1967-05-17 — Page 17

『

買一張五第

日无初月四年未丁胶夏

中文中學會考試題預習專欄

WAH KIU YAT PO

之之訴及食 背面:

,成取會格,請註須移至屉司中人包全

數學科

(廿九)

同申故數試仍。選用一其現今

*酱仲強•

第八次預習試題解答

(1)(a) ‡ a+b+2=0, ££ a(‡ + £}+b( ↓ + £) + C( £+1) 20

(解) d(b+c), flera), cla+fs _—_—_ _&(b+c)+&{c+a)+c2(a+b)

bc

ca

abc.

$$=a(b+c+a]+b2(c+a+b)+c^(a+b+c)-(a+b+c3)

=(a+b+c) (a+84)-(a2+ ffc* 3abc)-30&c

=(a+b+c) (a+brc2)=(a+b+c)(a+b+c=ab-bc-ca) -3abc =(a+b+c)(ab+be+ca)-3 abc —— 3abc_{@ a+b+a=0]

~Ja&C -3°(答)

abc

(註)可参看廿六次習題第一題得 (8+e)+b(cra)+C*(a+b)+zalc

能分解為(a+b+c) (al+bc+ca)

(6)若xyz=1. 求xy

•

解式

*

+

XYZ

期星

CITY HALL

港大入學試完畢

一出,此次計需之究施 古兹東共七年藏段戲人數一千一百卅四名

·香港大學在去年八名,比前年之及格 - 繪加一百廿四名。

有完由

生野 餘入在 增: 其步 容計

今薪 改

四劣,名称包生能

【展完成後,容納新生 學的數字亦已婚至五 紀錄,當其第一年藐 時,帶大容納新生人

『些資料之考生無法入 大容納新生入塾名額不佔讣苏名,他仍有 二一愆估計,今秋營 學院佔六十名,建築 難得三百餘名而已。學院佔九十名,工程

週年

四二

報慶

日七十月五年七六九一股公年六十五國民中

學,程庙香,九流百的今。

入場卷在 増成於本

加該

預校考 料七生

·今年共 秋擴 入展 學

可望七月間放榜

華僑日報

高級程度科自考試之【出生、高中學生

舉辦大中小學生徽文徵畫比賽

定五月二十日截稿請踴躍參加

【月之役,始許說考。

有關上項域文微恐之器做如下,

題目:香港地方:纹制之河榷,如何善用海棉空地

題目:中文縻與英文排列爲官式文,改善香港市容秩序清潔

大專生薇文〔限三千字

「此外,投考者又必須

高中生镫文八中四F4以上,限二千字)

初中生被文一中三F3以下,一千字)

題目:我的老師,锬際音樂節見聞

|並不列爲喜級程度料

自由棗:港九新界社會百態(新項目

商小八五六年)學生微文〔八百字]

自及格《閩之內

級,高良考成寜

以初所

自由畫,浓九新界的都

題目,學校風光(新項目

·初小八四年以下丷自由:我所愛的人,我所愛的寵物,我所愛的任何食物

獎金

初小姐:冠軍十元,軍二千元,每十元。 北自由- 冠軍三千元,軍二十元,季十元

·自由盡,行軍三十元,既二十元,季取千元

初中及初级小學生自由報比賽,再憑優勝者若千名致送考。

D高小組;冠軍三十元,二十元,季第十元。

C初中祖:冠第三十元,二十元,季军千元

B高中祖,冠軍七十元,年五十元,季軍四十元」

A大專組 - 冠軍一百元,亞軍七十元,季軍五十元。

各組發文及審即日開始收處,靈止期爲月二十日,那批爲源,謝期不收。 忠设者在文稿內附同身份證或學生證號碼子所屬學校班級人自由碰务滋。 所有文,恨,一經選碌,其發表鐧及嗆刺緞旣篩踉俄日報所有,落選稿件概不退

以下表格各慈役同學可剪出境象,貼在作品後面,打無路,只要照高,亦可接

别

q題 姓名 學校組 地址

別及

齡年

望有被取錄的機會。格者有一千五十 *東條件,然後才可一六各,甜果,及

2.(a)\£ (bx)x+(c-a)x+(a+b) = o`z i☎E ¶ 試証a成AP(等差級數)

(正)因一元二次方程式之两根相等,其判别式為零,

D= (c-a)2-4(b−c)(a−b)=o

c-zac+a-4(ab+bc-ac−f')==0)

C2+2ac+a-4b(a+c) + 4f2 = 0

(a+c)*—4b(a+c) + (2b)==0

(a+c-2b)"

By c-b-b-a, tha, b, c & A.P.

