lamdethi.sty định dạng các loại đề thi

8/14/2019 Lamdethi.sty nh dng cc Loi Thi

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lamdethi.sty nh dng cc loi thi vbi tp

Nguyn Hu inKhoa Ton - C - Tin hc

HKHTN H Ni, HQGHN

Mc lc

1. Gii thiu. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22. i hi gi lnh km theo . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

3. Ty chn ca gi lnh . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

4. nh dng bi thi . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

4.1. Mt bi v li gii . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

4.2. Cc lnh nh dng cho cc loi cu hi . . . . . . . . . . . . . . . . . . . . . . . . . . 64.2.1. Cu hi t lun . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

4.2.2. Cu hi trc nghim . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

4.2.3. Cu hi trc nghim dng c bit . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

4.2.4. Cu hi trc nghim trong bng . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

4.2.5. Cu hi in ch trng . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

4.2.6. Cu hi ng sai . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

4.3. S dng tp ngoi v lnh gi vo . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124.3.1. Cc tp cu hi . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

4.3.2. Cc lnh ly bi trc tip tp . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

4.3.3. Cc lnh ly bi tp . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

4.4. Nhn trch dn cho cc bi tp . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

4.4.1. Chn s ngu nhin . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

4.5. Mu thit lp cho tng loi cu hi . . . . . . . . . . . . . . . . . . . . . . . . . . 18

4.5.1. t lun . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

4.5.2. trc nghim . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

4.5.3. in ch v ng sai . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

4.5.4. Trc nghim theo bng . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

4.5.5. c th thit k . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

4.5.6. Bi tp cho cc chng cun sch . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

1


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http://nhdien.wordpress.com - Nguyn Hu in 2

5. a hnh v bng vo cu hi, li gii. . . . . . . . . . . . . . . . . . . . . . . 22

6. Hng pht trin gi lnh . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

A. Ph lc v cc gi lnh . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

A.1. Gi lnh enumitem.sty . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

A.2. Gi lnh shortlst.sty . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

A.3. Gi lnh float.sty . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28A.3.1. To ra mt mi trng ng mi . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

A.3.2. Nhng lnh lin quan n gi lnh. . . . . . . . . . . . . . . . . . . . . . . . . 29

A.3.3. S dng gi lnh . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

A.4. Gi lnh nonfloat.sty . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32A.5. Gi lnh ifthen.sty . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

A.6. Cc gi lnh son thi hoc cu hi kim tra khc. . . . . . . . . . . . . 33

Ti liu tham kho . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

1. Gii thiu

1. Trc y ti c vit ra gi lnh dethi.sty cng vi vic s i examdesign.clsto ra trc nghim v mt s loi thi vi s so trn c cu hi ln phngn tr li. Ti dng gi lnh ny nhiu ln v rt tt, nhng y l mt lp vnbn nn khng dng chung vi cc nh dng vn bn khc. Ngha l khi dngli cu hi lm sch, tp hp thnh tuyn tp th li phi sa nh dng v gykh khn cho ngi dng.

2. Gi lnh answer.sty v alterqcm.sty l nhng gi lnh kt hp c vi

cc vn bn khc. Do ti c thit k mu lm cc thi t lun v cc bitp trong nhiu chng ca mt cun sch. Ti dng v chy kh tt, khi bi v li gii trong cng mt khung son tho nhng lc a ra th tchc li gii ra v tr ty ca vn bn.

3. Mt gi lnh n gin probsoln.sty ca Nicola L.C. Talbot cng nhmmc ch to ra cc thi theo ty chn mt s trong cc bi c. Gi lnhc th ly t tp ngoi vo mt s cu hi lp ra mi. Gi lnh c ti ach

http://tug.ctan.org/tex-archive/macros/latex/contrib/probsoln/

Bn c th xem ti liu hng dn.4. Hon ton da vo gi lnh probsoln.sty cha li cho thch hp vi

mi trng ca ta v thm rt nhiu mc ch khc nhau khi s dng. Do cthay i kh nhiu khc vi bn gc nn ti tm t l lamdethi.sty, ti cgng bao lu cc lnh ca gi lnh probsoln.sty v thm lnh cho thch hp viyu cu mi. Gi lnh kt hp vi mi nh dng ca LATEX.

Hin ti gi lnh mi lm c:


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Nn dng chc nng ny bit bi a ra l bi no.4. final Phng n cui cng in ra khng c cc nhn trn km theo.

4. nh dng bi thi

4.1. Mt bi v li giinh dng c bn mt bi tp

\begin{baitap}{}\begin{problems}\end{problems}\begin{loigiai}

\end{loigiai}\end{baitap}

1. Mi bi c gn cho mt nhn, c im ca nhn lchi lm hai phn c du hai chm gia. Phn trc du chm ta cth cho cng nhm k t nh diffeasy lot bi o hm loi d sauny ly ra hng lot cng loi, phn sau arctan cho khc nhau, duy nhttrong tp d liu. V vy ta c th cho s cng c nh diffeasy:2 bis 2 trong tp ca ta.

2. , chp nhn tt c lnh ca LATEX tr mi trng ver-

batim. Vi ty chn km theo c th ch hin ra bi hoc khng.3. Tt c lnh v mi trng, k c hnh nh u chovo y c. Vi ty chn km theo c th ch hin ra li gii hockhng.

