh a)1 1a b+; 2 21 1a b+ ; 3 31 1a b+b) a2 +
b2;(a b)2 ; a2 b2 ;a3 + b3 ; a3 b3 ;a4 + b4 ;a5 + b5, a5 b5a6 + b6,
a8 + b8 HD : a5 + b5 = (a + b)(a4 a3b + a2b2 ab3 + b4)c) A = 3a
+5a4 3b + 4a3 + b3 4b4d) B = 2 22 23a 5ab 3b4ab 4a b+ ++e) C = 10
10a b +
15 GV : LM NGUYN THAO TR NG THPT L C NGHI P GI I TON TRN MY TNH
BT- C P THCS PH N V : PH NG TRNH - HPH NG TRNH b t ph ng trnh My
tnh Casiofx 500 MS570 MS500ES 570 ESc chc nng gii phng trnhbc hai ,
bc ba , h phng trnh bc nht hai n v h phng trnh bc nht 3 n . My tnh
VinaCal 570 MS cn c chc nng gii h phng trnh bc nht 4 n 4 phng trnh
. My Casio 500 VN PLUS cn c chc nng gii bt phng trnh bc 2 bc 3 - Cc
dng my 570 cn c th dung lnh Solve tm mt nghim gn ng ca phng trnh bc
cao .Tuy nhin c nhiu bi ton v gii phng trnh v h phng trnh , hc sinh
cn nm vng phng php bin i , t n ph .Bi tp :Bi 1: a)Gii phng trnh a b
1 x 1 a b 1 x + - = + - - theo a v b b)cho a = 250204 v b =
260204Bi 2: Cho 0x 1003 2005 1003 2005 = + - - l nghim ca pt x3 +
ax2 +bx + 8 = 0 (a,b Z) Tm a , b v cc nghim cn li Bi 3: Gii phng
trnh (x2 + 2x ) +6x2 + 12x = 2009Bi 3: Tm cc nghim ca ptx4 + 10x3 +
21x2 = 20x + 25Bi 4: Tm nghim khng nguyn ca pt sau (chnh xc n
0,001)22x x 5 3x4 0x x x 5+ -+ + =+ -Bi 5: Gii phng trnh 130307
140307 1 x 1 130307 140307 1 x + + = + - +Bi 6: Gii pt x 178408256
26614 x 1332007 x 178381643 26612 x 1332007 1 + - + + + - + =Bi 7:
Tm nghim nguyn dng (x,y) ca pt x4 x2y + y2 = 81001Bi 8: Gii h pt 2
2x3, (70731)yx y 192807= - =Bi 9: Gii hpt 3 32 2x y 14, 81328x xy y
4,1148 + =- + =Bi 10: Gii hpt x 2y x y 5x y x 2y 23x 5y 11+ -+ =- +
+ =Bi 11 : Gii hpt 2 22 2170 302x 2y x 2y34 15 3x 2y x 2y 10+ = +
--- =+ -Bi 12 : Gii hpt 2 22(x x 1)(y y 1) 1y 35y 012x 1+ + + + =+
+ =-Gii cc phng trnh phn nguyn (Cn dng ti my Casio 500 VN Plus gii
quyt cc bt phng trnh ) Khi nim : K hiu[ ] x c gi l phn nguyn ca x
+[ ] [ ] x x x 1 < ++ [ ] x 1 x x - 0(**)Vi x > 0 ta c [ x ]
x2 < ([ x ] + 1)2 T pt (*) 22[x] 2010[x] 2009 0([x 1] 2010[x]
2009 0 - + + - + > 22[x] 2010[x] 2009 0([x] 2008[x] 2010 0 - + -
+ > 1 [x] 2009[x] 1V [x]>2006 [ x] {1 ; 2007 ; 2008 ; 2009 }T
pt (*) kt hp vi (**) x=2010[x] 2009 -[x] = 1 x = 1(nhn)[x] = 2007 x
= 2007,999253 (nhn)[ x] = 2008 x = 2008,499 689 (nhn)[x] = 2009 x =
2009(nhn ) . Vy pt c 4 nghim Gii bi 15: Gii phng trnh x2 2010 [ ] x
2011 =0 (*)L thuyt phn nguyn : x 1 < [ x] x T pt(*) x2 2010 x
2011 x2 2010 [ ] x 2011 < x2 2010(x 1) 2011(**)Do x2 2010 [ ] x
2011 =0nn (**) 24 21 x 2011 x 2010 2011 0x 4, 97512.10Vx 2010,
000498 x 2010x 1 0- - - - - - > [x] { 1; 2011 ; 2010}(*) x
=2010[x] 2011 +[x ] = 1 x = 1 (nhn)[ x ] = 2010 x = 2010,500187
(nhn)[x ] = 2011 x = 2011 (nhn). Vy pt c 3 nghim Bi 17: Tm cc s t
nhin x sao cho tch cc ch s ca n bng x2 2005 + 11680 ( d b QG nm
2005)Bi 18: Cho a v b l hai s t nhin . Khi chia a2 + b2 cho a + b c
thng l q v d l r Tm cc cp s (a , b) sao cho q2 + r = 2005 ( d b QG
nm 2005)Hai bi 17 v 18 l hai bi s hc , nhng gii n cn ti chc nng gii
bt phng trnh bc 2 ca my tnh Casio 500VN Plus nn c a vo phn Phng
trinh Bt phng trnh 17 GV : LM NGUYN THAO TR NG THPT L C NGHI P GI I
TON TRN MY TNH BT- C P THCS PH N V : SH C I PHP CHIA HT PHP CHIA C
D K thut tm s d trong php chia hai s t nhin bng my tnhCASIO fx
500MSa) Trng hp s chia c t 10 ch s tr li Nguyn tc : S d ca A chia B
bng A B .[phn nguyn ca A:B]VD : Tm s d ca 987654321 : 12345Ghi vo
mn hnh 98765432112345v n=(Kt qu : 80004,40024)a con tr ln mn hnh
biu thc s duthnh du- v nhp tip 80004 S d 4941(Mn hnh biu thc hin th
: 987654321-12345 80004 )b) Trng hp s chia c nhiu hn 10 ch s Bc 1 :
Ct thnh nhm u c 9 ch s (k t bn tri ) v tm s d nh phn (a) Bc 2:Vit
tip sau s d phn cn li (ti a 9 ch s) ri tm s d ln 2 (nu cn na th tnh
lin tip nh trn) VD : Tm s d trong php chia 98765432188766: 12345Bc
1: tm s d ca 987654321 : 12345( c d l 4941 )Bc 2:Tm s d ca
494188766: 12345(lm nh phn a)c s d l 6071 Kt lun : D trong php chia
98765432188766: 12345l 6071 II TM CLN , BCNN CA HAI HAYNHIU S 1)
Thut gii : Do my c ci sn chng trnh rt gn phn s nn ta c thut ton tm
CLN , BCNN ca hai s A , B nh sau :A aB b= (ti gin) CLN (A , B) = A
: av BCNN(A,B) = A .b 2) Bi tp: Bi 1:Tm CLN v BCNN ca hai
s1545489742 v 3828909209Nhp : 1545489742b/ ca3828909209Kt qu :
20425059CLN(1545489742 ; 3828909209) = 1545489742 : 2042 =756851
BCNN(1545489742 ; 3828909209) = 1545489742 . 5059 Vi php tnh ny s
xy ra s c trn mn hnh . Ta lm nh sau : Nhp 15454897425059 Kq
:7,818632605. 1012(Kt qu ny cho bit 9 s u v kt qu c 13 ch s) Ghi kt
qu 781863260 (v s 5 cha chc chnh xc) . a con tr ln mn hnh biu thc
xa s 15 (Hin th : 454897425059 ) kq: 2,301325048 . 1011(Mc tiu l tm
chnh xc ca nhng ch s cui ) Ghi tip kt qu 7818632604(v s 8 cha chc
chnh xc) Xa tip s 4 u tin trn mn hnh biu thc(Hin th : 54897425059 )
kq : 2,777260478. 1010 Ghi tip kt qu 78186326047(v s 8 cha chc chnh
xc) Xa tip s 5 u tin trn mn hnh biu thc(Hin th : 4897425059 ) kq
2477604778 Ghi tip kt qu 7818632604778Vy BCNN(1545489742 ;
3828909209) = 7818632604778Trong trng khi nhp phn s AB m kt qukhng
