[Date Prev][Date Next][Thread Prev][Thread Next][Date Index][Thread Index]

Re: [VietTUG] Bác nào hướng dẫn giúp em cài Tex trên Mac OS X với!!!




Em đính kèm file để anh xem giúp em luôn ạ. 
\documentclass[10pt]{article}
\usepackage[utf8]{vietnam}
\usepackage{amsmath,amssymb,amsfonts}
\usepackage{array}
\usepackage{ntheorem}
\usepackage{answers}
\usepackage{fancyhdr}
\usepackage{multicol}
\usepackage{enumerate}
\usepackage{graphicx,picinpar}
\usepackage[unicode]{hyperref}
\usepackage[top=1.cm, bottom=1.cm, left=2.5cm, right=2.cm] {geometry}

\theorembodyfont{\rmfamily}
\theoremseparator{.}
\Newassociation{sol}{Solution}{ans}
\newtheorem{ex}{B?i}
\renewcommand{\solutionstyle}[1]{\textbf{#1}.}

\newcounter{dapan}
\def\dapan{\stepcounter{dapan}{\bf\Alph{dapan}}}
\newcommand{\boncot}[4]{\setlength{\parindent}{0.2cm}
\par\indent\setcounter{dapan}{0}\begin{tabular}{p{4.2cm}p{4.2cm}p{4.2cm}p{4.2cm}}
\dapan.\ #1.&\dapan.\ #2.&\dapan.\ #3.&\dapan.\ #4.
\end{tabular}
}
\newcommand{\haicot}[4]{
\setlength{\parindent}{0.2cm}
\par\indent\setcounter{dapan}{0}\begin{tabular}{p{8.8cm}p{8.8cm}}
\dapan.\ #1.&\dapan.\ #2.\\
\dapan.\ #3.&\dapan.\ #4.
\end{tabular}
}
\newcommand{\motcot}[4]{\setlength{\parindent}{0.4cm}
\par\setcounter{dapan}{0}
\indent\dapan.\ #1.\\
\indent\dapan.\ #2.\\
\indent\dapan.\ #3.\\
\indent\dapan.\ #4.
}

\newcounter{made}
\def\made{\stepcounter{made}{\arabic{made}}}
\def\chumade{c?a m? ??}
\newcounter{madapan}
\def\madapan{\stepcounter{madapan}{\arabic{madapan}}}
\newcounter{socau}
\def\socau{\stepcounter{socau}{\arabic{socau}}}

\renewcommand{\baselinestretch}{0.5}

\def\NN{{\mathbb N}}
\def\QQ{{\mathbb Q}}
\def\RR{{\mathbb R}}
\def\ZZ{{\mathbb Z}}

\def\vec{\overrightarrow}
\columnsep=24.6pt

\begin{document}
\pagestyle{plain}
\centerline{\bf\large PH??NG PH?P T? S? TH? T?CH}
\vspace{0.5cm}

\begin{ex}
Cho t? di?n $ABCD$ c? th? t?ch $V$. Tr?n c?c c?nh $AB, AC$ v? $AD$ l?n l??t l?y c?c ?i?m $B', C', D'$ sao cho $B'$ l? trung ?i?m c?a $AB, AC'=\dfrac 13AC$, $AD'=\dfrac 14AD$. T?nh th? t?ch t? di?n $AB'C'D'$.
\end{ex}

\begin{ex}
Cho t? di?n ??u $ABCD$ c?nh $a$. L?y c?c ?i?m $B', C'$ tr?n c?c c?nh $AB$ v? $AC$ sao cho $AB'=\dfrac a2, AC'=\dfrac{2a}3$. T?nh th? t?ch t? di?n $AB'C'D$. ?S: $V=\frac{a^3\sqrt 2}{36}$.
\end{ex}
\begin{ex}
Cho h?nh ch?p $S.ABC$ c? $SA=a, SB=b, SC=c, \widehat{ASB}=60^0, \widehat{BSC}=90^0$ v? $\widehat{CSA}=120^0$. T?nh th? t?ch h?nh ch?p. ?S: $\frac{abc\sqrt 2}{12}$.
\end{ex}

