Commutative & Associa
Complement Law
identity Law
尙教僑華 真三第張六第
日六十月十年申戊
WAH KIU YAT PO
日僑華 四期星
日五月二十年八六九一曆公年七十五國民華中
一九六九市中文中学試題預語
1969
會考試題預習
數學科
(F) ·喬仲强·
現代數學科(五)
MODERN MATHEMATICS; (5):
第四次預習題解答
(1)設
•2x-3=0的两根,求下式之值:
(b)(2+ £)(p+2), (c)
(d)
(1) Proof of AUB » BUATI
由根與係數之關係。
BU
(k) (~~~1)(p+ £) = ±2++, "A+! Cat!)
= (<+ 6}~2« §a—[(«+p)-2«p]~200
(-3)2 - 100-1882.
= (x+8)2 = 4x8 = 22 - 4 (-3) — 16
5164(因較大,故被取正號)
<+ «p+ p3) = (<~p} [(x+8)_dp]
· [2-(-3)] = 4 × 7=28.
=-10, (a+1)(p+ d) = ~,
82
(2)飯/_p為5x+2元~4=0的两根,求作方程式使其
## $ $ (a) 2×+ß 2ßtd *}} iz−) (a) 1}} <+8=-—,
(b) (2+1), (b+1).
ap=-=
*****484v = (2×+5)+(36-
)=3(+8)=3(-3)
**** =(2++B)(28 + x)=2a+5+B+28
(∞+p) +αß=2(-3) + (-9) = ·
*****1* 3 y ( 3
By 254+
(書) 平成方程式の根和(は
#4 (4+1)(+1) = 1, 18 = 1+ (70)
· = 4+ [=0 Bp 4y2=10y+1=0.
(Hik=)(a) 4 = 2d)
= y+3 › ‹ Bp x=y+z+4>]1⁄2?$${
5(y+ — }) + 2(y + z ) − 4 -
5ly + y + — ) + 2y + — — 4 = 0
5y+4y+#+2y+/-
by + by
+1.
5(y-1)*+ 2(74)-
5+2(Y - 1)—4(y-
10y=12=0
+zy-2-4y2+8y-4=0 Bp 44 -10y
:(a) 25y +304-12=0; (b) 44*_104+1 = (註)本題(解=),稱為方程式之变换(Transformami
tion of Equation)
(3) 38 x-(x-2)x−(m+3)==0 74% £=9+8 ¥**£
25 求m值,並解方程式以核验之
(46) b h i k £ 138] 15, 19 x+ =m-2 <ß=-(m+3)
' = (a+p)=2&p= (m−2) + 2(m+3}
・種類筒
分解
1025
(m~5) (m+3)
(若税=5代入原方程式・得 コピー3x-8-0
3)± (-3)=4×18-3)
(ش)
६६६ &EE
AMMMC
Since the membership table of "AUB and Bua aze identical, they are equal
(2) Proof of (AUB)UC - AU(BUC)
(AUB) U C
MM!
MAT MAT MAT
AU (BUG)
Since the membership table of (AUB)UC and) AU(BUC) are identical, they are equal,
(3) Proof of “nu » A and ano
६६६
Bence. Ang Awand And
(4) Proot orfan
40 A
Hence AD ATTA
Worked examPLAU
4) Examples of simplications
1. Simplify (AOB'NC® JUI ANBING ND} {ADO" } L.H.S.- ((ARÁNC®)u(AMB®NG' WD)JU (ANC
Simplify
..... Associative Law
Absorption Law
-XLANO`JU ((ANC) B.)
Commutativa d Associative Lawa) Absorption Law
(XUYUZ')N(X'NI '72); H.S.-((XYZ}}^{(XUY)^2}
Law
Associative & De
•* ́ ̄ {( XUY ] 11 2] 1 {{ XUY ]u z }\
Commutative Law
Distributive Law
-3x-3-3(X+X+
2m +10
分解
(x+3)=0
x= - 3 (重根)
2E0
卡(重根)
(6) at?
(8) Examples of proving identities
Prove that LAUK'UC! } {} %17 ( BNC ) { ► Proof
(AU{BAC)"}NŢAU(BIC)].
De Morgan's law.
AU{{BMC) N(BAC)}
AUD
Distributive Law
Complement Law aldantity Law
2.[Prova"that [ M(A •UB]]U{ BO{ A 3U B' )
Prapf
AM(AUB)|U BR( A ?U B` }}
( An A *- Ju ( An B ) ]u {{BA® )U{ BI B' }}
(dol AAB)(U
Distributive Law.
Identity. Law
(BIA)U (B. Complement Law
BOLAVA BA
Commutative Law Distributive Law Complement Law Identity Low
(C) Examples involving subsets
Prove that if AB and A<C. then As(BAC). Proof
Assume X-FA Then -x-9 ATEGO XEC 10. BDC
Therefore,
АС ВЛС
that AcB 1f B'CA
AP6umption A B W. Acc.
Defination For intersection of 2 sets
Definition or a
subset of a given sev.]