l mt phn s ti gin do t v mu ca phn s rt gn c nhiu ch s , kt qu trn
mn hnh l s thp phn. tm CLNta dng thut ton EuclidIII . THUT TON
EUCLIDETIM CLN CA HAI S Thut ton Euclide v tim CLN cua hai st nhin
a v b( a > b) Ta c a = b.q + r CLN(a , b) = CLN (b , r) 18 GV :
LM NGUYN THAO TR NG THPT L C NGHI P GI I TON TRN MY TNH BT- C P
THCS Tip tc : b = r.q1 + r1(0 r1 < r) CLN (b , r) = CLN (r , r1
) Qui trnh tip tc cho khi tm c d cui cng bng 0 s ny l c ca s kia Lc
ny CLN ca a v b chnh l s chia cui cngVD : Tm CLN(135,105) bng thut
ton Euclide n :105SHIFTSTO A 135: ALPHA A 1 ALPHA A SHIFT STO B
ALPHA A: ALPHA B 3 ALPHA B SHIFT STO C ALPHA B:ALPHAC (2) .D cui
cng bng 0, s chia cui cng nh trong C l 15 CLN (105,135)=15Trong thc
t , sau khi tm c s d r,thay v my mc thc hin thut ton Euclid nh VD
trn , ta nn quay v cch tm CLN (b ; r) theo cch mc I .Bi tp :Bi 1
:Tm CLN v BCNN ca hai s a v b a) a = 345621440 ; b = 234540b) a =
192919385 v b =204995305 .c) a = 480480090 ; b = 256256048 d) a =
2029745404 ; b = 314873978Bi 2 : Tm CLN (a ,b ,c) vi a = 5 621 630
802, b = 200 350 214v c = 757 481 361IV NG D THC1 . nh ngha Cho s
nguyn p > 0 . Nu hai s nguyn a v b cho cng mt s d khi chia cho p
th ta ni a ng d vi b theo mun p . K hiu a b(mod p) VD : 201 chia 2
d 1; 191 chia 2 d 1 201 191 (mod 2) 2005 chia 1967d 14;14 chia 1967
thng l 0 v d 14 2005 14 (mod 1967 ) (Rt quan trng gii ton)
987654321 4941 (mod 12345)2 Tnh cht:Tnh cht1(mod p)(mod p)b (mod p)
a ma b m nn Tnh cht2: a b (mod p) an bn (mod p) Tnh cht3:a b (modp)
na nb (mod p)Hin nhin a b (mod p), bc(mod p) a c (mod p)Ngoi ra cn
mt s tnh chtkhc nhng cha cn s dng trong lot bi tp nyBitp : Bi 1 :
Tm s d ca php chia 2004376 cho 1975 Bi 2 : Tm ch s hng chc ca s
232005 Bi 3 :Tm hai ch s cui cng ca tng A = 22001 + 22002 + 22003
Gii : Bi ton ny ca v bi ton tm s d trong php chia A cho 100 . Phi
thc hin gii thut sau :A = 22001 (1 + 2 + 22) = 7.2200122001 =
22000.2 = [(225)4]20.2 225 32 (mod 100) (Ch gii :225 = 33554432 ,
khi chia cho 100 d l 32- Tt nhin 32 chia 100 cng d 32) (225)4 324
76 (mod 100)hay 2100 76 (mod 100)(p dng tnh cht1) (2100)4 764 76
(mod 100)(Ch gii : 764 = 33362176 nn 764 76 (mod 100) )19 GV : LM
NGUYN THAO TR NG THPT L C NGHI P GI I TON TRN MY TNH BT- C P THCS
Hay 2400 76 (mod 100) (2400)5 765 76 (mod 100) (Ch gii : 755 =
2535525376) Hay 22000 76(mod100) 2.22000 2 . 76(mod 100) ( tnh
cht3) 22001 152 52 (mod 100) 7. 22001 7. 52 = 364 64 (mod 100). Vy
2 ch s cui cng ca tng A l 64 Bi 4 :Tm s d trong php chia 20032005
cho 2007 :Gii : 20032 16 (mod 2007) (20032)5 = 200310 165 922 (mod
2007) 200320 9222 1123 (mod 2007) 200340 1123 2 733 (mod 2007)
200320 . 200340 1123 . 733 289 (mod 2007) hay 200360 289 (mod 2007)
tng t 2003100 289 . 733 1102 (mod 2007) 2003200 11022 169 (mod2007)
2003800 1694 1627 (mod 2007) 20031000 1627 . 169 4 (mod 2007)
20032000 42 = 16 (mod 2007) Mt khc ta c th tmc20035 983 (mod 2007)
Nn 20032005 16 . 983 = 15728 1679 (mod 7) Bi 5:Tm :a) 3 ch s cui ca
tng B = 32004+32005+32006b) 4 Ch s cuitngC = 52009 + 52010 +
52011c) 4 ch s cui ca 19962010Bi 6: Tm s d khi chia
2009.2010.2011.2012 cho 2012.2013Bi 7: Cho A = 21 + 22 + 23 + .+
22010. Tm s d khi chia A cho 2011 nh l Fermat :Nu p l mt s nguyn t
th ta c : ap a (mod p) c bit nu (a , p) = 1 th ap 1 1 (mod p)Bi 8:
Tm ch s thp phn th 2009 sau du phy ca php chia 1 cho 23.Gii bng my
tnh Casio fx-500MSTnh bng my 123=0.04347826Ghi li 123=0.0434782Tnh
110^7-23434782=14Tnh 1423=0.608695652Ghi li
123=0.043478260869565Tnh 1410^8-2360869565=5Tnh 523=0.217391304Ghi
li 123=0.0434782608695652173913Tnh 510^7-232173913=1n y ta suy ra:
123=0.(0434782608695652173913) (c 22 ch s phn thp phn)Ta li c:
20097 (mod 22) ch s th 2009 ca php chia bng ch s th 7 phn tun hon
ch s cn tm l 2.Li bnh: Bi ton ny trc kia dng my tnh Casio FX 500-
570 MS , ES , vic tm ra chu k s thp phn v hn tun hontn nhiu thi
gian , t khi c my tnh Casio 500 Vn Plus , vic gii bi ton ny ht sc n
gin vdng my ny cho ra chu k ca s thp phn v hn tun hon .20 GV : LM
NGUYN THAO TR NG THPT L C NGHI P GI I TON TRN MY TNH BT- C P THCS V
MT DNG TON KHC Bi 1 : Tm hai s t nhin a v b tha 20103 = a2 b2 Bi2:
Tm hai s t nhin a v b tha 20113 = a2 b2 Bi 3: Cho A = 34562 + 12342
. Tm mt cp s t nhin a , b sao cho 2 2Aa b2= +Bi 4: Cho a = 1232 +
3452 , b = 5672 + 6782 . Tm mt cp s t nhin m, n sao cho ab = m2 +
n2 (4 bi trn c trcht Chuyn cp Tnh Gii ton trn MT Casio ca gv Trng
Cng Cng Trng THCS L Li Di linh Lm ng . ) Bi 5: Tnh tng cc ch s ca
A2 bit A = 999999.98 (2007 ch s 9) Bi 6: Tm cc s nguyn dng m,n,p
bit 2m + 2n + 2p = 557120 Bi 7: Tm s t nhin M nh nht c 12 ch s, bit
rng khi chia M cho cc s 1256; 3568 v 4184 uc s d l 973.Bi 8: Cho
tng S = 30 + 31 + 32 + 33 +.. + 330 Tm ch s tn cng ca S . S c l s
chnh phng hay khng ?Bi 9: Mt tp hp cc s t nhin lin tip bt u t 1 c
vit ln bng. Nu ngi ta xa i mt s th trung bnh cng ca cc s cn li bng
60217.Tm s b xa.Bi 10: Ngi ta bn 2 con tru, 5 con cu mua 13 con ln
th cn tha 1000 ng . em bn 3 con tru , 3 con ln ri mua 9 con cu th
va . Cn nu bn 6 con cu , 8 con ln mua 5 con tru th cn thiu 500 ng .
