1969-01-09 — Page 15

FERENCE LIRRAK

10-JAN-3969

育教室:頁三第張四第日一廿月一十年申戊曆夏 WAH KIU YAT、PO

郭日橋

CITY HALL

四特星

日九月二十年六六十一座公“十十平民劳中

一九六九中文中學會考試預習

數學科 (十)

喬仲强®

(4) (a) =

21 E

正因

(1) (a) π of

第九次預習題解答

(正)毅如右图,工為內切园:

= (5-a) tan

DEF導各位上切美則

ID=IE=IF-V

BABI=— AB×ID==CA ABCI BCKIE=ar

CAT=CARIF=br

19 26BC=AABIT ABCI+ 4CAZ

@$# #$%$#$65 #253 #5734392 85%$ #34»«»

擦第四張第二頁)

文中學會考題預習

30

TAČE A ABC Å ¥IB÷A#

)由正美律

現代數學科(+)

現題談andB 汉由餘弦定律

ain A cac C

ADX

ce=?,则因切线等長.

AF

除以之

(4)-(2)

2x+

AAID.

(5)卷

AABC等腰三角形

thet

誰AABC鴬直角三角形

(1) A

〇(田工馬内心)

tik == (5-0) tan^

詠華

求△ABC的面積

***

内切园半徑並由是求出各角

0.71

9.01 3.60

-=(a+b+c)=√(4:35+ 11.91+9,06) = 12.66

ian f=4

8=121/2′′

40.36

NO;

LOG

Sal17025

5-a= 8.31 0.91.96.

#2.60 0.5563

6. 8.6平方單位

=121°12′ =40:36

(2)証明

af c

正弦定律

R

tha=13, &

径儿及各学半径,

(a)

22.4535

12268

12268

MORSE

2.33

0.7243

0:1243

+5-8

(8-4) (5-c) [#25=a+b+c|

~(s-b)(s-c)

•2(b+c)s+tc

#}»R_14>]____(b+c_a) s-bc_

====(a+b+c) 4X

(b+c=a) x ( b + c − a) --

(b+c)-a-abr-o

= b+c=a

软△ABC為直角三角形 注意】本题光移

(畢氏定理水逆定理) Q..E. D..

the乃使两边文分數其分

a, b, c, s x = x. 於指數的運算,我們常用的有下面的

0.91966

(am)n

- a

T. 2047

M (ab)

(0+1243

7.97512

10,3492

0.1243

如果加,都是正整數時,我們是很容易明白而導出 述公式的,但是指數不是正整數時,它的意義怎样 对於上述三大定律,仍否適用?是我們的需要討論的 我們先說「特種指数的意義:

I.零指數

·0.55636 T 5680

五、買指數

Ik

an

15. 求外接园半径尺,内切园半

14). ==(a+b+c) === (13+16+15)=al,

21-13=8

14-7,

21-15=6.

s (s-a) (s—b) (s~c) =] 21 × 8 × 7x..

| 3×7 × 23 × 7 × 2 × 3 —

x3x7=84

35=(a+b+c).

24 - 10/

1=14

(葛高线)

=(Ta)" [4(TAT=A&4]

如果指數是有理數時,對於指数的三大定律,仍可適

af x at

a

免於指數是無理數或虛數時,其意義若何?则超 越初等代数範圍之外,我們暫不置論不遇凰体

塞指数(Q“=1),我們要假足2=0的,因為0=1 是「不定值」,同時閎拎指數的運箕,我們常用下述定 理:ax=a*,則x=y”這一個定理的真確性,也 *18* 2+1, Ka=0 & BŽ 1= 0=0.04.1 対数定律是根據指数定律引導而来,可以对数的 ***** | # # loga a=1, 12 log, | top & 7 定值(因為能够適合 17=|一式的,就是不定的)xly, a 是無意義的(因為能通合1*=a,且a+1是有可能的) 例)對數的定義者何?証明 logaN=2&N. ds & Bu off (loga b) (logs C)( log a) == 1.

(*) Ara, N, & a=1 mA a2=NZ 15, 25

x 13 x ix a šta, NJ, & x=los. N

(a*—N)↔ (x = log1N)

(EE) Ek loge N = x, q] a^=N

loge (ax) = loze N

李泰

rearranging and negating the parts. These

conditionals together with the given

conditional upon which they are constructed

ire listed as follows

ConditionalUp

Converse P

Inverse

P

Contrapositive q

We shall have a further study on thes- conditionals when we have considered the equivalence of statements,

Examples:

Determine the truth value of the statements

represented by the following sentences:

(1) If Hong Kong is in Europe, than 2+2=5

(2) If Hong Kong is not in Europe, then

2+2 422

(3) If bread is stone, then seawater is selted

(4) If 6 is a multiple of 3, then doga are

not animals.

Solutions:

(1) Since both p and q are false, the

conditional is true.

2) Since both p and q are true, thà

conditional is true.

(3) Sindep is false and q is trum, the

conditional is true,

4) Since p is true and q is false, the

conditional is false,

Reminders!

{1) "If p, then qa does not mean the same thing as "p implies q", "p implies qu only when "If p, then q" is always true Later we shall consider in detail the implication "p

Work for the week

1. Write the negation or each of the following

(1) 5-2-0

(2), 163315

(3) 2075

(4) All cats are black.

(5) *******

(6) 45 × 2 × 30

Boel is black

Which of the following sentences express trus

statements and which express false statements? (1) If 16 - 5 = 3, then 4(5 + 3) = 32

(2) If triangles are squares, then horses are

birda

(3) If 240, then 6 is a prim,

(4) If cowboys are poets, then animals can fly, (5) If Paris is in Asia, then snow la white.

(列三)下列两式有無錯誤?

(4)着

(b) logx log yash

(解)

=m

(答)

blogx - log y = log &)

第十次預習題

小化簡下列各式

(c).

(a=a) (k-

(d)

+ lng 5 - tog [1000 (7-341 $ # #KUL 10 75/5. Log 12mm

zah

普通

Thirr Ms hel

证

(c)T(M=h) (4-4) (b−4)

\œ) ★ ± − ( _—_ - ✨) (~~~

依对数定律,

x log,

lage N

X = logen

loge th

arbites (+)

* = log. lixe

_ (logab) (lose <) (log, a)___ logst,

例)解指數方程式

4× (2×)2+3×(2*)–

(e) (log2 3 + log49) ( logzz +logq4]

f) 3 log, 23+ log,, / 3 = log, x, * X * /] £= /£ A

●效數字

(2)解下列方程式式

2×27*-5× 97 +

'b log (35-X3) = 3′′

log (5-x)

(C = √1000X.

畢)

(e)

* 3 − λ = × > $ IX <k X, Y M Z 4y2+3y-1-0) 分解 "(4×2′′-1) (2×4 1)=

2

WŽ 2+1=0, @

5

log (x-4)+ log (7x-84)=2) (38010)

| log (x+y)= log(x=XY+ y2);

(3) 3 13 α. Czœm!**], ti

(指數相等)

(2)

5)

証

無窮項之值

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