1972-06-07 — Page 23

華僑日報 All

三期星

日七月六年二七九一腊公年一十六國華中

・雨遷各加入(火+1)(+2)(+3)頃

1-2-3 +23:4 + ··· + k (k+1}{k+2}

+ {K+1)(k+2)(R+3)=¢k{k+1){k+2)(k+3)

+(k+1)(k +2) (k +3) ·

育敏備真三第張六第 日六廿月四年子壬靨夏 VAH KIU YAT PO

報日僑華

中中會考高級數學(一)試題中中會考高級數學(一)解答

AW04(1 $

張正邦。

AMS, JE, KS

高級數撃(試巻 試題解答 (a) Xi, X2, X3, X4 th Kay

Tub

21

XX

X2

BIBL=

・「太

A-NA AR

(凍,等號成)

f(x) ***

2x^

化我邸纷分式。

..

www

1 x − Z JX X + X z z O

LÆG ÁTLÁ JE PREST

1 2 3 2 3 4

• mêm • 15m 1 23 • Zném + 1)(n + 2500 * }}

√ X1 X2 < x1 + x2

(11) + / X1 X2 X3 X4. <^1 + X 2 + X 3 + X2

X1+X=+X3+X4.

(k+1)(k+2)(k+3){ $k+1]

† (k +1)( k + 2) (k +3) (k + 4)

(k+v[(k+D+i][(k+1)+2][(k+1)+3}

當九=+1時,原式亦成立,故得證

(3) @ BRO X BK, a √2x+√2x-JT-ZX

=1,求X之值.

√2X+ √2X-J1-27

12X

育教僑華 英中會考數學科答案(四)

堅道英文書院。

Kathematic

(Lesson 33)

20. A√√√ - V=X

- At

(x+4) BARBARA

*** (*-*) * ((x)) . #RAH.

II 36A - £124A LÆRER - BADGAON PAPERZSAR •

問此爾必î幣的实归了所彻忠诚然期或向了

*MMAG7+-*ABIWA --AQQIAAAQIE

* BE -- EN KI JE

22x+2x 1-2√2x 1-4√2x +8X

ふう結果、故

平方之

2X

5X

· X1 + X 2 + X 3 + X4 → JX, X2 +JXX4

= XX + 16

2

z

平方之

· X1 + X2 + X3+X4

4

By..

X 1 X 2 X 3 X 4

12011-12=x3=X4 $22/

何门

(E)

網點式中,紘三個連續的係數让你

& 15

200 -

6

16) HA

RENATA

+-

无分錢修理

b

成立

(*) 1a) **; # 2 sen10 - (? + √3) cos @ = 2 + √3

羅心栳 BRYCI

<A < 180°-

* F 100 120

(b) to X Y X-5

(2X-1)2

2

FOIMEEN241 · **

y≤72

***¥p€ #5 - £200 £ 47 q 2 - api ik be

(t) SHEA B C

D

ε

(J)

( ZADEFZ72

ÉF cos ARB CAZKA.

Risth 24 no 7B ↑ 20)

202**•_

K TO A

batan Cia (A 49 + C) # /

RAB CRERA.

Rip Atan 8 ian Clan Atas 8 tan C..

A B C = 180° *

妈在面角三角形ABC中,AB露时维

BC-3AC 今在 BC上取兩點P

QBP-PQ = QC - R2

¿BAC ¿PAC LOAC 180

* ENSTARZEŻE KOR >

X-5

(2x-1)2

Bpy (2x-1)2 = X-5

展開並依x之降需集項,得

44x2= (49+1) x + (y + ≤ ) = 0

因X慈實數,故此方程式立判别式大於 零或等於零,即

(-(49.+1)]2 = 4(47) (4+8)=0;

16 y2 +84 +1-1672-804 zo.

8x

·X (25X-8)=0

12 X=0 (16*); X = 0 代入原方程產生虛數.

X = 25

fà. £v (x − a) 172.

AKAR GE EnX (X-Q) BA (How] 餘數為3.

BK +(x) = (x − a) Q(X)+R. 1+Q(x)

・多項式

ABX F(x)=(+(x)] <= (x-a)2 (@(x)]2 * ZR(X-a)Q(x) +R2;

F(x)被(カーム)除で、根檪数定理 19 = F(Q),

~ 27 $2 = (a-a)2(Q (a)]+ZR (a-a)

xQ(a) +R2 = R2

(*) (a). ncr + mcx =ntice

12727.

4572

mfx tmft-t

(-) (@) 2x + x2-3x+2

A

y co 20

0° << 180°

X* (*~1)

X-I

y = cos(+ 30°), 0" << 180′′

為一恆等式,求及f(x),並將

並利用这兩制線解方程

cos - nad 2 an 20

中中會考宗教知識科試題

(基督教)

(兩小時完卷)

マメ+83-3メ+マ化部分式。

*(X-1):

23x-3x2+ x2-3x + 2 = A +217-

14(x-1)

v% x*(x-1)£u): 2x4+x3-3x+2=Ax4

+(カーリf(x)

§ X=1 KÒλ(2): 2+1 - 3 + 2 = A + vi

A = = 2

(0)

因是恆等式,故才可適合任何實数

***TZER

- BEN-ARIULNES: FED

豐、麻辣影佥窖诚讵八越。

(-) #TUSAS: CR1 M12 73 70 3 ) (+9)

VX A=2 Hix (2): 2x2+x23x+2=27

Lai 楚你

(c)

(3)

爱好 藏品

t $19

埏光购;(此必沿至部份深,等家數依數日次序,刀府在武链,

観察(3),知+(x)沸かえ二次

f(x)=Px2+8x+ɣ-

(4)

