Kì thi chọn học sinh giỏi cấp Thành phố - Môn thi: Giải Toán trên máy tính Casio - Phòng GD Hạ Long

Kì thi chọn học sinh giỏi cấp Thành phố - Môn thi: Giải Toán trên máy tính Casio - Phòng GD Hạ Long

BÀI 1 :

Sắp xếp các số sau theo thứ tự từ nhỏ đến lớn :

A = ( 649+ 13 . 180) - 13 . ( 2 . 649 . 180)

B =

C =

Bài làm :

.

.

.

.

 Không viết vào phần

 gạch chéo này

.

BÀI 2 :

a) Cho biết tỷ số của y - 13 và 7x + 5 là hằng số và y = 20 khi x = 2 .

Hỏi khi y = 2004 thì x bằng bao nhiêu ?

b) Bốn người góp vốn để thành lập công ty TNHH . Sau 5 năm, tổng số tiền lãi nhận được là 9 902 490 255 đồng và được chia theo tỷ lệ giữa người thứ hai và người thứ nhất là 3 : 2 ; tỷ lệ giữa người thứ ba và người thứ hai là 5 : 4 ; tỷ lệ giữa người thứ ba và người thứ tư là 6 : 7 . Hỏi số tiền lãi mỗi người nhận được là bao nhiêu ?

Bài làm :

.

.

.

.

.

.

.

.

.

.

BÀI 3 :

Cho dãy số : U = với

1) Tính 8 số hạng đầu tiên của dãy này .

2) Lập một công thức truy hồi để tính U theo U và U .

3) Viết qui trình bấm phím liên tục để tính Utrên máy .

4) Tìm tất cả các số tự nhiên n để U 3 .

 

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Phßng Gi¸o dôc H¹ Long Thi chän häc sinh giái cÊp thµnh phè
 ----------------------------- N¨m häc 2004-2005
 ------------------------
Ch÷ ký gi¸m thÞ
1: ...................
2: ...................
M«n thi : Gi¶i To¸n trªn m¸y tÝnh CASIO
 Ngµy thi : 05/12/2004
Thêi gian lµm bµi : 150 phót ( kh«ng kÓ thêi gian giao ®Ò )
Sè ph¸ch
Hä vµ tªn thÝ sinh : ................................................................
Ngµy sinh : ................................. Sè b¸o danh : ...................
N¬i sinh : ..............................................................................
Häc sinh tr­êng : ..................................................................
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Sè ph¸ch
§iÓm bµi thi : B»ng sè :................. ( B»ng ch÷ : ....................................................)
Ch÷ ký cña hai ng­êi chÊm thi : : 1: .................................... 2 : ...............................
* Ghi chó : - ThÝ sinh lµm bµi ngay vµo ®Ò thi nµy . §Ò thi gåm 02 tê . ( Tê sè 1 )
 - C¸c bµi ë nh÷ng b­íc dïng m¸y tÝnh ®Ó t×m kÕt qu¶ ph¶i viÕt qui tr×nh bÊm phÝm .
 - NÕu kh«ng nãi g× thªm, h·y tÝnh chÝnh x¸c ®Õn 6 ch÷ sè .
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Bµi 1 :
S¾p xÕp c¸c sè sau theo thø tù tõ nhá ®Õn lín :
A = ( 649+ 13 . 180) - 13 . ( 2 . 649 . 180)
B = 
C = 
Bµi lµm :
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 Kh«ng viÕt vµo phÇn
 g¹ch chÐo nµy
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Bµi 2 :
a) Cho biÕt tû sè cña y - 13 vµ 7x + 5 lµ h»ng sè vµ y = 20 khi x = 2 . 
Hái khi y = 2004 th× x b»ng bao nhiªu ?
b) Bèn ng­êi gãp vèn ®Ó thµnh lËp c«ng ty TNHH . Sau 5 n¨m, tæng sè tiÒn l·i nhËn ®­îc lµ 9 902 490 255 ®ång vµ ®­îc chia theo tû lÖ gi÷a ng­êi thø hai vµ ng­êi thø nhÊt lµ 3 : 2 ; tû lÖ gi÷a ng­êi thø ba vµ ng­êi thø hai lµ 5 : 4 ; tû lÖ gi÷a ng­êi thø ba vµ ng­êi thø t­ lµ 6 : 7 . Hái sè tiÒn l·i mçi ng­êi nhËn ®­îc lµ bao nhiªu ?
Bµi lµm : 
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Bµi 3 : 
Cho d·y sè : U = víi 
1) TÝnh 8 sè h¹ng ®Çu tiªn cña d·y nµy .
2) LËp mét c«ng thøc truy håi ®Ó tÝnh U theo U vµ U .
3) ViÕt qui tr×nh bÊm phÝm liªn tôc ®Ó tÝnh Utrªn m¸y .
4) T×m tÊt c¶ c¸c sè tù nhiªn n ®Ó U 3 .
Bµi lµm :
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Bµi 7 : 
Cho ®­êng trßn t©m O b¸n kÝnh R = 3,15 (cm) . Tõ mét ®iÓm A ë ngoµi ®­êng trßn kÎ 2 tiÕp tuyÕn AB vµ AC ( B vµ C thuéc ®­êng trßn t©m O ) . BiÕt AO = a = 7,85 (cm) .
a) TÝnh gãc BOC vµ diÖn tÝch cña phÇn mÆt ph¼ng giíi h¹n bëi hai tiÕp tuyÕn AB ; AC vµ cung nhá BC .
b) ViÕt qui tr×nh bÊm phÝm liªn tôc trªn m¸y ®Ó tÝnh ®­îc gãc b»ng gãc BOC vµ diÖn tÝch ®· nãi ë trªn .
Bµi lµm : 
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