V d

\begin{baitap}{diffeasy:arctan}\begin{problems}$y = \arctan x = \tan^{-1}x$\end{problems}

\begin{loigiai}\[\tan y = x\]diff w.r.t. $x$:\begin{eqnarray*}\sec^2y\frac{dy}{dx} & = & 1\\\frac{dy}{dx} & = & \frac{1}{\sec^2y}\\

& = & \frac{1}{1+\tan^2y}\\& = & \frac{1}{1+x^2}


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\end{eqnarray*}\end{loigiai}\end{baitap}

Vi lnh

\begin{enumerate}\item \useproblem{diffeasy:arctan}\end{enumerate}

cho ra kt qu

1. Ly o hm hm s sau y = arctan x = tan1 xCn thm vo lnh

\showanswers\begin{enumerate}\item \useproblem{diffeasy:arctan}\end{enumerate}

Cho kt qu ch li gii

1. Li gii.:tany = x

diff w.r.t. x:

sec2ydy

dx= 1

dy

dx=

1

sec2y

=1

1 + tan2y

=1

1 + x2

Cho c li gii v bi

\begin{enumerate}\hideanswers\item \useproblem{diffeasy:arctan}\showanswers

\useproblem{diffeasy:arctan}\end{enumerate}


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1. Ly o hm hm s sau y = arctan x = tan1 x

Li gii.:tany = x

diff w.r.t. x:

sec2

y

dy

dx = 1dy

dx=

1

sec2y

=1

1 + tan2y

=1

1 + x2

Nh vy, vi lnh \useproblem{} v kt hp vi

\hideanswers v \showanswers ta c th cho ra phn cu hi hoc tr li.

4.2. Cc lnh nh dng cho cc loi cu hi

Nh vy ta c th vit bi v li gii trc tip trn mt vn bn cc cu hiri dng chng, cn cu no khng dng s khng hin ra. Nhng nh vy sri trn mt vn bn, gi lnh c kh nng a tt c cc cu hi vo cc tpsau gi ra s dng. Ngha l phn cu hi c son ring ra tng tp ty khi s dng th gi vo. nh dng macro sau y c th dng trc tip nhnh dng trn v c th lu vo tp gi ra.

4.2.1. Cu hi t lun

Ging ht nh v d trn ch c khc l cc mi trng c rt gn li

\baituluan{}{%Cu hi 1}{%Tr li}%Ht cu hi 1

Cch dng ging trn v cch dng khc na phn sau.

4.2.2. Cu hi trc nghim

Son mt cu hi trc nghim cng c hai phn

\baitracnghiem{}{%Cu hi 1


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}{%Tr li}%Ht cu hi 1

1. Ging phn trn.

2. Cu hi c tnh n bt u phng n. Cc lnhv mi trng ton u dng c.3. C hai lnh dnh cho phng n:

\dung[]{}\sai[]{}

Bnh thng khng c ty chn [] th cc phng n ch chim 1/4 theochiu ngang. Nu c 4 phng n ngn th mc nh khng coa ty chnny. Nu mt trong 4 phng n rng hn 1/4 th ta cha 2 phng n 1 ctvn cho [2] vo mi lnh ng sai. Cn cc phng n qu di cho [4] tt cxp hng dc lun. Nu vn bn chy 4 phng n ngn m vn xung dng

th cc bn chnh li 1 cht nh\setlength{\shortitemwidth}{0.1\textwidth}S 0.1 c th tng hoc gim 1 cht nh 0.15.

V d

\baitracnghiem{giaitich:1}{%Cu hi 2Gii phng trnh $2^{3\frac{x-1}{x }}\cdot 3^x=\sqrt{9}$ vch ra nghim khng nguyn ca n.}{%Phng n tr li

\sai {$\frac{3}{2}$;}\dung {$-3\log_32 $;}\sai {$\frac{5}{7}$;}\sai {$\log_23 $;}}%Ht mt bi

Cu 1. Gii phng trnh 23x1x 3x = 9 v ch ra nghim khng nguyn ca

n.A. 32 ; B. 3log3 2; C. 57 ; D. log2 3;

4.2.3. Cu hi trc nghim dng c bit

A. Mt cu hi trc nghim thng c phn dn gii v phn cc phng n,nhng thc t nhiu ra phn dn gii h chung vo mt cm sau ch ccc cu phng n. nh dng loi ny ch bng lnh

\bangtracnghiem*{}{%Cu hi 1


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}

Nh vy ch c phng n tr li. Ta xt v d

\baitracnghiem*{chontu:1}{%Cu hi 1\dung{\underline{h}our }

\sai{\underline{h}igh }\sai{\underline{h}ouse }\sai{\underline{h}ome}}%Ht mt bi\baitracnghiem*{chontu:2}{%Cu hi 2\sai{ b\underline{a}sic }\sai{n\underline{a}tion }\dung{c\underline{a}ncer }\sai{p\underline{a}tience}}%Ht mt bi

\baitracnghiem*{chontu:3}{%Cu hi 3\sai{ stopp\underline{ed} }\sai{work\underline{ed} }\dung{want\underline{ed} }\sai{lik\underline{ed}}}%Ht mt bi

Vi cch dng trong tp

\def\dschontu{chontu:1,chontu:2,chontu:3}

\loadselectedproblems[btchontu]{\dschontu}{cauhoitienganh378}\tieude{Chn t (ng vi A hoc B, C, D) c phn gch dic pht m khc vi nhng t cn li trong mi cu sau.}

\begin{enumerate}[leftmargin=*,align=left,label={\bf Cu \arabic*.}]\foreachproblem[btchontu]{\item\label{prob:\thisproblemlabel}\thisproblem}\end{enumerate}

Chn t (ng vi A hoc B, C, D) c phn gch di c pht m khc vinhng t cn li trong mi cu sau.