Hi mi con cu , con tru , con ln gi bao nhiu ?Bi 11: Cho s t nhin
n(1010n2010) sao cho an = 20203+21ncng l s t nhin.a) Khi y an phi
nm trong khong no?b) Chng minh rng an ch c th l mt trong cc dng
sau: an=7k+1 hoc an=7k-1kNc) Tm cc s t nhin n(1010n2010) sao cho an
= 20203+21ncng l s t nhin.Bi 12 : Cho A = 300300 A c tn cng l bao
nhiu ch s 0 ?Hi ch s khc 0 k trc ch s 0 l 2 ch s no ?A c bao nhiu
ch s ?Bi 13 : Tm cc s t nhin n (50 000 n 100 000) sao cho vi mi s
than = 32290 7n +cng l s t nhin Bi 14: Tm s a c 6 ch s khc nhau v
khc 0 , bit rng 2a , 3a , 4a , 5a , 6a cng to thnh t chnh cc ch s
.Bi 15: Tm s t nhin nh nhtm khi em nhn vi 333667 ta c mt s biu din
bng cc ch s 5 Bi 16:Phn tch 7110 + 442221 thnh bnh phng ca mt tng
Bi 17: Tm cc ch s a , b sao cho 917a30b4 chia ht cho 2009 v 4 Bi
18: Tm s t nhin nh nht c 6 ch s bit rng khi chia n cho 15 v 17 c s
d ln lt l 7 v 5Bi 19: Phn tch cc s sau ra tha s nguyn t : 252633033
v8863701824Bi 20: Tm tt c cc cp s nguyn dng (m , n) c 3 ch s tha mn
2 iu kinsau: Hai ch s ca m cng l hai ch s ca n v tr tng ng ; ch s
cn li ca m nh hn ch s ca n ng 1 n v. C hai s m v n u l s chnh phng
Bi 21 :21 GV : LM NGUYN THAO TR NG THPT L C NGHI P GI I TON TRN MY
TNH BT- C P THCS a) Tm a , b bit : 329 1A110513151ab= =+++b) Tm a ,
b bit : B = 1719 113976213151ab=++++c) Tm a , b , c bit : 463 1C
33137254bac= = ++++v bcl phn s ti gin Bi 22: a) Tnh tng cc s chia
ht cho 3 trong khong t 10000 n 100000b) Tnh tng cc s va chia ht cho
3 va chia ht cho 5trong khong t 10000 n 100000c) Tnh tng cc s chia
ht cho 6 trong khong t 10000 n 100000d) Tnh tng cc s chia ht cho 35
trong khong t 10000 n 100000Bi 23: Tm s ch s ca 20082008 Bi 24:S
2,0092009 c bao nhiu ch s , trong bao nhiu ch s phn nguyn ? phn thp
phnMT S BI TP V PHP M (Xem thm v qui tc cng v qui tc nhn trang 160
sch Gii tch lp 12 )Bi 1 : Thy c 3 loi sch 3 Ton, 6 L v 4 Ha . Thy
mun xp cc quyn sch ln cng mt k v cc loi sch cng loi phi gn nhau .
Hi c bao nhiu cch xp?Bi 2 :Lp 9A c 44 hc sinh , chon 3 em lm cn s
lp : 1 LT , 1LPHT , 1LPL . C bao nhiu cch chn ?Bi 3 : Cho cc s
1,2,3,4,5 . a) Lp c bao nhiu s c 4 ch s i mt khc nhau ?b) Lp c
baonhu ch s i mt khc nhau v chia ht cho 2 ?Bi 4Cho cc s
0,1,2,3,4,5c my cch lp :a) Cc s c 4 ch s i mt khc nhau b) Cc s c 4
ch s i mt khc nhau m chia ht cho 2 22 GV : LM NGUYN THAO TR NG THPT
L C NGHI P GI I TON TRN MY TNH BT- C P THCS PH N VI : DY S- CNG TH
C TRUY H I V l do s phm - y khng nh ngha dy s l g , nhng thng qua
cc v d v dy s Fibonaco , hc sinh nm c khi nim v dy s .Dy s v thit
lp cng thc truy hi ca dy t cc s hng u tin v cng thc s hng th n ca
dy s l bi ton kinh in ca gii ton trn my tnh cm tay . 1. Dy
FibonaciCho dy s : 1 ; 1 ; 2 ; 3 ; 5 ; 8 ; 13 ; 21..Nhn xt g v dy s
trn ?S hng th nht ca dy k hiu U1 = 1 S hng th hai ca dy k hiu U2 =
1 Ta thy ngoi hai s hng u tin U1 = 1 v U2 = 1 , s hng th 3 : U3 =
U1 + U2 = 1 + 1 = 2 Tng t : U4 = U2 + U3 = 1 + 2 = 3 .Cng thc Un+1
= Un-1 + Un c gi l cng thc truy hi ca dy s trn .Phng trnh 2 21 1 0
l =l + l - l - = c gi l phng trnh c trng ca dy (phng trnh sai phn
)Phn ny khi hc sinh i thi cp QG mi ni r chi tit v phng trnh c trng
ni chung rt kh .T phng trnh sai phn trn ngi ta tm c cng thc s hng
th n ca dy Fibonaci :n n1 5 1 52 2Un5 + - - = , mt biu thc cha cn
cng knh nhng vi mi n N u cho ra mt gi tr Un tng ng l mt snguynQua
dy Fibonaci , hS phn no hiu c khi nim dy s , s hng ca dy , cng thc
tng qut v s hng th n ca dy s v cng thc truy hi .2. Mt s bi tp v dy
s .Bi 1. Cho dy s( ) ( )n nn5 7 5 7U2 7+ vi n = 0,1,2,3. .a) TnhU0,
U1, U2, U3, U4
b) Lp cng thc truy hi tnh Un+2 theo Un v Un+1 v chng minh cng
thc c) Vit qui trnh n phm lin tc trn my tnh CASIO FX 500MS tnh Un+2
theo Un v Un+1Gii : a) D dng tnh c U0 = 0
, U1= 1 , U2=10, U3= 82, U4= 640b) Gi cng thc tnh Un+2 theo Un+1
v Un l Un+2 = aUn+1 + bUn + c. Ta c :U2 = aU1 + bU0 + c10 = a. 1 +
b . 0 + c U3 = aU2 + b U1 + c 82 = a.10 + b.1+ c U4 = aU3 + bU2 + c
640 = a. 82 + b. 10 + c Ta c h phng trnh sau :a 0b c 1010a b c
8282a 10b c 640+ + + + '+ +
Chn chng trnh gii phng trnh bc nht 3 n v nhp cc h s vo my gii ta
c :(x = 10 , y = 18 , z = 0 )hay a = 10 , b = 18 , c = 0T ta c cng
thc truy hi tnh Un+2 theo Un+1 v Un nh sau :Un+2 = 10Un+1 18Un
Chhng minh cng thc :tan =( )n5 72 7+ bn = ( )n5 72 7 Un = an bn 23
GV : LM NGUYN THAO TR NG THPT L C NGHI P GI I TON TRN MY TNH BT- C
P THCS Un+1 = n n(5+ 7)a (5 7)b ( ) ( )2 2n 2 n nU 5 7 a 5 7 b+ + n
nn n n nn n n nn n n nn 1 n(32 10 7)a (50 10 7)b(50 10 7)a (50 10
7)b 18(a b )10(5 7)a 10(5 7)b 18(a b )10 (5 7)a (5 7)b 18(a b )10U
18U+ + + + 1 + ] (Phn ny c nhiu cch chng minh khc nhau)c ) Vit qui
trnh n phm :C nhiu cch vit qui trnh n phm cho nn nn ty theo hc sinh
s dng dng my no , gio vin hng dn hc sinh cch n phm theo dng my .