\begin{ex}
Cho kh?i ch?p t? gi?c ??u $S.ABCD$ c?nh $a$, g?c gi?a m?t b?n v? ??y l? $\alpha$. G?i $M$ l? trung ?i?m c?a c?nh $SC$, m?t ph?ng $(MAB)$ c?t $SD$ t?i $N$. T?nh th? t?ch h?nh ch?p $S.ABMN$. ?S: $V=\frac 1{16}a^3\tan\alpha$.
\end{ex}

\begin{ex}
Cho h?nh ch?p ??u $S.ABCD$ c? ??y $ABCD$ l? h?nh vu?ng, ???ng ch?o $AC=2\sqrt 2$, ???ng cao $SH=\sqrt 3$. T?nh th? t?ch kh?i ch?p ??nh $S$, ??y l? thi?t di?n c?a h?nh ch?p t?o b?i m?t ph?ng $(P)$ qua $H$ v? vu?ng g?c v?i $SC$. ?S: $V=\frac{2\sqrt 3}5$. 
\end{ex}

\begin{ex}
Cho h?nh ch?p $S.ABCD$ c? ??y $ABCD$ l? h?nh vu?ng c?nh $a$, $SA$ vu?ng g?c v?i m?t ph?ng ??y v? $SA=2a$. G?i $B', D'$ l? h?nh chi?u c?a $A$ l?n $SB$ v? $SD$ t??ng ?ng. M?t ph?ng $(AB'D')$ c?t $SC$ t?i $C'$. Ch?ng minh $SC\perp (AB'D')$ v? t?nh th? t?ch h?nh ch?p $S.AB'C'D'$. ?S: $V=\frac{16a^3}{45}$.
\end{ex}

\begin{ex}
Cho h?nh ch?p $S.ABCD$ c? ??y l? h?nh b?nh h?nh. G?i $I, J$ l?n l??t l? trung ?i?m c?a $SA, SD$ v? $(P)$ l? m?t ph?ng ch?a c?nh $BC$ v? $IJ$. T?nh t? s? th? t?ch c?a hai kh?i ?a di?n do m?t ph?ng $(P)$ chia ra.?S:$\frac35$
\end{ex}

\begin{ex}
Cho h?nh ch?p $S.ABCD$ c? ??y $ABCD$ l? h?nh vu?ng c?nh $a$, ???ng cao $SA=h$. G?i $N$ l? trung ?i?m c?a $SC$. M?t ph?ng $\alpha$ ch?a $AN$ v? song song v?i $BD$ l?n l??t c?t $SB, SD$ t?i $M$ v? $P$. T?nh t? s? $\dfrac{SP}{SD}$ v? th? t?ch h?nh ch?p $S.AMNP$. ?S: $\frac 23, V=\frac{a^2h}9$.
\end{ex}

\begin{ex}
Cho h?nh ch?p t? gi?c ??u $S.ABCD$ c?nh b?ng $a$. M?t ph?ng $(\alpha)$ qua $A$ vu?ng g?c v?i $SC$ c?t $SB, SC, SD$ l?n l??t t?i $B', C', D'$. Bi?t r?ng $\dfrac{SB'}{SB}=\dfrac 23$.
\begin{enumerate}[a.]
\item T?nh t? s? th? t?ch c?a hai kh?i ch?p $S.AB'C'D'$ v?i $S.ABCD$. ?S: $\frac 13$.
\item T?nh th? t?ch kh?i ch?p $S.AB'C'D'$. ?S: $\frac{\sqrt 6a^3}{18}$.
\end{enumerate}
\end{ex}

\begin{ex}
Cho ch?p t? gi?c ??u $S.ABCD$. G?i $M, N, P$ l?n l??t l? trung ?i?m c?a $AB, AD$ v? $SC$. T?nh t? s? th? t?ch c?a hai ph?n h?nh ch?p ???c chia b?i m?t ph?ng $(MNP)$. ?S: $1$.
\end{ex}