Cu 1. A. hour B. high C. house D. homeCu 2. A. basic B. nation C. cancer D. patienceCu 3. A. stopped B. worked C. wanted D. liked

B. Dng trc nghim c bit gch di v cc phng n ta dng lnh

\bangtracnghiemb{}{%Cu hi 1}{%Tr li


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}%Ht cu hi 1

Phn ni dung v phng n nh sau:

1. vn nh trn2. c gch di vi cc phng n sai hoc ng dng

lnh:\dungh phng n t ng theo hng;\saih phng n t sai theo hng;

3. l cc phng n ng v sai\dung{}\sai{}

V d

\baitracnghiemb{suatu:1}{%Cu hi 1\saih{That} is \saih{the} man \dungh{which}

told me \saih{the} bad news.}{%Phng n tr li\sai{That}\sai{the}\dung{which}

\sai{the}}%Ht mt bi\baitracnghiemb{suatu:2}{%Cu hi 2

My \saih{younger} brother \saih{has} worked

in \saih{a} bank \dungh{since} a long time.}{%Phng n tr li\sai{younger}\sai{has}\sai{a}\dung{since}}%Ht mt bi\baitracnghiemb{suatu:3}{%Cu hi 3\saih{It is} \saih{the English} pronunciation

that \dungh{cause} me \saih{a lot of} difficulties.}{%Phng n tr li\sai{It is}\sai{the English}\dung{cause}\sai{a lot of}

}%Ht mt bi

Thc hin


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\def\dssuatu{suatu:1,suatu:2,suatu:3}\loadselectedproblems[btsuatu]{\dssuatu}{cauhoitienganh378}

\tieude{Chn phng n (A hoc B, C, D) ng vi t/ cm tc gch di cn phi sa cc cu sau tr thnh chnh xc.}

\begin{enumerate}[leftmargin=*,align=left, label={\bf Cu \arabic*.}]\foreachproblem[btsuatu]{

\item\label{prob:\thisproblemlabel}\thisproblem}\end{enumerate}

Chn phng n (A hoc B, C, D) ng vi t/ cm t c gch di cn phisa cc cu sau tr thnh chnh xc.

Cu 1. ThatA

is theB

man whichC

told me theD

bad news.

Cu 2. My youngerA

brother hasB

worked in aC

bank sinceD

a long time.

Cu 3. It isA

the English

B

pronunciation that causeC

me a lot ofD

difficulties.

4.2.4. Cu hi trc nghim trong bng

Son mt cu hi trc nghim bng cng c hai phn

\bangtracnghiem{}{%Cu hi 1}{%Tr li}%Ht cu hi 1

1. Ging phn trn.2. Cu hi c tnh n bt u phng n. Cc lnh

v mi trng ton u dng c.3. C hai lnh dnh cho phng n:

\chon {}\khong{}

V d

\bangtracnghiem{bangtn:2}{exp$(\ln x) = x$ pour tout $x$ appartenant }{%Phng n tr li\chon{$\mathbb{R}$}\khong{$\big]0~;~+ \infty\big[$}\khong{$\big[0~;~+\infty\big[$}}%Ht mt bi


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Cu hi Tr li

1. exp(ln x) = x pour tout x appartenant R0 ; +

0 ; +

4.2.5. Cu hi in ch trng

Ta c th dng phng n * cho bt c mi trng no cng c, ti dnglnh cho cau hi t lun

\baitracnghiem*{}{}

1. Ging phn trn.2. Vn bn c khong chng in vo vi

lnh \blank{}V d

\baitracnghiem*{diencho:so1}{%Cu hi 1How much \blank{wood} would a \blank{woodchuck} chuck,

if a \blank{woodchuck}would \blank{chuck}, wood?

}1. How much would a chuck, if a

would , wood?

4.2.6. Cu hi ng sai

Ta c th dng phng n * cho bt c mi trng no cng c, ti dnglnh cho cau hi t lun

\baitracnghiem*{}{}

1. Ging phn trn.2. Vn bn c khong bng nhau in vo

vi lnh \answers{ng} hoc \answers{Sai}V d


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\baitracnghiem*{dungsai:1}{%Cu hi 1\answers{ng} This sentence is not false.

}\baitracnghiem*{dungsai:2}{%Cu hi 1

\answers{ng} Roger \& Trng Me chroniclesone mans attempt to get into

Disneyland so that he can visit Toontown.}

1. This sentence is not false.2. Roger & Trng Me chronicles one mans attempt to get

into Disneyland so that he can visit Toontown.

4.3. S dng tp ngoi v lnh gi vo

4.3.1. Cc tp cu hi

Mi loi cu hi ti ghi vo mt tp ring. C th ghi chung vo mt tpcng khng nh hng g, d qun l v s cha ta ghi vo cc tp khcnhau. V d km theo bao gm cc tp:

cauhoituluan.tex Tp nhng cu hi t lun. cauhoitracnghiem.tex Tp nhng cu hi trc nghim. cauhoibangtn.tex Tp nhng cu hi trc nghim bng. cauhoidiencho.tex Tp nhng cu hi in ch trng. cauhoidungsai.tex Tp nhng cu hi ng sai.

Bn c th b tr cc tp cha cc cu hi d, cc cu hi kh, ... khi thitlp thi ch ly mt s cu trong thi.

4.3.2. Cc lnh ly bi trc tip tp

A. Ly mt vi cu hi trong mt tp c bao quanh mi trng enumerate

\begin{enumerate}\selectrandomly{}{}

\end{enumerate}

1. tn tp c cha cu hi, nu khng phi th bo li.2. s cu hi cn a ra trong s cc cu hi trong tp.

V d a ra 2 cu hi ca tp in ch trng:

\begin{enumerate}\selectrandomly{cauhoidiencho}{2}

\end{enumerate}


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1. Nht tr How much would a chuck, if awould , wood?

2. s first work was the Tractatus- Philosophicus.

C th ly mi tp vi cu mt cch ngu nhin ri cng a vo mitrng trn cho ta b cu hi cn lp.

B. C th a ra ton b s cu hi ca tp:\begin{enumerate}\selectallproblems{}\end{enumerate}

l cc tp c cu trc trn, k c tp cha nh dng chung nht phn u. V d

\begin{enumerate}\selectallproblems{cauhoidungsai}\end{enumerate}

1. This sentence is not false.2. Roger & Trng Me chronicles one mans attempt to get into

Disneyland so that he an visit Toontown.3. Ladenswallows fly faster than unladen swallows, unless they

carry coconuts.4. Monty Python and the Holy Grail is a very funny movie.5. All animals are created equal, but some animals are more

equal than others.