Tuy nhin cch bm phm lin tc trn my CASIO fx 570 MS (nu ES th nh phm
Calc) kt hp vi b m l ngn gn v hin i nht .Bi 2: Cho dy s n nn3 5 3
5U 22 2 _ _+ + , ,vi n = 0 ,1 ,2 ,3 . a) Tnh 5 s hng u U0 ,U1 , U2
, U 3 , U4 b) Lp cng thc truy hi tnh Un+1 theo Un v Un1c) Lp qui
trnh n phm lin tc tnh Un+1trn my tnh CASIOBi 3 : Cho dy s Un = 2n
2n5 1 5 12 2 + - + Vi n = 1,2,3.....a) Tnh U1 , U2 ....U5b) Thit lp
cng thc truy hi tnh Un +2theo Un+1 v Un c) Vit qui trnh n phm lin
tc trn my tnh CASIO FX 500MS tnh Un+2 theo Un v UnBi 4: Cho U1 = 2
; U2 = 5 ;Un+1 = 4Un + Un1. Tnh U10 Bi 5: Cho dy s xc nh bi cng thc
2nn 12n4x 5xx 1+++ nN , n 1 a) Bit x1 = 0,25 . Vit qui trnh n phm
lin tc tnh c cc gi tr ca xn b) Tnh x100 Gii : a) Ta c: 2 2n nn 12 2
2 2n n n n4x 5 4x 4 1 1x 4x 1 x 1 x 1 x 1++ + + ++ + + +Khai bo x1
= 0, 25(t ng gn vo vo nhAns)n tip 4+ ( 1 (Ans2x +1 ) v bm lin tc b)
Sau 7 ln bm phm c kt qu x7 = x8 = x9 = 4,057269071 (dy dng ) Kt lun
x100 = 4,057269071Bi 6: Cho dy sxn+1 = 4n4n3 10x 3++ . Tnh x 5;
x100 bit x1 = 1,25Bi 7: a) Cho dy s sau : 1 2 3 nn lan x 2; x 2 2;
x 2 2 2 ....x 2 2 ....... 2 + + + + +1 4 442 4 4 43. Tnh x50b) Cho
dy s :1 2 3 nn lan x 3; x 3 3; x 3 3 3 ....x 3 3 ....... 3 + + + +
+1 4 442 4 4 43. Tnh x10024 GV : LM NGUYN THAO TR NG THPT L C NGHI
P GI I TON TRN MY TNH BT- C P THCS c) Cho dy s1 2 3x 5; x 5 5; x 5
5 5 .... + + + . Vit cng thc tng qut tnh xn+1 . Tnh x10; x50Bi 8:
Cho dy s n22005U n(n N*)n= + . Tm s hng nh nht ca dy .Gii :3 3n2 2n
n 2005 n n 2005 2005U 3 . . 32 2 2 2 4 n n= + + =( Bt ng thc Cauchy
cho 3 s khng m) Un 3200534 . Du = xy ra khi 3 32n n 2005hay n
2005.2 n 4010 15,8872 2 n= = = = ;Vy n c th bng 15 hoc 16 .Kim tra
vi n = 14 , n = 15 , n = 16 , n =17 (V c kh nng dy ang gim ri li
tngln ) Vi n = 14Un 24,22959184Vi n = 15Un 23,91111111Vi n = 16 ,
Un 23,83203125Vi n = 17 , Un 23, 93771626Vy U16 l s hng nh nht cn
tm Bi 9: Cho dy s Un c xc nh bi cng thc {Un} = n n8 5 15 1 8 5 15
110 105 2 5 2 - + - - - + Hy thit lpcng thc truy hi Tnh Un+2 theo
Un+1 v Un . Vit qui trnh n phm lin tc tnh U15 Bi 10:Cho phng trnh
bc hai x2 6x + 2 = 0 . Gi x1 x2 l hai nghim ca phng trnh . t Un = n
n1 2x x + .Hy thit lp cng thc truy hi tnh Un+2 theo Un+1 v UnBi 11:
Cho dy s Un xc nh bi cng thc :1n 1nn 1U 0, 20092010UU1 U--= =- Tnh
U2010 chnh xc vi 13 ch s thp phn Bi 12: Cho dy s Un xc nh bi cng
thc Un = n11.2 2n 82- -. Xc lp cng thc truy hi tnh Un+2 theo Un+1 v
UnBi 13:Cho dy s sp th t U1 ; U2 ; U3 . Un-1 ; Un; Un+1 BitU5 = 588
, U6 = 1084v cng thc truy hi Un+1 = 3Un 2Un-1. TnhU1 ; U2 : U3 ; U4
v U20 Bi 14: Cho dy s Un xc nh bi cng thc truy hiUn+1 = 5Un 2 Un-1
Bit U5 = 407 ; U6 = 1857 . Tnh U1 ; U2 ;U15 ; U16 Bi 15:Cho dy s :(
) ( )n nn9- 11 - 9+ 11U =2 11 vi n = 0; 1;2;3; a. Tnh5 s hngU0;U1;
U2; U3 ; U4 .b. Trnh by cch tm cng thc truy hi Un+2
theo Un+1 v Un . c. Vit quy trnh n phm lin tc tnh Un+2 theo Un+1
v Un . T tnh U5 v U10 ( thi gii Ton trn my tnh cm taycp Quc gia ln
th 10 nm 2010- cp THCS)Bi 16:Cho dy s Un , bit U1 = 1 , U2 = 6v cng
thc truy hi Un+2 = 6Un+1 7Un.Xc nh cng thc tng qut ca s hng th nBi
17: Cho dy s Un c xc nh bi cng thc truy hi : Un+1 = 7Un-1 6Un - 2(n
3, n N ) 25 GV : LM NGUYN THAO TR NG THPT L C NGHI P GI I TON TRN
MY TNH BT- C P THCS Bit U1 = 0 , U2 = 14, U3 = 18a. Lp qui trnh bm
phm tnh Unb, Tm cng thc tng qut ca dy s Un c. Chng minh rng vi mi s
nguyn t p th Up chia ht cho p Kin thc b sung: nh l nh Fermat Nu p l
mt s nguyn t, th vi s nguyn a bt k , ap a s chia ht cho p. Ngha l :
ap a (mod p)26 GV : LM NGUYN THAO TR NG THPT L C NGHI P GI I TON
TRN MY TNH BT- C P THCS PH N VII : TON KINH T Vi khi nim :Khi ta c
tin khng s dng n , ta c th gi ngn hng sinh li . C hai cch tnh li
:Li n :Li c tnh theo t l phn trm trong mt khong thi gian c nh trc
.Li kp:Sau mt n v thi gian (thng , nm) li c gp vo vn v tnh li.Bi
ton 1: Mt ngi gi vo ngn hng a ng sau n thng vi li sut hng thng l r
. Ngi y rt ly ra ton b s tin c gc ln li . Gi T l s tin lnh ra , tnh
T theo a,r,n .(bit rng mi thng khng ly li ra)L lun i n cng thc :+ S
tin ngi y sau 1 thng l a + a.r = a( 1+ r)+ S tinngi y sau 2 thng l
a( 1 + r) + [a ( 1 + r)].r = a (1 + r) ( 1 + r) = a (1 + r)2+ S
tinngi y sau 3 thng la (1 + r)2 + [a (1 + r)2] .r= a(1 + r)2 (1 +
r) = a(1 + r)3 . Sau khi qui np khng hon ton ta c cng thc nT a(1 r)
= +Bi ton 2: Mt ngi hng thnggi tit kim vo ngn hnga ng vi li sut l r
mt thng . Bit rng ngi khng rt li ra . Lp cng thc tnh s tin T ca ngi
nhn c cui thng th n .L lun i n cng thc+Cui thng th nht , s tin
trong ti khon ca ngi l a+ ar = a(1 +r) + Hng thng gi a ng nn u thng
th hai , s tin trong ti khon l a(1 +r) + a =a [(1 + r) +1 ]= 2 2a
a[(1 r) 1] [(1 r) 1]1 r 1 r+ - = + -+ -+ S tin cui thng th hai l :2
2a a[(1 r) 1] [(1 r) 1] rr r + - + + - 2 3a a[(1 r) 1](1 r) [(1 r)
(1 r)]r r= + - + = + - ++ S tin u thng th 3 ca ngi l 3a[(1 r) (1
r)]r+ - ++ a = 3a[(1 r) (1 r) r]r+ - + + = 3a[(1 r) 1]r+ -+ S tin
cui thng th 3 ca ngi l 3 3a a[(1 r) 1] [(1 r) 1].rr r+ - + + -= 3
4a a[(1 r) 1](1 r) [(1 r) (1 r)r r+ - + = + - +. + Tng t , s tin ca
ngi u thng th n l
na[(1 r) 1]r+ -vs tin cui thng th n ca ngi rt ra l T =na[(1 r)
1]r+ - (1+ r) Hay n 1aT [(1 r) (1 r)r+= + - +27 GV : LM NGUYN THAO
TR NG THPT L C NGHI P GI I TON TRN MY TNH BT- C P THCS A . CC BI TP
V TON KINH TBi 1 : Anh Lnh H Xungtrng s c c250 000 000 ng , sau khi
khi trch ra 20% s tin chiu i bn b v lm vic thin , anh gis tin cn li
vo ngn hng vi li sut 0,2 % / thng. D kin 10 nm sau anh rt c vn ln
li cho con gi anhvo i hc . Hi lc ny anh Xung rt ra c bao nhiu tin ?