(b) 3k a(b+c) x2 + b (c=a)x+c(a+b)=02

誠証 0lc成HP(調和级數)

(*) D(*]»X)= b' (c-a)-4ac(b−c) (a-t)=0.

li*(c*-2ac+2)− 4ac(a8+bc-ac-2)=0

保項

b(c+aac+a)-4abc(a+c}+4ac.

[b(a+c} = 2 b(a+c), 2a6+ (2ac)2=0.

[b(a+c)-24c]=0.

英文中學會考試題預習專欄。

Q.E.D.]

€

化學科

(廿九)

·王錦釗·

CHEMISTRY (29).

`b(a+c)-2ac=0)

2:即fac = 調和中項 a b c & H.P.

(3)分解因式(a) x*+ x -2ax+1-a

(解)原式。

(*+28+1)-(C+żax+a)

-(x2+1)-(x+a)

(C+1+x+a)(*+/-x-a)(答)

(七)分解因式:X+3ax+(3-6]x+a(K-b]

( @6) /RX = X +32x+2ax+(a-f}x+a[a

((x+3ax+2a)+(a−b)(x+aj

=x(x+2) (x+2a)+(b)(x+2+

X+Q) 2ax+a-f2)

-(x+a) [(x+ a) - b

(4)解方程式:(

(二)答)

(解)去分母 (b-c)x(x-c)+(c-a)(x-2) (x-c)

A [-c(6-c)-(c-a) (cra}~a{a-b)]x+ac

去找果(-fc+cc2+a*-i+ab)x+ac(c-a)=0.

b(a−c)x=ac(a–c)

因此值不能使分母骂界故為真棍”答:七

Q.ED

(注)a-c=0(即a=c)則本題為恒等式,又可為任何數 *值(除零外),均能適合

解方程式(x+3xy-by+2x+3=0-

2x + xy +ay-4-0

(2)

(解)因13,(2)两式中含有导之項的傈数成比例,故可用加減法 消去之(2)×3+(1) 得 7X+20÷9=0

SÃ¥_____ (x-1) (7x+8)=0_____x=1 ; x=-

(1)*x=1, 4x(2) (u)差x=一号,代入(2),

-63y+989-

35y=34

(5)(2)月一日(等比數)其無窮項之和為4,其各項之立方 开成级數無躬項之和為192.求此银數

解)設原来级數為 aaar

則S=a+al+as+

10公比馬

48

=48(1++)..(3)

#WX18_a=4(1-2) Hλ(3)& N(1-4)=4&(1+2+X}

分解:(+2)(ar+1)=

号

但無窮等比級數之和;孝头绝对值須 以仁支代入得

故不合

故原来级數為

lead dioxide, effervescence occurs, and a yel green gas, chlorine, is given off.

Pb02 * LHC1 = PhC12 • 2420 • Clą

The residue is white insolubie Lead cadanlde The chlorine may be tested by its action on starch paper dipped in lead iodide solution, Chlorine will displace iodine from the solution which will tum starch blue

·2K1 + C12 = ZAG1 +

When conc. hydrochloric acid is added to black copper. oxide, the oxide dissolves forming a blue solution. The acid neptralises the copper oxice, base, to form a salts

chloride, and water.

Cu o + 2H

Boluti

Na So

The same test cas

ZNA

ado qilute.

up chlorids.

phate insolu)

u out on sodium nitrate

(will show no apparent reaction.

Q.za/(1jdilute hydrochloric acid

(ii)benzene

(iii)carbon disulphide

(iv)alcohol

(b)(1).

(11)

(iii)

3

Name of residue

lead monoxide

sodium nitrite

zinc exide

Colour when cold]

yellow

pale yellow white.

(c)(1) At first a placket. of Insoluble lead sulphida,

ds formed.

On heating the mixutre to dryness, a colourless, vapour which condenses to a colourless liquid, water, is given off first. Then brown fumes of nitrogen dioxide are given off as the nitric acid: decomposes. In addition, a colourless gas, oxygen, which rekindles a feintly glowing splint 1e aler given off

(11) When red lesa oxada ia boiled with dilute nitric

acid, a dark brown precipitate of lead dioxide. will be formed.