4.3.3. Cc lnh ly bi tp

A. Mt lnh nhp cu hi dng sau ny khng nhp vo vn bn, ch khino dng n bng lnh khc mi ly vo.

\loadrandomproblems[]{}{}1. Ta t mt tn cha cc cu hi trong b nh,

dng n sau ny nh bttuluan, hoc bttracnghiem,...2. S nguyn, ln nht bng s cu hi c trong ,

khng th bo li.3. Mt trong cc tp cha cu hi trn.

V d ly 4 cu trong tp t lun (ta bit tp c hn 3 cu hi)

\loadrandomproblems[bttuluan]{5}{cauhoituluan}C th ly hng lot tp vo mt lc nh

\loadrandomproblems[bttracnghiem]{12}{cauhoitracnghiem}\loadrandomproblems[btdiencho]{6}{cauhoidiencho}


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\loadrandomproblems[btdungsai]{5}{cauhoidungsai}\loadrandomproblems[btbangtn]{11}{cauhoibangtn}

Khi ta c th dng lnh theo nhn ly ra

\begin{enumerate}[leftmargin=*,align=left,label={\bf Cu \arabic*.\ }]

\item\useproblem[bttuluan]{logic:4}\item\useproblem[bttracnghiem]{mc:bay}

\end{enumerate}

Cu 1. a) Pht biu nh ngha th no l hng t v cng thc tn t trong lthuyt h tn t. b) Cho v t ba bin P(x,y, z) x.y = z trn trng sthc. Xc nh gi tr chn l ca mnh : (x)(y)(z)P(x,y, z)v (z)(x)(y)P(x,y, z). Din gii mnh thnh cu ni thngthng.

Cu 2. Gii h phng trnh x(x + 3y) = 18,

y(3y + x) = 6

v ch ra i lng n(x2 + y2), y n l s nghim ca h phngtrnh.

A. 10; B. 1; C. 20; D. 6;

Bn c th t tm thi draft cho ty chn gi lnh s nhn thy nhn.

B. Khi ly ra ri ta c th gi cc bi ra nh tp d liu nh

\begin{enumerate}[leftmargin=*,align=left,label={\bf Cu \arabic*.\ }]

\foreachproblem[btdungsai]{\item\thisproblem}\end{enumerate}

Cu 1. This sentence is not false.Cu 2. Roger & Trng Me chronicles one mans attempt to get

into Disneyland so that he an visit Toontown.

Cu 3. Laden swallows fly faster than unladen swallows, unlessthey carry coconuts.

Cu 4. Monty Python and the Holy Grail is a very funny movie.Cu 5. All animals are created equal, but some animals are more

equal than others.

C. Thc ra cc bi c gi hnh thc khng c tp cha th chngtrnh dng tp [default] cha chng. C th dng cu trc gi lit k tt ccc d liu gi cha cc cu hi


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\begin{enumerate}[leftmargin=*,align=left, resume,label={\bf Cu \arabic*.\ }]

\foreachdataset{\thisdataset}{%\foreachproblem[\thisdataset]{\item\thisproblem}}\end{enumerate}

D. c bit chng cn dng lnh trn ch cn \input cauhoituluan.tex ridng lnh


\item\useproblem{logic:2}\item\useproblem{logic:1}

\end{enumerate}

Bi v tp d liu cha mc nh cha cc bi gi vo l [default]. Tt c cclnh phn trc c dng tp c s d liu u dng thay vo [default] lc nh


\foreachproblem[default]{\item\thisproblem}\end{enumerate}

Ch tp ny [default] c tt c cc bi dng cch gi khng gn vo tp d

liu c th no.

4.4. Nhn trch dn cho cc bi tp

1. Khi dng lit k ta c th dng nhn cho s bi tp nh

\begin{enumerate}[{\bf Cu 1.}]\item \label{logic:4}\useproblem[bttuluan]{logic:4}\item \label{logic:1}\useproblem[bttuluan]{logic:1}\item \label{logic:2}\useproblem[bttuluan]{logic:2}

\item \label{logic:3}\useproblem[bttuluan]{logic:3}\end{enumerate}

Sau dng \ref{logic:4},\ref{logic:1},...2. Dng lit k tt c cc bi tp ln \selectalllabels{cauhoituluan} mi

bi c gn mt nhn chnh l nhn ca bi ton. V d

\setlist{labelwidth=40pt, itemindent=45pt,topsep=0pt,


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partopsep=0pt,parsep=0pt,leftmargin=0pt,align=right}\begin{enumerate}[label={\bf Cu \arabic*.\ }]\selectalllabels{cauhoituluan}\end{enumerate}

Cu 1. a) Cho P1, P2 v Q l nhng mnh . Hy ch ra s tng ng sau

y(P1 P2) Q (P1 Q) (P1 Q).

b) S dng s tng ng trn chng minh mnh sau y: " Nu n khngchia ht cho 3 th n2 khng chia ht cho 3".

Cu 2. a) Pht biu nh ngha 4 phn ca l thuyt tin L.b) Cho cng thc A, B,C ty . Chng minh rng

((A B) (B C)) A C.

Cu 3. Cho cng thc

(A B) ((B C) ((A B) C)).

Hy thc hina) a cng thc v dng chun tc hi.b) Ch ra cng thc l hng ng.

Cu 4. a) Pht biu nh ngha th no l hng t v cng thc tn t trong

l thuyt h tn t.b) Cho v t ba bin P(x,y, z) x.y = z trn trng s thc. Xc nh gitr chn l ca mnh : (x)(y)(z)P(x,y, z) v (z)(x)(y)P(x,y, z). Dingii mnh thnh cu ni thng thng.

Cu 5. Trong mn hc gii tch ton hc ngi ta nh ngha hm lin tcnh sau: "Hm f(x) c gi l hm lin tc ti x0 D nu cho trc mt s > 0 ty th ta c c mt s > 0 tng ng sao cho vi mi x D thamn |x x0| < th |f(x) f(x0)| < ".a) Hy vit li nh ngha theo cc k hiu ca h ton tn t.

b) Hy lp mnh ph nh cho nh ngha trn (ngha l hm khng lintc ti im x0)

Tham kho cc nhn \ref{all:logic:1} l Cu1. , \ref{all:logic:3} l Cu3. , \ref{all:logic:2} l Cu 2.