Bi 2 : Bn gi 1000 USD vi li sut n c nh theo nm . Sau 6 nm , s tin l
1330 USD . Hi li sut tit kim l bao nhiu ?Bi 3 : Gi s ngn hng tnh li
kp . Hy tnh li sut tit kim nu sau 6 nm bn vn nhn c s tin l 1330 USD
( vi vn ban u l 1000 USD)Bi 4 : Bn gi ngn hng 10 triu ng ( VN) vi
li sut hng thng l 0,8% , sau 12 thng bn lnh ra ton b s tin c gc ln
liv em i thnh USD du lch nc ngoi . Hi trong ti bn c c bao nhiu USD
(bit 1 USD = 18 495 VN)Bi 5:Mt ngi hng thng gi vo ngn hng mt s tin
l1000 000ng vi li sut0,8% / thng . Bit rng ngi khng rt tin li ra .
Sau 12 thng ngi y rt c vn ln li v i mua vng . Hi ngi y mua c bao
nhiu ch vng bit gi vng l2134 000 / ch .Bi 6: a. Mt ngi gi tit kim
250.000.000 (ng)loi k hn 3 thng vo ngn hng vi li sut 10,45% mt nm.
Hi sau 10 nm 9 thng , ngi nhn c bao nhiu tin c vn ln li. Bit rng
ngi khng rt li tt c cc nh k trc .b. Nu vi s tin cu a, ngi gi tit
kim theo loik hn 6 thng vi li sut 10,5% mt nm th sau 10 nm 9 thng s
nhn c bao nhiu tin c vn ln li. Bit rng ngi khng rt li tt c cc nh k
trc v nu rt tin trc thi hn th ngn hng tr li sut theo loi khng k hn
l 0,015% mt ngy ( 1 thng tnhbng 30 ngy ).c. Mt ngi hng thng gi tit
kim 10.000.000 (ng) vo ngn hng vi li sut 0,84% mt thng. Hi sau 5 nm
, ngi nhn c bao nhiu tin c vn ln li. Bit rng ngi khng rt li ra.(
thi gii Ton trn my tnh cm taycp Quc gia ln th 10 nm 2010- cp
THCS)Bi 7 : Bn vay 5000 USD t ngn hng mua xe, li sut l 1.2% / thng.
Nu mi thng bn phi mt s tin bng nhau th mi thng phi tr bao nhiu tin
nu bn mun tr trong vng 3 nm ?B . TON V TNG TRNG DN SBi 1 :Dn s x Hu
Lc hin nay l 10000 ngi . Ngi ta d on sau 2 nm na dn s x Hu Lc l
10404 ngi . a) Hi trung bnh mi nm dn s x Hu Lc tng bao nhiu phn trm
?b) Vi t l tngdn s hng nm nh vy , hi sau 10 nm dn s x Hu Lc l bao
nhiu ngi ? Bi2:Dn s 1 nc l 65triu , mc tng dn s l 1,2 % mi nm . Tnh
s dn nc y sau 15 nm Bi 3 : Dn s mt nc n vonm 1986 l 55 triu ngi ,
mc tng dn s l 2,2 % mi nm . . Tnh dn s nc y vo nm 199628 GV : LM
NGUYN THAO TR NG THPT L C NGHI P GI I TON TRN MY TNH BT- C P THCS
PH N VIII : NH NG BI TO N LIN QUANN HM SV TH Bi 1:Trn mt phng ta
Oxy cho 3 im A(2 ; 2) , B(3 ; 5 ) , C(4 ; 3) a) Tnh chu vi , din
tch tam gic ABC b) S o cc gc A;B;Cc) Tm ta trng tm ca tam gic ABCBi
2 : Cho 3 ng thng : (d1) : y = 2x 3 ; (d2) : y = 3x +2 ; (d3) : y =
x + 6i mt ct nhau ti 3 im A, B,C .Gi G l trng tm ca tam gic ABC .
Tnh din tch tam gic GBC .8642-2-5 5 10d3d2d1q x ( ) = -x+6h x ( ) =
-3x+2f x ( ) = 2x-3ACBBi 3 : Cho hai ng thng (d1) y = -2,45x + 5,5v
(d2): y = 3,75x + 5,5 . tnh gc nhn to bi hai ng thng trn.Bi 4:
8642-2-10 -5 5g x ( ) = 0.43x+5.24f x ( ) = 1.2x+3.4Cho hai ng thng
y = 1,2x + 3,4 v y = 0,43x + 5,24 a) Tnh gc nhn to bi hai ng thng
trn b) Gi l gc to bi ng thng d2 v trc Ox . Tnh gi tr biu thc sau :A
= 3 45 42cos 3sintg 2 cotga + ba - + b 29 GV : LM NGUYN THAO TR NG
THPT L C NGHI P GI I TON TRN MY TNH BT- C P THCS Bi 5:( thi gii ton
trn MTBT Quc gia nm 2007)Cho hai hm s 52253+ x y(1)v 535+ x y (2)
.a) V th ca hai hm s trn mt phng ta Oxyb) Tm ta ca giao im) , (A Ay
x Aca hai th ( kt qu di dng phn s hoc hn s )c) Tnh cc gc ca tam gic
ABC , trong B , C th t l giao im ca th hm s (1) v th hm s hai vi
trc honh ( ly nguyn kt qu trn my )d) Vit phng trnh ng thng l phn
gic ca gc BAC ( H s gc ly kt qu vi hai ch s phn thp phn )Bi 6: Cho
Parabol (P) : y = ax2 bx +c . Xc nh a,,b,c Parabol (P) i qua cc
im
13 3 2551 2 199A 2; , B ; , C ;3 4 48 5 15 _ _ _ , , ,Bi 7: Cho
3 ng thng (d1) : 3x + 2y = 14, (d2):x + y = 2 v (d3) : x 6y = 18 i
mt ct nhauCho Gi A l giao im ca d1 v d2 , B l giao d2 v d3v C l
giao im d3v d1 K trung tuyn AM v phn gic trong AD ca tam gic ABC .