تا

191

(1)ETTEM DE ME (1) -17, JE PRO 22

KBS (5) WD •,

feb MSC_6_) • Həya (7)

11

人口

亞受苦婚禮凇的一个先知,新起

(9) AKU BEANS SANTA MA

JAVA - 1o & & ( 10 ) OV L 30 =

(E) *$£ to; as an 12 (a me s·

DALAM A

4

( **AE & WILDER · 1, X 1941 5 10 1911

( t test

(1) A 3; na ka dhe vë d£40 EPA DE ME 2209 · at å

LYRIŲ PASAUR - 29 STENDUR y

2 # #

(—) DOWN ATAENS21 (# HK in 75 · 09-743 75 A4 A)

1

2

EBA

=)

AU

623 Bate

Tel

8011 &

ZUMA

*LAMPU 117

Minat

fal STAL 11% = FITUAR COM THE CONTE

(2

#1 ·

HERBA BETALAR WAST DOA - ZORNEGAANI

JJ

12 BB43 @ 1) #0&R **** LIKEABHAINA

一楼

i ma is an ta dá - MERCANVAS WA

Iri

$1

12

BELA NE IN SKUPINE F • %* [_9_) •

*ዳ፡

ANKARA I — 12 · 20: ·

JEDALA KERN WI#7 - #7 HIER #3+

CI) HERBA DNE 25 AN AD X INCHWERINE 34 •

(*) BAVODUNBURNUDE DO NA MO -

}

(+)

{+}

(+)

+ (x-1)(PX2+8x+r) | +

= < x2 + ex2 + = x2 +

|x-r

DE #METEO 21% KK, 19 / ++3=

L

P=1,8=1, Y=-2

vi P., q. r àleλ_(4): f(x)=x2x-2. 因此()可化成

2x4+x3=3x+2=x+ 2x2++x=-3x + 2 = 2 + x2+x-2-

74(x-1)

=

2/1 + x/2 + 1/3-2/3/4

(b)運用敷蹄納法明下式

1-2-3 + 2·3·4+ ··· + N (n + 1)(n+2)

m(n+1)(n+2)(n+3) 當時,左邊=1.2.3=6,

右=本(1)(2)(3)(4)6 $n =20$ £ −1·2·3 +2·3·4 =30/

右=本(2)(3)(4)(5)=30

n = k = 1·2·3 +2·3·4++kk+ik:

Л

→本(+1(+2) (K+3) 成立

Y! (n−X)}\\ {Y~1)!(n~r+i)!

m!

(**

-Y+1

(F-1)! (n-x)! (7+~/

!

7

n+)

(Y-1)!(m~X)! Y(X+1)

r! (M-Y+D!

(n+D!

Y! (n+1 − r)!

(5) Given: PQRS je a equare of

aide 4 inches ihgaribad in a circle centre 0.

V is the mid-point of the aide PS

QV is produced to mesT

RS produced at X.

PX is joined.

To prove (1) XPà de aniacscelas

triangle

(11) PX is a tangent to the

circle at P.

To calculate the area of the trapesi tam Paks.

Solution

/ PSK - /SPQ

·

___PSX + ___PSH

• PSX -

L_PVQ

90° (PQKS is a

■quara)

180*

180° (adj.28 on a st.

line)

90o LPSR

LXVS (vart. opp. (6)

PV - V5

(Given)

A PVQ - ASVX (4.5.4.)

Y未完未完明日續)

!

(Y+1)! (1~Y~1}!

15a

(૩)

x + nort

Bp zr = 2+1

Hi ==

(4)

·(5)

Bp ZY 7-3

1-2, nall.

(4) nt!

ZY=U=4 - Y=2

Cy

VAN Ž VÍRA (4)

THI

bharata (1);

#!

( !

(2−1)! (11−2+1)!

b

(b)試證

(6)菜土腐有男士 36人女工24人。

某天發花紅,每人得真鈔票一張 其中有两張為偽幣!

開張偽幣均為工所得之武率

河? 又問雨中一高男工所得 ・張所得を無率為付き

男女工人合共36+4. 任何二人取得偽幣之方法有C2

60x59

1770 2

: 60 his

佳何二男工取得方法有 36C

36235-630 $4.

630種

∴兩張偽幣均為男士所得之或然率

* 630 = 24

1770

一男一女工取得方法有

3601 · 24 C1 = 36·24 = 864 #§

故此種情形之或然率為864 144

1770

295

(五)(a)若在(1+x)”之展開式中某三個 Ža§ 8a91% BZ #a, sa Bisa #nka a ¿11.

22 (1+x)^2 xt.frá Y+I

zh, r+2 F 24k a, saß isa

BP ~ Cr1 = a, ncr = 5a, ~(x+1=159

Tr=Q! (n=r+i)!

-

ww

a

}

b+c- a' cta-b a+b-c 調和級必要反-

分條件為

a b. c by-to $$. BY

2(b+(-a)·(a+b-c).

(b+(-a)+ (a+b-c).

24c

Zac

aca+b-c)+c(6+(-a);

a2-2ac+c2+ab+bC

Zac

a2+zac+ c2+ab+bc-4ac

sac

(a+c)2+b(a+c).

2 ac

a+c

-40C

(a+c) ~[2 (2a (()-b]

и

a+c

C+Q-b

a+b-c

因此,

btc-a 為調和級數之必要及-

分條件為

zac

b

a+c

Ep a, b, c Ex-

- (1)

級數

=sa-.

(2)

稅下期續

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