3. C th dng lit k cc cu hi gi ra ri dng



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\foreachproblem[bttuluan]{\item\label{prob:\thisproblemlabel}\thisproblem}\end{enumerate}

T sau ta dng nhn

\begin{enumerate}\foreachproblem[bttuluan]{\item[\ref{prob:\thisproblemlabel}]\thisproblem}\end{enumerate}

4.4.1. Chn s ngu nhin

Ta c lnh sinh s gi ngu nhin khi dng lnh \loadrandomproblems.1. \PSNrandseed: \PSNrandseed{} S l s nguyn khc khng.

sinh ra cc s ngu nhin mi khi bin dich ta c th t u vn bn

\PSNrandseed{\time} hoc sinh ra theo s trong nm sau tip theo t\PSNrandseed{\year}

2. \PSNgetrandseed: \PSNgetrandseed{} Cha s va truynhp vo . V d

\newcount\myseed\PSNgetrandseed{\myseed}

3. \PSNrandom: \PSNrandom{}{} sinh ra mt s t nhint 1 n n, ri ghi vo . V d sinh ra mt s trong khong 1 n 10

ri ghi vo \myreg:

\newcount\myreg\PSNrandom{\myreg}{10}

4. \random: \random{}{}{} Sinh ra sngu nhin trong khong cn di n cn trn v ghi vo s m. V d

\newcounter{myrand}\random{myrand}{3}{8}

5. \doforrandN: \doforrandN{}{}{}{} Chnngu nhin n gi tr trong cch nhau bi du phy. Mi ln thchin lp th thc hin bng lnh V d gi bi t hai ca danh sch tp:

\doforrandN{2}{\thisfile}{file1,file2,file3}{%\loadrandomproblems{1}{\thisfile}}


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4.5. Mu thit lp cho tng loi cu hi

4.5.1. t lun

C 3 phng n th hin . trong tp dethituluan.tex trin khai ba kh nngny cc bn c th bin dch 1. Ch in ra thi

\loadrandomproblems[dttuluan]{4}{cauhoituluan}\hideanswers\begin{enumerate}[leftmargin=*,align=left,

resume,label={\bf Cu \arabic*.\ }]\foreachproblem[dttuluan]{\item\label{prob:\thisproblemlabel}\thisproblem}\end{enumerate}

2. Ch in ra li gii

\showanswers\begin{enumerate}[leftmargin=*,align=left,

resume,label={\bf Cu \arabic*.\ }]\foreachdataset{\thisdataset}{%

\foreachproblem[\thisdataset]{\item[\ref{prob:\thisproblemlabel}]\thisproblem}}\end{enumerate}

3. In y c v li gii

\hideproblems\showanswers\begin{enumerate}[leftmargin=*,align=left,


\foreachproblem[\thisdataset]{\item[\ref{prob:\thisproblemlabel}]\thisproblem}}

\end{enumerate}

4.5.2. trc nghim

1. Ch in ra

\loadrandomproblems[bttracnghiem]{15}{cauhoitracnghiem}\hideanswers\begin{enumerate}[leftmargin=*,align=left,


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resume,label={\bf Cu \arabic*.\ }]\foreachproblem[bttracnghiem]{\item\label{prob:\thisproblemlabel}\thisproblem}\end{enumerate}

2. In ra cu hi v p n

\showanswers\begin{enumerate}[leftmargin=*,align=left,


\foreachproblem[\thisdataset]{\item[\ref{prob:\thisproblemlabel}]\thisproblem}}\end{enumerate}

3. In ra p n ngn gn

\hideproblems\showanswers\begin{multicols}{3}\begin{enumerate}[leftmargin=*,align=left]\item[\sf Phng n] A\quad B\quad C \quad D

\foreachdataset{\thisdataset}{%\foreachproblem[\thisdataset]{\item[\ref{prob:\thisproblemlabel}]\thisproblem}

}\end{enumerate}\end{multicols}

4. In ra phiu kim tra theo

\lamtieude\begin{center}{\bf PHIU KIM TRA TRC NGHIM}\end{center}

H v tn \dotfill Lp \dotfill\hideproblems\showanswers\lamphieu\begin{multicols}{3}\begin{enumerate}[leftmargin=*,align=left]\item[\sf Phng n] A\quad B\quad C \quad D

\foreachdataset{\thisdataset}{%


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\foreachproblem[\thisdataset]{\item[\ref{prob:\thisproblemlabel}]\thisproblem}}\end{enumerate}\end{multicols}

4.5.3. in ch v ng sai

1. ch in ra

\loadrandomproblems[btdiencho]{5}{cauhoidiencho}\loadrandomproblems[btdungsai]{5}{cauhoidungsai}\hideanswers\noindent {\bf in vo ch trng}\begin{enumerate}[{\bf Cu 1.}]\foreachproblem[btdiencho]{

\item\label{prob:\thisproblemlabel}\thisproblem}\end{enumerate}\noindent {\bf Tr li ng sai}\begin{enumerate}[{\bf Cu 1.}]\foreachproblem[btdungsai]{\item\label{prob:\thisproblemlabel}\thisproblem}\end{enumerate}

2. In ra v p n

\begin{center}{\bf BI V P N }\end{center}\showanswers\noindent {\bf in vo ch trng}\begin{enumerate}\foreachproblem[btdiencho]{\item[\bf Cu \ref{prob:\thisproblemlabel}.]\thisproblem}\end{enumerate}\noindent {\bf Tr li ng sai}\begin{enumerate}\foreachproblem[btdungsai]{\item[\bf Cu \ref{prob:\thisproblemlabel}.]\thisproblem}\end{enumerate}

4.5.4. Trc nghim theo bng

1. In ra bi


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\newcounter{problem}\renewcommand{\PSNitem}{\refstepcounter{problem}%\theproblem. }\renewcommand{\endPSNitem}{ }\loadrandomproblems[btbangtn]{11}{cauhoibangtn}\setcounter{problem}{0}