% \begin{ex}
% Cho h?nh ch?p $S.ABC$ c? $G$ l? tr?ng t?m tam gi?c $ABC$. M?t m?t ph?ng $(\alpha)$ b?t k? c?t $SA, SB, SC$ v? $SG$ t?i $A', B', C', D'$. Ch?ng minh r?ng $\dfrac{SA}{SA'}+\dfrac{SB}{SB'}+\dfrac{SC}{SC'}=3\dfrac{SG}{SG'}$.
% \end{ex}

\begin{ex}
Cho t? di?n $ABCD$ v? $M$ l? m?t ?i?m b?t k? b?n trong t? di?n. K? hi?u $h_A, h_B, h_C, h_D$ l?n l??t l? kho?ng c?ch t? c?c ??nh $A, B, C, D$ ??n c?c m?t ??i di?n v? $m_A, m_B, m_C, m_D$ l? kho?ng c?ch t? ?i?m $M$ ??n c?c m?t t??ng ?ng. Ch?ng minh r?ng $\dfrac{m_A}{h_A}+\dfrac{m_B}{h_B}+\dfrac{m_C}{h_C}+\dfrac{m_D}{h_D}=1$.
\end{ex}


\begin{ex}
Cho h?nh ch?p ??u $S.ABCD$ c?nh ??y l? $a$. M?t ph?ng qua $AB$ v? trung ?i?m $M$ c?a c?nh $SC$ h?p v?i ??y g?c $\alpha$.
\begin{enumerate}[a.]
\item T?nh th? t?ch h?nh ch?p $S.ABCD$. ?S: $\frac{a^3}2\tan\alpha$.
\item G?i $I$ l? trung ?i?m $AB$ v? $J\in[BC]$ sao cho $JB=2JC$. M?t ph?ng qua $IJ$ v? vu?ng g?c v?i ??y chia h?nh ch?p th?nh hai ph?n. T?nh th? t?ch m?i ph?n ??. ?S: $\frac 4{21}$.
\end{enumerate}
\end{ex}

\begin{ex}
Cho ch?p tam gi?c $SABC$. G?i $M, P$ l? trung ?i?m c?a $SA, BC$; $N$ l? ?i?m thu?c $AB$ sao cho $\dfrac{AN}{AB}=\dfrac 13$.
\begin{itemize}
\item [a.] T?m $Q$ l? giao ?i?m c?a $SC$ v?i m?t ph?ng $(MNP)$.
\item [b.] CMR m?t ph?ng $(MNP)$ chia ch?p th?nh 2 ph?n c? th? t?ch b?ng nhau. 
\end{itemize}
\end{ex}

\begin{ex}
Cho h?nh ch?p $SABC$, l?y $M\in SA, N\in SB$ sao cho 
$$\dfrac{SM}{MA}=\dfrac 12; \dfrac{SN}{NB}=2.$$
\begin{itemize}
\item [a.] D?ng thi?t di?n qua $MN$ v? song song v?i $SC$. 
\item [b.] T?m t? l? th? t?ch 2 ph?n c?a kh?i ch?p b? chia b?i thi?t di?n. ?S: $\frac 54$. 
\end{itemize}
\end{ex}