The statment means that (1) if ACB then
BLA'
and (11) if BCA
Proof of (1) i
Assume bi B
Then biß:
Since DEB, A Since bEA. BE A '>
then ACB.
Assumption Def. of the complement of s
Ac By
*. Def. or nes complement of a set
Therefore, firibɩ B', bɛG! Hence BZAZ
Proor of (11)
Assume 4 BLA Then
Def. or a SUDDER of a given set
Assumption Def, of the complement of a set
BCA
Then
a
a B
If at B a B
Why
Therefore A<B
Work for the week!
(1) Simplify (ANB)U CAN BOC "
Def or a subset ofa:g)
(2) Simplify :((Xu?}{(x*UT ® )]u(xny).
[1]{Prove that{{A!UB! }n{A'u_B}^\ {ÂU B }} ~ŢA \\ B_
(4) Prove that¦{ANBAC)U{A'NC)U(BSC)]={0}
(5) Prove that if¡A=8.Ethen]BA
(a-b), (b+c), (c-a) M 75 hk
b-c-ok cato
(畢)
Lori) 14
(1) AB CD 11
=4(9+6541+41+9¬614+41)=25(id2)
2x1.
8p" d= = (3+14),
B==(3-547)
**¥**<=4(3+140)+4(3-(41)
*x19 x-(-3−2)x−(−3+3)=0
Ep x+5
根平和
(4) % m. n bjjuR.
比两根亦為有理數
2+(-5)2 = 25 (142).
JE 3mx-(2x+3x)x+2n=0
(3z i4 −) * * *171 £ A=[ (am+3x)] −4x3m=2?
4m+12mn49n-a4mx.
== 4 m2-12mn+In
故其根笃有理數
(am-3n)" %-2278★
Q:E.D.
($£14➡) /1⁄2 *$** † 3mx-2mL−3RX+21=
MI(3x-2)-2(3x-2)=0
(3x-2)(mx-n)—6
善或發,其根為有理數(紅)
(5) || 2x-3x-3+k{x2+x+2)=02@ +8 +84 at
在
并核驗之
(解上因两根相等,其判别式△口原方程式可化骂
化簡
(k+2)x+(R−3)x+(2k-3)=0
Am (k-3)=4(k+z)(zk−3)—o
1分解
1-6k+9-8k -4k+24=0
7k+10k-33=0
(k+3)(7π-11)=0 [k=-34/
(86 Tju) k=-3.04, 1»J£÷ ÁÀ
#m z 11.
(解)原方程式分母
----- =1,两根同值而異號,試
a(x+b»m) + f (x+a+m]=(x+Q=m) (x+b_m)
2x+a[f=m}+bx+bla-m
-11-1
• X2+(4¬m+b=m}X
+(a=m) (b-m)
X-2mX+ (a+m}{f^m)—a{b_m)− f(a-m)=0. 誠又為其一根到,亦為其根由根與係數飔係
答:
(7)設 a& c均笃实數,方程式
(x-f)(x−c)+(x−c)(x−q)+(x~a}(x-4) = 0 = @ #B +
it amb C
【正)原方程式去括号
x2−(b+c)x+b¢+x=(c+a}x+ca+x^(a+b}x+ab=
10 3x*~2(a+b+c)x+( &c+ca+ab) = 0
因两根相等,故判别式為零
A=[2(a+b+c)] -4x3(berca+ab)=0}
4(a+b+c)2 = 12 (bc+ca+ab)=0
(a+b+c)~36bc+ca+ab)==0
a2+of+c2-abibc-ca=o
20+2b+26'-zab-abc-zca—0
(a−2ab+b') + (b=zbc+c)+(c2—zca+ a®)—
(a−b)2+(b−c)2+(¢
*#*#34 MB,(Intersection of
PAB1 - 43 14-0 典AB相切P桌且典CD相切,
(分析)設日為訴求园之园心,則QP為半程,故QP1AB所以 祇是典AB切柊ㄗ奌园心(略法與CD相切之另一搏)是 無限多的,其軌跡是過P奌而與AB要直之直线,只祇典AB, CD相切(略切奌善P),則Q奌典ABCD等距離,故軌跡 為其交角之平分线,作出此两軌跡之直线,其支奌Q即為好
(8 km) AB CD# ok
PABATI.
求作一团Q使與
AB W☀?), 1 cotto
(142) 1.442807
R4B0C M4A #lel
||2.LE PAT ABWE
Kathuk
5. viata c. Qṛb
半程作园,
4 Ul Q & Q wc, QP 3 $48442, 8];Q; Qmańžnk.
PROD ME
(4)求作一园,切一已知因於性設奌,且典-已知直线相切协
(**:*8*8 +0 4 * to atom (4)
(2)袁作一园,tan已知直线於定奌,且與一已知园相切 (3)求作一团,切一已知园於在設桌,且澳另一己知园相 (4) 0, PAB) 1 KAB TEAM-10 POBA*ck
TØRKDAS CA=2AD
(5) A30BD - 2* & DB A 13-1 BC M◊ BA¬ZAC
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