\hideanswers\renewcommand{\arraystretch}{1.5}

\begin{longtable}{| p{0.7\textwidth} |c|}\hline\centering \textbf{Cu hi} & \textbf{Tr li}\\\hline %\foreachproblem[btbangtn]{\addtocounter{problem}{1}\theproblem.\thisproblem}&\\\hline\end{longtable}

2. Thay \hideanswers bng \showanswers cho bi v li gii.

4.5.5. c th thit k

1. Ta c th thit k dng cu hi

\newproblem{tab:1}{%Kt qu $(3+2)\times5$ l? &

25 \ifshowanswers\selected\else\notselected\fi &13 \notselected &10 \notselected &}{Brackets come first}%

\newproblem{tab:2}{%Kt qu $-1+2\times3$ l? &3 \notselected &-7 \notselected &5 \ifshowanswers\selected\else\notselected\fi &

}{Multiplication comes first}%2. Ri thit k cu hi tr li

\begin{longtable}{lrrrl}\bfseries Cu hi & \bfseries A & \bfseries B &\bfseries C & \ifshowanswers \bfseries Reason\fi\\\selectrandomly{cauhoithietke}{2}\end{longtable}


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Cu hi A B C

1. Kt qu (3 + 2) 5 l? 25 13 102. Kt qu 1 + 2 3 l? 3 -7 5

3. Quan st k v d ny cho ta sng to hnh thc cc cu hi khc na.

4.5.6. Bi tp cho cc chng cun sch

Ti lm mt tp baitapcuasach.tex Cch thc lm cc bi tp theo chngca sch, hoc l in theo tng chng hoc l cui cng ta in theo tng phnca bi tp trong tng chng. Cc bn tham kho tp ny v ny sinh cc tng mi.

5. a hnh v bng vo cu hi, li gii

Do cc cu hi c ly ra gn nh c nh ton b ni dung cu hi v trli, nn bng v hnh phi lun ti v tr c t vo, v vy gi lnh khngchp nhn mi trng di ng kiu nh table hay figure m c ty chn v tr.Nhng cc mi trng bnh thng nh tabular, longtable, ... v cc lnha nh vo nh \includegraphics[height=2cm,width=3cm]{tex1.eps} uc, khng thay i g. Vn l ta phi nh s cc hnh hoc bng bnglnh \caption{...} khng c, y l lnh cho mi trng ng.

khc phc hn ch trn ta dng gi lnh float.sty v dng theo mu sau c s v dng nhn c:

1. i vi bng

\begin{table}[H]\centering%\tabcaption{Ch thch bng}%\label{tab:Commands}%\begin{tabular}{c c c}*&*&*\\*&*&*\\\end{tabular}\end{table}

Ngoi ra c s bng c th dng gi lnh longtable.sty vi mi trnglongtable nhng vi \caption{...}

\begin{center}


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\begin{longtable}{c c c}*&*&*\\*&*&*\\\caption{Ch thch bng}%\label{tab:Commands}%\end{longtable}

\end{center}

2. i vi hnh

\begin{figure}[H]\centering%\includegraphics[width=0.8\linewidth,clip=]{input.eps}%\figcaption{Ch thch hnh}%\label{fig:input}%\end{figure}

Ly nhn bng \ref{fig:input} v \ref{tab:Commands}.3. V d

\baituluan{Viduhinh:1}{%Cu hi 4Hai ng gic u $ABCDE$ v $AEKPL$ trong khng gian sao cho$\widehat{DAK} =60^o$. Chng minh rng hai mt phng $ACK$v $BAL$ vung gc.\begin{figure}[H]\centering%\includegraphics[width=5cm, height=5cm]{hinh12mat.eps}\figcaption{Ch thch hnh}%\label{fig:input}%\end{figure}}{%Tr liNu ta quay $AEKPL$ quanh trc $AE$,bt u v tr trng nhau vi $ABCDE$, th gc $\widehat{DAK}$tng cho n khi $AEKPL$ li nm trn mt phng cha $ABCDE$. .....}%Ht cu hi

\begin{enumerate}[label={\bf Cu \arabic*.\ }]\showanswers\item \useproblem{Viduhinh:1}\end{enumerate}

Cu 1. Hai ng gic u ABCDE v AEKPL trong khng gian sao choDAK = 60o. Chng minh rng hai mt phng ACK v BAL vung gc.


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A

B

C

D E

KP

L

T

O

Hnh 1: Ch thch hnh

Li gii. Nu ta quay AEKPL quanh trc AE, bt u v tr trng nhau viABCDE, th gc DAK tng cho n khi AEKPL li nm trn mt phng chaABCDE. .....

6. Hng pht trin gi lnh

1. C th ni rng gi lnh ny lm tt c cc loi thi ca cc gi c. Ngoira n cn thch ng vi s thay i ca cc dng thi khc. c bit, Ti ssu tm cc dng thi th, cc bn c dng thi no c th gi cho ti cng thit k.

2. Ti liu ny gi km cc tp sau y:

baitapcuasach.texcauhoibangtn.texcauhoidiencho.texcauhoidungsai.texcauhoithietke.texcauhoitracnghiem.texcauhoituluan.texdethibangtn.texdethidiencho.texdethidungtructiep.tex

dethithietke.texdethitracnghiem.texdethituluan.texhinh1.texlamdethi.stylamdethihelp.pdfshortlst.stytitledot.sty


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3. y l cng c thu thp v s dng cc thi thnh tuyn tp rt nhanh,chnh xc v s dng chng. Nu c thi gian ti s thu thp trn mng mts lm thnh sch.