% \begin{ex}
% Cho h?nh ch?p t? gi?c ??u $S.ABCD$ c?nh ??y b?ng $a$, g?c gi?a m?t b?n v? ??y l? $\alpha$. T?nh th? t?ch kh?i ch?p theo $a$ v? $\alpha$. ?S: $V=\frac 1{6}a^3\tan\alpha$.
% \end{ex}
% 
% \begin{ex}
% Cho h?nh ch?p $S.ABC$ c? ??y l? tam gi?c vu?ng c?n t?i $A, BC=2a, SA=SB=SC=3a$. T?nh th? t?ch kh?i ch?p v? kho?ng c?ch t? ?i?m $C$ ??n m?t ph?ng $(SAB)$.
% \end{ex}
% 
% \begin{ex}
% Cho h?nh ch?p $S.ABCD$ c? c?c c?nh b?n $SA=SB=SD=\dfrac{a\sqrt 3}2$, ??y l? h?nh thoi c?nh $a$ c? g?c $\widehat A=60^0$. T?nh th? t?ch h?nh ch?p. 
% \end{ex}
% 
% \begin{ex}
% Cho h?nh ch?p $S.ABCD$ c? ??y $ABCD$ l? h?nh thang, $\widehat{BAD}=\widehat{ABC}=90^0, AB=BC=a, AD=2a, SA=2a$ v? $SA$ vu?ng g?c v?i ??y. G?i $M, N$ l?n l??t l? trung ?i?m c?a $SA, SD$. Ch?ng minh $BCNM$ l? h?nh ch? nh?t v? t?nh th? t?ch h?nh ch?p $S.BCNM$ theo $a$. ?S: $V=\frac{a^3}3$.
% \end{ex}
% 
% \begin{ex}
% Cho h?nh ch?p $S.ABC$ c? ??y $ABC$ l? tam gi?c vu?ng c?n t?i $B$ v?i $BA=BC=2a$, h?nh chi?u vu?ng g?c c?a $S$ tr?n m?t ph?ng $(ABC)$ tr?ng v?i trung ?i?m $E$ c?a $AB$ v? $SE=2a$. G?i $I, J$ l?n l??t l? trung ?i?m c?a $EC$ v? $SC$. G?i $M$ l? ?i?m tr?n tia ??i c?a tia $BA$ sao cho g?c $\widehat{ECM}=45^0$. G?i $H$ l? h?nh chi?u vu?ng g?c c?a $S$ tr?n $MC$. T?nh th? t?ch h?nh ch?p $EHIJ$ theo $a$. ?S: $V=\frac{5a^3}{24}$.
% \end{ex}
% 
% \begin{ex}
% Cho kh?i t? di?n ??u $ABCD$ c?nh $a$. 
% \begin{enumerate}[a.]
% \item T?nh th? t?ch c?a kh?i ch?p theo $a$. ?S: $\frac{a^3\sqrt 2}{12}$.
% \item Ch?ng minh r?ng b?n tr?ng t?m c?a c?c m?t l?p th?nh m?t kh?i t? di?n ??u. T?nh th? t?ch kh?i t? di?n ??. 
% \end{enumerate}
% \end{ex}
% 
% \begin{ex}
% Cho ch?p t? gi?c ??u $S.ABCD$ c? c?nh ??y b?ng $a$ v? $\widehat{ASB}=\alpha$. T?nh di?n t?ch xung quanh v? th? t?ch c?a h?nh ch?p. ?S: $S_{xq}=a^2\cot\frac{\alpha}2$
% \end{ex}
% 
% \begin{ex}
% Cho h?nh ch?p $S.ABCD$ c? ??y $ABCD$ l? h?nh vu?ng c?nh $a, SA=a$. H?nh chi?u vu?ng g?c $H$ c?a ??nh $S$ tr?n m?t $ABCD$ thu?c ?o?n $AC$ v?i $AH=\dfrac{AC}4$. G?i $CM$ l? ???ng cao c?a tam gi?c $SAC$. Ch?ng minh $M$ l? trung ?i?m c?a $SA$ v? t?nh th? t?ch h?nh ch?p $S.BMC$ theo $a$. ?S: $V=\frac{a^3\sqrt{14}}{48}$.
% \end{ex}
% 
% \begin{ex}
% Cho h?nh ch?p $S.ABCD$ c? ??y $ABCD$ l? h?nh vu?ng c?nh $a$, m?t b?n $(SAB)$ vu?ng g?c v?i ??y v? $SAB$ l? tam gi?c ??u. T?nh th? t?ch h?nh ch?p.
% \end{ex}
% 
% \begin{ex}
% Cho h?nh ch?p tam gi?c $S.ABC$ c? ??y $ABC$ l? tam gi?c ??u c?nh $a$; hai m?t b?n $(SAB)$ v? $(SAC)$ vu?ng g?c v?i ??y; $SA=\dfrac{3a}2$
% \begin{enumerate}[a.]
% \item  T?nh th? t?ch h?nh ch?p. ?S: $\frac{a^3\sqrt 3}{8}$.
% \item T?nh di?n t?ch xung quanh v? di?n t?ch to?n ph?n c?a h?nh ch?p. ?S: $S_{xq}=\frac{a^2(3+\sqrt 3)}2, S_{tp}=\frac{3a^2}4(1+\sqrt 3)$.
% \end{enumerate}
% \end{ex}
% 
% \begin{ex}
% Cho h?nh ch?p $S.ABC$ c? ??y l? tam gi?c $ABC$ vu?ng t?i $B, BA=3a, BC=4a$. M?t ph?ng $(SBC)$ vu?ng g?c v?i ??y. Bi?t $SB=2a\sqrt 3, \widehat{SBC}=30^0$. T?nh th? t?ch h?nh ch?p $S.ABC$ v? kho?ng c?ch t? $B$ ??n $(SAC)$. ?S: $V=2a^3\sqrt 3, h=\frac{6a}{\sqrt 7}$.
% \end{ex}
\end{document}