Tnh din tch tam gic ADMBi 8:Cho 3 ng thng (d) : -6x + 8y =2 (d) :
7x 3y = 23v (d) : -x 5y = 13i mt ct nhau ti 3 im. Tm ta trc tm ca
tam gic to ra bi 3 im trn (chnh xc vi 5 ch s thp phn)Bi 9 :Cho 3 ng
thng3x 5y = 14; 4x+9y = 3 v 7x 4y = 30 i mt ct nhau ti 3 im H ,I ,
V .Trong im H nm gc phn t th II ,I nm gc phn t th nht v V nm gc phn
th th IV .Tm ta trng tm ca tam gic HIV . Bi 10: Cho ng thng (d) : y
= 4x 8 ct trc Ox v Oy ln lt ti A v B a) Gi al gc to bi ng thng vi
trc Ox , tnh s o ca ab) Phn gic gc AOB ct AB ti I . Tnh IA , IB Bi
11Cho hai ng thng: (1d)3 1 32 2y x+ + 25 1 5( ) :2 2d y x 1) Tnh gc
to bi cc ng thng trn vi trc ox (chnh xc n giy)2) Tm giao im ca hai
ng thng trn (tnh ta giao im chnh xc n 2 ch s sau du phy)Tnh gc nhn
to bi hai ng thng trn (chnh xc n giy)( QG nm 2008)Bi 12 : Cho ba hm
s 8- 27y x (1) , 338y x (2) v18629y x +(3)a) V th ca ba hm s trn mt
phng ta ca Oxyb) Tm ta giao im A(xA, yA) ca hai th ha m s (1) va
(2); giao im B(xB, yB) ca hai th ha m s(2) va(3); giao im C(xC, yC)
ca hai th ha m s(1) va(3) (kt qu di dng phn s hoc hn s).c) Tnh cc
gc ca tam gic ABC (ly nguyn kt qu trn my)d) Vit phng trnh ng thng l
phn gic ca gc BAC (h s gc ly kt qu vi hai ch s phn thp phn) ( thi
tnh TT-Hu 2008)30 GV : LM NGUYN THAO TR NG THPT L C NGHI P cabBACGI
I TON TRN MY TNH BT- C P THCS PH N Ix : HNH H C PH NG V l thuyt ,
ngoi cc cng thc c bn ca cp THCS t sch gio khoa , gii quyt tt cc ton
Hnh hc gii trn my tnh CASIO , gio vin trang b thm mt s cng thc thuc
cp THCS . I . MT S CNG THC TNH DIN TCH TAM GIC Ngoi cng thc S =
a1ah2Ta cn c cc cng thc sau :S =p(p a)(p b)(p b) (p l na chu vi ca
tam gic-cng thc Heron)S = 1 1 1absinC acsinB bcsinA2 2 2 S = abc4R
(R l bn knh ng trn ngoi tip tam gic)S =pr (r l bn knh ng trn ni tip
tam gic)II NH L SIN V COSIN TRONG TAM GICa b c2RsinA sibB sinC (nh
l sin)a2 = b2 + c2 2bc.cosA (nh l cos) b2 = a2 + c2 2ac. cosB c2 =
b2 + a2 2 ba. cosC(Hc sinh suy ra cc h qu t 2 nh l trn trong qu
trnh vn dng vo gii bi tp)III CNG THC TNH DI NG PHN GIC TRONG TAM
GIC Tuy nhin , ng phn gic ca tam gic cng c phng php dgn kin thc cp
THCS chng minh vi mt ng hng khc : AD AB.AC DB.DC = -Chng minh :Trn
tia AD ly im E sao cho AEC ABC =Ta c DA.DE = DB.DCDA.AE = AB.AC
DA.AE DA.DE = AB.AC DB.DC AD (AE DE ) = AB.AC DB .DC AD2 = AB.AC DB
.DC31 GV : LM NGUYN THAO TR NG THPT L C NGHI P AD = 2bcp(p a)b c+GI
I TON TRN MY TNH BT- C P THCS IV :CNG THC TNH DI NG TRUNG TUYNTRONG
TAM GIC K hiu ma , mb , mc ln lt l di cc trung tuyn ng vi cnh BC ,
CA , AB ca tam gic ABC . Ta c cc cng thc sau 2 2 2 2 2 2 2 2 22 2
2a b cb c a a c b a b cm ; m ; m2 4 2 4 2 4+ + += - = - = - chng
mnh cc cng thc trn , dng kin thc lp 10 chng minh rt on gin . Ty
nhin nu dng kin thc cp THCS chng minh tng i kh Cch 1: (Gii s tam
gic ABC nhn v AB < AC)Cn chng minh :2 2 22AB AC BCAM2 4+= -Ta c
AB2 +AC2 = AH2 +BH2 +AH2 +HC2 = 2AH2 + (BM HM)2 + (CM +HM)2 = 2AH2
+ BM2 2BM.HM + HM2 + MC2 + 2MC.HM + HM2 Do MB = MC = BC2Nn AB2 +AC2
= 2AH2 +2 HM2 + 2MB2 = 2( AH2 + HM2 ) + 2BC2 = 2 AM2 + 2BC2 2 2
22AB AC BCAM2 4+= -Cch 2 : Chng minh AM2 = 2 2 2AB AC BC2 4++ AM2 =
AH2 + HM2 = AH2 + BM2 BH2=(AH+BH) (AH BH) + BM2= AB.(AH BH ) +BM2 =
AB( AB BH BH ) +BM2 = AB(AB 2BH) + BM2 = AB2 + 2BC4 2AB.BH(1) + Tng
t AM2 = AC2 + 2BC4 - 2AC . CK (2)+ Cng (1) v (2) 2AM2 = AB2 + AC2
+2BC2 2( AB. BH + AC.CK) AM2 = 2 2 2AB AC BC(AB.BH AC.CK)2 4++ +(cn
chng minhAB.BH+ AC.CK = 2BC2)32 GV : LM NGUYN THAO TR NG THPT L C
NGHI P GI I TON TRN MY TNH BT- C P THCS + K ALBC Ta c : BL + LC =
BCTam gic ABL vung nn cosB = BLAB BL = ABcosBTam gic ACL vung nn
cosC = LCAC LC = AC.cosC BC = ABcosB + ACcosC = AB.BHBM+ ACCKMC m
BM = MC = BC2 BC = AB.BH+AC.CK BC2 AB.BH+AC.CK = BC.2BC BC2 2BI TP:
(Tng hp khng phn loi )Bi 1: Cho ABC ABC theo t s1,3 . Bit ABC c din
tch l 112 m2. Tnh din tch ABCBi 2:Tam gicABC c din tch S= 27,3529
m2 ng dng vitam gicABC c din tch S= 15,38600625 m2.Tnh t s hai ng
caoAHA'H' v ghi di dng phn s ti gin . Bi 3:Cho hnh thang ABCD
(AB//CD) Bit DAC ABC , AB = 4,93cmDC = 7,89 cm . Tnh di ng cho
AC.Bi 4: Cho hnh thang cn c 2 ng cho vung gc vi nhau . y nh di
13,724 cm , cnh bn di 21,867. Tnh din tch S ca hnh thang cho .Gii2
2 22 2 22 2 2AB IA IBAB CD 2ADCD ID IC + + ; + CD = 2 22AD AB 222
2AB CD AB CD AB 2AD ABS h2 2 2 _+ + + _ , , Vi AB = 13,724; AD =
21,867 Bi 5: Cho ABC vung ti A . Bit AC = 12,5543 cmv trung tuyn AI
= 9,7786 cm . Dng ng cao AH . Gi M , N ln lt l trung im ca AH , BH
. Gi K l giao im ca NM v AC . Tnh cc gc : ABC; ACB ; NAK(bng n v o
) v on thng AK (bng cm) Cho ABC vung gc A , Tnh cc gc B , C v ng
cao AH . Bit AB = 4,0312 cm , BC = 8,0151 cm ( 0 0B 59 48'16.53C 30
11'43.47, AH =3,48422708cm) Bi 6:Tam gic ABC c 0B 120 AB = 6,25cm ,