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A. Ph lc v cc gi lnh

A.1. Gi lnh enumitem.sty

Mc ch gi lnh ny l ch ng iu khin cc nhn v khong cch trongmi trng enumerate, itemmize, description. Gi lnh ny khng cng vi

gi lnh enumerate.sty, khi dng chung s bo li, trong n c y ccmi trng danh sch chun ri, ti dng gi lnh lm nhn v iu khin cccu hi vi khong cch thch hp. Khi a gi lnh \usepackage{enumitem}vo u vn bn, th c th thc hin lnh v mi trng. Khi dng mi trngc ty chn

\begin{enumerate}[]\item \end{enumerate}

Mc nh ca ty chn nh l khng c gi lnh, ta c th gn li1. Thng s cho cc khong cch ng ca danh sch:topsep, partopsep, parsep, itemsep

2. Khong cc theo chiu ngang:leftmargin, rightmargin, listparindent, labelwidth, labelsep, itemindent

3. C th t li khi thc hin mi trng

\begin{enumerate}[ leftmargin=*,itemindent=12pt, ...]\item

\end{enumerate}4. C th t li chung cho ton vn bn bng lnh\setlist{topsep=0pt, partopsep=0pt, parsep=0pt, itemsep=0pt, ...}

5. \setlist{noitemsep} b khong cch dng trong danh sch v cc dngst nhau hn.\setlist{nolistsep} tt c cc khong cch trong mi trng danh sch ucho bng 0. Dng lnh ny ko st cc cu hi trc nghim st nhau.

6. Dng vi ty chn phong ph nh sau

\begin{enumerate}[labelindent=\parindent, leftmargin=*,label=\Roman*., align=left, resume, start=8, widest=IV]

\item \end{enumerate}

(a) labelindent=\parindent Nhn ca danh sch c li vo i lngbao nhiu so vi mp tri ti liu cho pha bn phi, c th cho bng 0pt,2trucm, ...


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(b) leftmargin=* Hon ton tng t nh trn nhn li vo cho l tril bao nhiu. Ngoi cc s c th nh 2cm, c th cho bng * l gi tr mcnh.

(c) label=Roman*. nh s nhn ca danh sch, verb!label=Roman*.! chs la m I, II, III, ... cn verb!label=arabic*.! cho ch s thng dng 1, 2, 3,....Ta c th cho nhn label={\bf Cu \arabic*.\ } thm t vo trc s. ngdng iu ny cc bn xem mu ti dng c cc nhn thch hp.

(d) align=left Nhn c dong thng hng theo bn tri, v d 5 v 13th s 5 v 1 thng hng. Mc nh l thng hng bn phi.

(e) resume Cho php nh s tip tc mi trng trc , ng dng ttkhi thi c nhiu phn.

(f) lstart=8 Bt u nh s t 8, mc nh bao gi cng nh s t 1nu khng c resume.

(g) widest=IV rng ca nhn c th v d c 3 ch s c th dngwidest=000

Ta dng

\setlist{noitemsep}\setlist{nolistsep}\setlist{labelwidth=40pt, itemindent=45pt,topsep=0pt,

partopsep=0pt,parsep=0pt,leftmargin=0pt,align=right}

A.2. Gi lnh shortlst.sty

Gi lnh nhm mc ch nh s danh sch chy theo chiu ngang, ti dng gi lnh lm cc phng n cho cu hi trc nghim. Khi a\usepackage{shortlst} ta c 3 mi trng

1. shortitemize Danh sch chm trn en.

2. shortenumerate Danh sch nh s th t.

3. runenumerate Danh sch nh s chy lin lin tc bn trong mitrng v c ngoi mi trng ni lin tc.

Mt s lnh thng s iu khin mi trng ny nh sau:

1. \runitemsep khong cch gia cc danh sch c t li vi mc

nh\setlength{\runitemsep}{1em plus .5em minus .5em}.

2. \labelsep cch ch s v ch.

3. \labelwidth rng ca nhn.

4. \shortitemwidth rng ca mt ct danh sch danh sch. gi trmc nh l ly rng ca vn bn tr i cc ch nhn ri chia cho 4. Cth t li cho thch hp: \setlength{\shortitemwidth}{0.5\textwidth}


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A.3. Gi lnh float.sty

A.3.1. To ra mt mi trng ng mi

To ra mi trng c th dng mt s lnh khc nhau nh \newtheorem, ta bit dng lnh ny to ra mi trng nh l, Mnh , B , nh ngha, ...Tach l trc lnh ny thng c lnh \theoremstyle{...} iu khin nidung in nghing hoc khng nghing trong cc mi trng s dng sau ny.Hon ton tng t nh vy gi lnh float.sty c lnh lm mi trng ng\newfloat{} v trc l lnh \floatstyle{}v d

\floatstyle{plaintop}\newfloat{program}

Ri dng mi trng program nh mi trng ng vi tiu ch thch

phia trn on chng trnh. c cc t kha:1. plain Ch thch ng khng gi thay i so vi LATEX m di hnhchnh vo gia.

2. plaintop Ch thch hnh pha trn v tng t nh ty chn trn.

3. boxed Khi mi trng ng c ng khung, nhng ch thchngoi khung v pha di.

4. ruled Ch thch nm trong hai ng k ngang v cui khi cng cng k ngang nh.

V d

\floatstyle{ruled}\newfloat{Program}{htbp}{lop}[section]

ta c th dng Lnh v thng s y nh sau: required and one optional

Program A.1 y l chng trnh dng phong cch ruled.#include int main(int argc, char **argv){

int i;for (i = 0; i < argc; ++i)printf("argv[%d] = %s\n", i, argv[i]);

return 0;}

argument; it is of the form\newfloat{}{}{}[]


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1. t tn cho mi trng dng.

2. L cc quy nh n l ch ci hoc kt hp t khi: t ti u trang; b ti Cui trang; p ti Trang Di ng; h ti y, nu c th; H Ti y, dt khot nh vy.

Ch l ch c ty chn H l mi cn, cc ty chn khc ging nh mi trnghnh v bng ta thng dng.

