Laboratoriya ishi 5 Mavzu: Matematik formulalar bilan ishlash Ishdan maqsad

921-19 guruh talabasi
Qalandarova Mumtozbegimning
Latexga kirish fanidan


Laboratoriya ishi - 5


Mavzu: Matematik formulalar bilan ishlash


Ishdan maqsad: Dasturda matematik formulalarni yozishni o’rganish
Nazariy qism: Matematikada ko’p hollarda grek harflaridan foydalaniladi.Shu sababli biz
ham LATEXda matematik formula kiritishni grek harflarini kiritishdan
boshlaymiz. LATEXda grek harflarini kiritish buyrug’i “\” belgisi va shu belgining
inglizcha nomini yozish orqali kiritiladi(Masalan:

harfi \alpha kabi kiritiladi).Shu
o’rinda yana bir ma’lumotni aytib o’tish kerak.Grek harflari ro’yhatidan

(“omikron” deb o’qiladi) harfini bu usul bilan kiritib bo’lmaydi(Ya’ni \omikron
deb yozish no’to’g’ri hisoblanadi).Bu harfni kiritish uchun kursivda yozilgan
lotincha “o” harfi,yoki odatdagidek o harfini kiritish kifoya.Misol tariqasida bir
necha grek harflarining LATEXda yozilishini jadvalini keltiramiz.




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belgilar vazifasini tushuntiring





Topshiriq natijasi:

\documentclass[12pt,a4paper]{article}

\usepackage[utf8]{inputenc}
\usepackage[T1]{fontenc}
\usepackage[english]{babel}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{makeidx}
\usepackage{graphicx}
\usepackage{xcolor}
\author{Qalandarova Mumtozbegim}
\title{Matematika}

\begin{document}\huge

\maketitle
\begin{center}
\textbf{Matematik simvollar}
\end{center}
\pagecolor{green}
\begin{math}
\alpha\
\beta\
\gamma\
\omega\
\delta\
\pi\
\xi\
\psi\
+\
\odot\
\ominus\
\cdot\
\frac{3}{8}\\\\
\sqrt{x^3}=x\\\\
\textbf{Matematik oddiy formulalar}\\\\
\frac{a^2+a-3} {1-3a^{-1}}\cdot\left [\frac{(a+2)^2-a^2} {4a^2-4}-\frac{3}{a^2-a}\right] \,.
\end{math}\\\\
\huge\bf1.\;\( 1+\left(\frac{1}{1-x^{2}}\right)^3 \) \\\\
\bf2.\;$D=\sqrt{b^{2}-4ac}$\\\\
\bf3.\;$\pi\approx 3{,}14$\\\\
\bf4.\;$\int\limits_0^1 x^2 dx=1/3$\\\\
\bf5.\;{\color{blue}\(A=\pi{r^2}\)}\\\\
\textbf{Matematik Murakkab formulalar}\\\\
\bf6.\;$\prod\nolimits_{i=1}^ni=n!$ \\\\
\bf7.\;\( \frac{25}{36}=(\frac{1}{1+\frac{1}{5}})^2 \)\\\\
\bf8.\;\((x+a)^n=\sum_{k=0}^n\)\\\\
\end{document}