Bc = 12,5cm . ng phn gic ca gc B ct AC ti D .a) Tnh di on thng BD
b) Tnh t s din tch ca cc tam gic ABD v ABC c) Tnh din tch tam gic
ABD [S : a)BD 4,166666667 ;b) 1/3;c)SABD= 11,27637245 ]Bi 7: Cho
hnh ch nht ABCD , qua nh B v ng vung gc vi ng cho AC . Gi E , F, G
th t t l trung im ca cc on thng AH , BH , CD a) Chng minh EFCG l
hnh bnh hnh b) Gc BEG l gc nhn , gc vung hay gc t ?c) Cho bit BH =
17,25 cmBAC = 38040 Tnh din tch hnh ch nht ABCD d) Tnh di ng cho AC
?Bi 8 : Cho ABC vung gc A , Tnh cc gc B , C v ng cao AH . Bit AB =
4,0312 cm , 33 GV : LM NGUYN THAO TR NG THPT L C NGHI P A BC DIGI I
TON TRN MY TNH BT- C P THCS BC = 8,0151 cm Bi 9 :Cho hnh thang vung
ABCD (B C 1v) c AB = 12,35 cm , BC = 10,55cm , 0ADC 57 a) Tnh chu
vi hinh thang ABCD b) Tnh din tch hnh thang ABCD c) Tnh cc gc cn li
ca ADCBi 10:Mt hnh thang cn ABCD c hai ng cho vung gc vi nhau . Cho
bit y nh AB = 13,72 cm , cnh bn AD = 21,87 dm . Hy tnh DC , AC , gc
DBCv din tch hnh thang .Bi 11 :Cho tam gic u ABC c cnh l a . M l mt
im nm trong tam gic . Gi MH1 , MH2 , MH3 l khong cch t im M n cc
cnh ca tam gic .a) Chng minh tng cc khong cch t M n 3 cnh l mt hng
s .b) Cho a = 4,358 cm . Tnh MH1 + MH2 + MH3 Bi 12: Cho ABC , t im
D thuc cnh BC k cc ng thng song song vi cc cnh ca tam giac tao thnh
hai tam gic nh c din tch 6,25 cm2 v 12,4609 cm2 . Tnh din tch
ABC.Bi 13 : Cho hnh vung ABCD v mt im M nm trn cnh BC . Gi N l giao
im ca hai ng thng AM v DC . Cho bit AM = 3,14 cm AN = 7,12 cm .Tnh
din tch hnh vung .Bi 14 Cho t gic ABCD c hai ng cho vung gc vi
nhau. Bit AB = 8, 33 cm , BC = 7, 49 cm , DA = 4, 55 cm . Tnh di on
thng CDBi 15 Cho hnh thang ABCD (AB//CD) c ng cho BD hp vi tia BC
mt gc bng gc DAB . Bit rng AB = a = 12,5 cm . DC = b = 28,5cm a)
Tnh di x ca ng cho BD (chnh xc n 0,01 )b) Tnh t s phn trm gia din
tch ca tam gic ABD (SABD) v din tch tam gic BCD (SBCD) (chnh xc n
0,01) Bi 16 : Cho tam gic ABC vung A , vi AB = a = 14,25 cm , AC =
b = 23,5 cm . AM , AD theo th t l cc ng trung tuyn v phn gic ca tam
gic ABC .a) Tnh di cc on thng BD v CD .b) Tnh din tch tam gic ADM
Bi 17:Cho tam gic ABC c AB = 7,3456 cm , BC = 9,4753 cm v 0ABC 38
37'36" . Gi G l trng tm ca tam gic . Tnh din tch tam gic GBCBi 18:
Cho tam gic ABC , Gi G l giao im 2 trung tuyn AD v CE . Bit rng AD
= 5,8518 cm 0 0ACE 45 53'; DAC 22 33' . Tnh din tch tam gic ABCBi
19 : Cho tam gic ABC c AB = 8,93AC = 9,57BC = 13 , 456.Tnh cc gc ca
tam gic ?Bi 20:Cho tam gic ABC c BC = 17,89 0 ' 0 'B 24 39C 43 42 =
=Tnh din tch v chu vi ca tam gic .Bi 21: Cho ABC c AB = 15,72 AC =
21,81cm BC = 25, 63cm . Trung tuyn AD v phn gic AE . a) Tnh SABC v
s o cc gc A,B,C b) Tnh SADEc) Tnh di ng phn gic AE Bi 22: Cho hnh
ch nht ABCD c AB=20,345 cm v AD=15,567 cm. Gi O l giao im hai ng
cho ca hnh ch nht. K AH vung gc vi DB; ko di AH ct CD E.a.Tnh OH v
AE.b.Tnh din tch t gic OHEC.Bi 23 :Cho tam gic ABC c AB = 10 , AC =
12 , SABC = 30 (vdt) . Phn gic ca ABC v phn gic ca gc ACBct nhau ti
I . Tnh BIC.Bi 24:Cho hnh bnh hnh ABCD c hai ng cho AC v BD ct nhau
ti O v0DOA 150 =. Bit AC = 5,23 cm BD = 6,39 cm . Tnh din tch hnh
bnh hnh ABCD Bi 25: Cho tam gic ABC c AB = 4,53;AC = 7,48, gc A =
73o. a) Tnh cc chiu cao BB v CCgn ng vi 5 ch s thp phn.b)Tnh din
tch ca tam gic ABC gn ng vi 5 ch s thp phn.c) S ogc B(, pht,giy) ca
tam gic ABC.d) Tnh chiu cao AA gn ng vi 5 ch s thp phn.Bi 26: Mt
hnh thoi c din tch l 24 cm2 , hiu di hai ng cho l 2 cm . Tnh ln ca
cc gc 34 GV : LM NGUYN THAO TR NG THPT L C NGHI P GI I TON TRN MY
TNH BT- C P THCS hnh thoi.Bi 27: Cho hnh ch nht ABCD. im M nm trong
hnh ch nht c MA =1930,MB=1945,MC=2009. Tnh MD?Bi 28 :Cho ng gic u
ABCDE c di cnh bng 1, gi I l giao im ca 2 ng cho AD v BE.( ly chnh
xc n 4 ch s thp phn)a) Tnh AD.b) Tnh din tch ABCDEc) Tnh IB, ICBi
29: Tam gic ABC vung A c AB = c = 23,82001 cm; AC = b = 29,1945 cm.
Gi G l trng tm tam gic ABC, A`, B`, C` l hnh chiu ca G xung cc cnh
BC, AC, AB. GiSvS`l din tch tam gic ABC v A`B`C`.Tnh t sS'SvtnhS`Bi
30:Cho lc gic u ABCDEF. Bit di BE = 3,12345 cm. Tnh din tch a gic
BCDEF.Bi 31: Cho hai ng trn (O; R) v (O; R) ct nhau ti A v B . Tnh
din tch phn chung ca hai hnh trnbit R =1996 , R =2010 v d = OO = 70
.Bi 32: Vit cng thc tnh din tch S ca hnh thangbit dai 2 ng cho l1 v
l2 v on thng d ni trung imhai cnh y . (cho php dung cng thc Heron
tnh din tch tam gic)p dng bng s : l1 = 302,1930 m ,l2 = 503 , 2005m
, d = 304,1975 m ( d b Gii ton trn my tinh cm tay cp Quc gia nm
2005)Bi 33: Cho hnh vung ABCD, ly cc im M,N,P,Q sao cho cc tam gic
ABM , BCN,CDP, PAQl cc tam gic u. Bit hnh vung ABCD din tch4 2 3
+cm2, tnh din tch a gic KLMN.Bi 34:Cho tamgic ABC c di cc cnh AB =
13,25AC = 17,875 v BC = 24,375 . I l mt im bt k nm trong tam gic .