3. Phn m rng ca. nh l *.toc.

4. Trong chapter, section, part.V d trn l

\floatstyle{ruled}\newfloat{Program}{tbp}{lop}[section]\begin{Program}\begin{verbatim}\dots\ program text \dots\end{verbatim}\caption{\dots\ caption \dots}\end{Program}

A.3.2. Nhng lnh lin quan n gi lnhMt s lnh c bn phn trn nhc ti, cn mt s lnh khc lin quan:

1. \floatname Mc nh tn mi trng l tn ch thch lun nhFigure 1.1 hoc Table 1.2 ta cng c th i tn nh v du trn t\floatname{Program}{Chng trnh}

Chng trnh A.2 y l chng trnh dng phong cch ruled.#include int main(int argc, char **argv){

int i;for (i = 0; i < argc; ++i)

printf("argv[%d] = %s\n", i, argv[i]);return 0;

}

2. \floatplacement Mc nh hnh c ch ra khi c ty chn. Nu


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ton bi t ch thch mt kiu thi t \floatplacement{figure}{tp} kiuu trang di ng.

3. \restylefloat Lnh t li phong cch ca khi ng, nh ta t licho bng ng khung

\floatstyle{boxed}

\restylefloat{table}\begin{table}[H] \def\B#1{$\displaystyle{n\choose#1}$}\begin{center} \begin{tabular}{c|cccccccc}$n$&\B0&\B1&\B2&\B3&\B4&\B5&\B6&\B7\\ \hline

0 & 1\\1 & 1&1\\2 & 1&2&1\\3 & 1&3&3&1\\4 & 1&4&6&4&1\\

5 & 1&5&10&10&5&1\\6 & 1&6&15&20&15&6&1\\7 & 1&7&21&35&35&21&7&1

\end{tabular} \end{center}\caption{Pascals triangle. This is a re-styled \LaTeX\ \texttt{table}.%

\label{table1}}\end{table}

n n0 n

1 n

2 n

3 n

4 n

5 n

6 n

7

0 1

1 1 1

2 1 2 1

3 1 3 3 1

4 1 4 6 4 1

5 1 5 10 10 5 1

6 1 6 15 20 15 6 17 1 7 21 35 35 21 7 1

Bng 2: Pascals triangle. This is a re-styled LATEX table.

4. \listof \listof{}{} to ra danh schcc mi trng s dng c cng tn v thm tiu trn.Ging nhlnh \listoffigures v \listoftables


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A.3.3. S dng gi lnh

1. Khi dng gi lnh ny mi trng table v figure gi nguyn gi tr v tngthm kh nng, v d thm ty chn c nh [H], dng

\floatstyle{plaintop}

\restylefloat{figure}t li ch thch hnh v bng.

\begin{figure}[H]\centering\includegraphics[height=4cm,width=6cm]{banco1.eps}\caption{Dng graphicx}\label{fig:}\end{figure}

Hnh 2: Dng graphicx

2. Nhiu lp hoc mi trng c nh dng c vi ty chn [H] mkhng b bo li.

3. Sng to ra cc khi nh

\floatstyle{ruled}\newfloat{vidu}{htbp}{lop}[section]\floatname{vidu}{V d}\begin{vidu}$$A^2=B^2+C^2$$\caption{y l v d hay}

\end{vidu}

V d A.1 y l v d hay

A2 = B2 + C2


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A.4. Gi lnh nonfloat.sty

Trong ch LaTeX bnh thng th mi trng table v figure lun bngv hnh trong ch di ng, ngha l vi thng s

\begin{figure}[!ht]

\centering\includegraphics[height=2cm,width=3cm]{*.eps}\caption{}\label{fig:}\end{figure}

th hnh c th t ti v tr c lnh nu cn ch khng th chuyn sang utrang sau hoc v cui bi. Nhiu lp hoc gi lnh khng dng ch ngny, gi lnh nonfloat.sty p ng yu cu ny, nhng khng dng c mitrng table v figure na, m phi thay i mt cht. ng l l

\begin{table}[htbp]\caption{Table Caption}%\label{tab:supertitle}%\begin{tabular}{...}...\end{tabular}\end{table}

ta thay bng

\begin{minipage}{\linewidth}\centering%\tabcaption{Commands for Table and Figure Captions}%\label{tab:Commands}%\begin{tabular}{c l c }...\end{tabular}\end{minipage}

Hnh c thay bng

\begin{minipage}{\linewidth}\centering%\includegraphics[width=0.8\linewidth,clip=]{input.eps}%\figcaption{Figure Caption}%\label{fig:input.eps}%\end{minipage}


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A.5. Gi lnh ifthen.sty

Gi lnh lp trnh iu khin c ni y trong cun sch ca ti.

A.6. Cc gi lnh son thi hoc cu hi kim tra khc

1. answers.sty Gi lnh son cu hi v tr li lin tc nhng thi thchin c th in cu hi ring v tr li ring, ti hng dn lm sch theo ccchng. Trang web ca ti c bi ring v gi lnh ny. Bn tham kho nguynbn ti ti a chwww.ctan.org/tex-archive/macros/latex/contrib/answers/

2. probsoln.sty Gi lnh tao ra thi m ti s dng lm ra gilnh ny. Nguyn bn ch c cc mi trng n gin t lun, trc nghim ths. Gi lnh c ti a chhttp://tug.ctan.org/tex-archive/macros/latex/contrib/probsoln/

3. dethi.sty cng vi vic s i examdesign.cls to ra trc nghimv mt s loi thi vi s so trn c cu hi ln phng n tr li. Ch dnglm c lp, tuy rt mnh v c s dng nhiu. Gi lnh c nitrong trang web ca ti v km vo vi Chng trnh VieTeX.http://nhdien.wordpress.com

4. alterqcm.sty lm thi theo dng bng. C ti a chhttp://tug.ctan.org/tex-archive/macros/latex/contrib/alterqcm/

Ti liu

[1] Nguyn Hu in,Nguyn Minh Tun, LaTeX tra cu v son tho,NXBHQG, 2001.

[2] Nguyn Hu in, LaTeX gi lnh v phn mm cng c, NXBHQG, 2004.

[3] Cc a ch gi lnh c lit k trn

lamdethi.sty định dạng các loại đề thi

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