EMBAR

However, the "brown ring best carried out on sodium nitrate solution willw@lvea bmwn rina showing that it is a nitratez=- (Brown ring test: To the sodium nitrate soru010年 In a test-tube, add some freshly prepared ferrous c sulphate.. solution. Then pour a few dropa of "concentrated aulphuric acid dein the sidesof the

test tube, A brown ring will be formed at the junction of the concentrated acid and the alustan c)Bubble sulphur dioxide gas into on aciqiied solu

tion of potassiim permangarate. The purple permanen: ganate solution will be decolarised: Hydrogen: chloride gas will not decoloriae potassium permangan"i

21120

550,- K√50 ** 23⁄4n50, +2H259%

To test for hydrogen chloride, dip s glass rod}

watos solution of silver nitrate, acidified with:

'hitric acid, and put the rod at the mouth of a gas-

jar of hydrogen chloride, The drop of Tiquid at the end of the rod will turn milky. owing to the Cormation of silver chloride which is insoluble inJ aitric acid.

HC1 + AgNO; • AgC1· É HND ̧

d)When carbon monoxide is burnt in a gas jar of

oxygen, a gas, carbon dioxide, La formed which turns 1ime water milky

200 * 02 = CO2 |

When hydrogen: 3 sulphide is burnt in a gas gar or

oxygen, a gas, sulphur dioxide, is formed which

will decolorise potassium permanganate,. some sulphur)

may. also be deposited on the sides'of the gas jar

250

() neat sodium nitrace strongly in a hard-glass. test:

tube, and collect the gas evolved over water. The gas evolved is oxygen.

2Ma NO.3 = 2Na NO¿ . + Oz.

Two tests for the gas:-

(a) The gas will reignite a jantiy glowing. splint (b) Nitric oxide will turn brow on being brought

into contact with the ons.

Make

2NO +0 # NO

brown

fumes:

an aqueous solution with the residue from (1) and add: an aqueous solution of ammonium sulphate: When the mixture is heated gently, a colourless gas; nitrogen, is given off. which may be collected overa water

(NH 2 SO

2NH2

2H50

Tests for the gas i

(The gas answers negatively to all common teste forr gases)

(a) It will noe turn me water milky (6) It is neutral to litmus:

(111 )Make a mixture of ammonium sulpnate ano slaked Time

and heat the mixture gently. The gas evolved. ammonia, is collected by upward delivery.

+ (NH 2 SO

Tests for the gast-

(a) It will turn red litmus blue

2NH

(b) when a glass rod which has been dipped. an

concentrated hydrochloric acid is placed at the mouth of a gas jar containing ammonia dense white fumes of ammonium chloride will be formed. (iv)Add an aquedus solution of sodium nitrate to an

aqueous solution of ammonium aulphate, and heat, the mixture gently. A gas; nitrous oxide, will be óbtained which may be collected over wari water 2NaNO3 + (NH)2 SOL; = Na2SO4 + 2NH NO

2420

Teste for the gassio

N20

(a) It will reignite a atrongly glowing aplint.

(oxygen will reignite even a faintly glowing splint.)

(b) It is fairly soluble in cold water. (Oxygen is

only very sparingly soluble in cold water.

(8)談x為正數且一六

(解)将肝詔式自乘(x)

x+↓ =[7+1+2=13+2=15 (A)

(註)」因x画正数,故根式祇取正號。

2. 本题亦可由 x + =1 方程式求出x

(解)原式=(1+x)+3)

3

ab +ac+abç- 2ab

然後代入計算

第廿九次預習試題

(1)”ABCD高尹行四边形一直线 PQIAC而哭 DA DC 交於P及Q,試证

AABPadB9 又希 P9典DA, DC之延线相交時上述结果仍成立否?

(2) ABCD為直角梯形,LA和D一加上遇对角线交互E作ERLADE 「要趾為F就話 FE 甲分 - BFC

(3)P為定角口内一定是束遇P作二直线興LD 两论交於A及B,而使 |OA=08 I PALPB

(4)(a) 何謂自(Aingle)?又角之正自如何碟主之?

b) A ABC & A=45′′, 8=60,

=2吋,不許用表試s:AB=隻(H含)

吋並由是本Am75之值(用最能相式表示之)

5)(a)設如右图 試用面積關係就明

sin (A+B) = sin. A con 8+ cm Å sin E

B. 2A= 2 min A con Å

(6)設如右图利用上述閺强薅用其他方法 試求器之值/

(6) ▲ ABC at

試話C=