Gi M,N,P theo th t l trung im ca IA,IB,IC .Tnh chu vi v din tch tam
gic MNPBi 35: Cho 2 ng trn (O ; 3,6) v (O ; 2,8) ct nhau ti A v B ,
bit OO = 5,22 a) Tnh di dy chung ABb) Tnh cc gc AOB, BO' A, OAO'Bi
36:Cho hai ng trn (O,R ) v (Or) tip xc ngoi ti A . BC l tip tuyn
chung ngoi cahai ng trn trn ,(B(O) v C(O) .Gi O l ng trn tip xc
ngoi vi 2 ng trn (O) v (O) ng thi tip xc vi vi ng thng BC ti K .a-
Tnh BC theo R v r b- Cho R= 9 , r = 4 , tnh rBi 37:Cho hai ng trn
(O ; 3,82) v (O ; 2,68 ) ct nhau ti A v B (O,O nm tri pha vi dy
chungAB)bit di dy chung AB = 4,588. K ng knh AOC v AOD ca hai ng
trn a) Chng minh 3 im C,B,D thng hng v tnh di on CD b) Tnh cc gc ca
tam gic ACD Bi 38:Cho ng trn (O,R) v dy AB ca ng trn . ng knh CD (D
nm trn cung nh AB) i qua trung im I ca AB . Bit R = 4,52 , AB =
6,7574 a)Tnh IC , IDchnh xc n 0,00001b)ChoACD . Tnh gi tr biu thc M
= 3 212 10tg 1 cos(cot g ).(tg )+ Bi 39:Cho ng trn tm (O,4 ) A l 1
im nm trn ng trn v AA l mt dy vung gc vi ng knhBC .Cho 0BCA 31 36'
, tnh di dy AABi 40: Tam gic ABC c cnh AC = b = 3,85 cm ; AB = c =
3,25cm v ng cao AH = h = 2,75 cma) Tnh cc gc A , B ,C v cnh BC ca
tam gic .b) Tnh di ca trung tuyn AM ( M thuc BC)c) Tnh din tch tam
gic AHM .35 GV : LM NGUYN THAO TR NG THPT L C NGHI P GI I TON TRN
MY TNH BT- C P THCS Bi 41: Cho ng trn ng knh AB = 2R, M v N l hai
im nm trn ng trn sao cho: AM MN NB = =. Gi H l hnh chiu ca N trn AB
v P l giao im ca AM vi HN. Cho R = 6,25 cm.a) Tnh: MBPb) Cho hnh v
quay mt vng xung quanh trc BM. Tnh din tch xung quanh v th tch hnh
do tam gic MBP to thnh (chnh xc n 2 ch s sau du phy)( QG 2008)Bi
42: Cho tam gic uABC ni tip ng trn (O ;2010cm). trn AC ly imE sao
cho AE 2EC 5=Tia BE ct ng trn ti D . Gi F l giao imca hai ng thng
AD v BC .Tnh di BF(Bi ny lin quan n chng Gc vi ng trn ch luyn thi
khi thi vng QG)Bi 43: Cho ng trn (O ; R) . Mt hnh vung v mt tam gic
u ni tip ng trn sao cho mt cnh ca hnh vung song song vi mt cnh ca
tam gic . Tnh din tch phn chung ca hnh vung v tam gic( tnh theo R
sau cho R nhn 1 gi tr ty v dng my bm ly kt qu )Bi 44: Cho tam gic
ABC vungti A c AB = m , BC = 2AB . Bn ngoi tam gic dng cc hnh vung
CDEB , tam gic u ABF v tam gic u ACG. Tnh din tch t gicGDEF theo
m.Bi 45: Cho tam gic nhon ABC c AB =c v AC = b . Cc ng cao BD v CE
ct nhau ti H a) Chng minh BH.BD +CH.CE = BC2 b) Cho b = 6,60668 v c
= 8,606160A 60 = . Tnh BH.BD +CH.CE(Bi ny lin quan n chng Gc vi ng
trn ch luyn thi khi thi vng QG)Bi 46: Cho ng trn tm (O ,R) -hai bn
knh OA , OB vung gc vi nhau. Trn cung AB ly im A sao cho 0AOA' 30
=K bn knh OB vung gc vi OA.Lp cng thc tnh din tch phn chung ca hai
tam gic AOB v AOB . p dng bng s : Cho R =2011(Bi ny li t bi s 8 d b
nm 2005 thi Casio cp QG) Bi 47: Hnh trng khuyt Hypocrat Cho tam gic
ABC vung ti A, AB = c v AC = b V ng trn ng kinh BC (ABC ni tip ng
trn ng knh BC ) , min ngoi tam gic v hai na ng trnng knh AB v AC.
Tnh din tch hai hnh li lim Bi 48: Cho tam gic ABC cn ti A , I l tm
ng trn ni tip tam gic . Bit IA = m = 3,7159 v IB = n = 2,6593 . Tnh
chnh xc vi 4 ch s thp phn :a) di cnh AB b) Din tch tam gic ABCBi
49: Cho a gic u 12 cnh A1A2A12 ni tip ng trn (O,R) . M l im nm trn
ng trn .Tnh tng S = MA12 + MA22 + + MA122 . p dng bng s : Cho R =
19,96 (Bi 47-48-49 li t quyn Mt s vn pht trin Hnh hc 9 ca tc gi V
Hu Bnh)Bi 50: Cho hnh thang vung ABCD ( B C 1v = =) c hai ng cho
vung gc vi nhau . Bit AD = a = 9,5895, BC = b = 6,9194 . Tnh di AB
v din tch hnh thang ABCD .(Bi ny ng sng tc vo 18h ngy 7/3/2010
trong lc vc vc phn mm Cabri ) Bi 51: Cho tam gic ABC vung ti A c
hai trung tuyn BD v AE vugn gcvi nhau . Bit AB =2010 . Tnh BC ( thi
Casio Huyn D nm 2008)Bi 52:Cho hnh thang ABCD (AB//CD) v AC BD .
Bit BD = 36,27 cmv ng cao hnh thang bng 27,36 cm. Tnh din tch hnh
thang( thi Casio Huyn D nm 2009)Bi53. Tm t gic c din tch ln nht ni
tip trong ng trn ( O , R) c nh ( trnh by c cch gii) Tnh chu vi v
din tch t gic bit R = 5, 2358( m)( thi QG -2010)36 GV : LM NGUYN
THAO TR NG THPT L C NGHI P HOA DBCSHOBCADSOFHGABCSGI I TON TRN MY
TNH BT- C P THCS PH N x: HNH H C khng gian Bi 1 Cho Hnh chp t gic u
S.ABCD. Bit 0ASC 84 42' = , AC = 12, 546 cm .Tnh Sxq , Stp , V ca
hnh chp Bi 2 : Cho hnh chp t gic u S.ABCD . Bit ng cao SO= 6,786 cm
, Gc gia cnh bn SA v ng cho AC l 42018 . Tnh Th tch v din tch xung
quanhBi 3 :Cho hnh chp tam gic u S.ABCbit SA = SB = SC = 7,689 cm
Gc to bi cnh bn SA v trung tuyn AF bng 53024.Tnh din tch xung quanh
v th tch hnh chp .Bi 4: Cho t din S.ABC c cc mt l cc tam gic u cnh
a . Tnh th tch t din theo a . Cho a = 14 7 3 +Bi 5: Cho hnh chp t
gic u S.ABCD c cc cnh bn v cnh y bng nhau v bng a .Tnh th tch hnh
chp theo a . p dng : cho a = 18,5792Bi 6:Cho hnh chp t gic u S.ABCD
c cc cnh bn v cnh y bng nhau . Bit trung on l d , tnh th tch hnh
chp theo d. p dng : d = 18 3Bi 7: Khi sn xut v lon sa hnh tr , cc
nh sn xut thit k lun t mc tiu sao cho chi ph nguyn liu lm v lon t
nht . (Stp nh nht) . Tnh gn ng din tch ton phn ca lon khi mun c th
tch lon l 1 dm3 . Khi bn knh y bng bao nhiu .Bi 8: Cho hnh ch nht
ABCD cha va kht 3 ng trn trong n ( hnh v) , bit bn knh ng ca ng trn
bng 20 cm a. Tnh din tch phn hnh phng nm ngoi cc hnh trn trong hnh
v . b. Cho hnh ch nht ABCD quay mt vng xung quanh trc l ng thng i
qua tm ca cc ng trn . Tnh th tch vt th c to nn bi phn hnh tm c cu a
( thi QG -2010)Bi 9: Cho tam gic ABC vung tai A ng cao AH . Bit AB
= 30 cm HC = 32 cm . Quaytam gic 1 vng quanh BC , tnh th tch hnh
khi to thnh .37 GV : LM NGUYN THAO TR NG THPT L C NGHI P GI I TON
TRN MY TNH BT- C P THCS 38 GV : LM NGUYN THAO TR NG THPT L C NGHI P
L I K T Tp ti liu ny ban u d kin ln n 100 trang A4 . Nhng gi cht do
c nhiu vic ring t nn khc sau son tho rt s si , ch yu l tng hp cc bi
tpging dy bi dng trong cc nm va qua .Trong tp ti liu c s dng nhiu
ngun khc nhau :- Cc thi cc cp trong cc nm hc trc-Trang web
http://casiovn.com- Cc sch ca cc tc gi T Duy Phng V Hu Bnh - Ti
liutham kho ca mt s ng nghip ly t th vin thi (Violet) .Trong qua
trnh son tho cn st rt nhiu li chnh t , ti liu s c chnh sa b sung
trong thi gian ti .Rt mongc s gp ca qu v ng nghip sau ny chnh sa li
ti liu hon chnh hnThc hin :LM NGUYN THAO Trng THPT Lc Nghip on Dng
Lm ng Tel : 0982849402Email : [email protected]
