<?xml version="1.0" encoding="utf-8"?>
<?xml-stylesheet type="text/xsl" href="../../css/rss2full.xsl"?>
<rss version="2.0"  xmlns:atom="http://www.w3.org/2005/Atom" xmlns:openSearch="http://a9.com/-/spec/opensearchrss/1.0/" xmlns:georss="http://www.georss.org/georss" xmlns:thr="http://purl.org/syndication/thread/1.0">
<channel>
<title>zuvluguu</title>
<link>https://mat-2011.coo.mn/</link>

<atom:link href="https://mat-2011.coo.mn/feeds/posts/" rel="self" type="application/rss+xml" />
<description>matematikiin hiucheeliin tuslamj</description>
<pubDate>Sun, 05 Apr 2026 04:17:20 +0800</pubDate>
<generator>BlogMN feed writer</generator>
<language>mn-mn</language>
<copyright>Copyright (c) 2026 zuvluguu (https://mat-2011.coo.mn/). All rights reserved.</copyright>
<image>
		<url>//coo.mn/images/logo_s.png</url>
		<title>zuvluguu</title>
		<link>https://mat-2011.coo.mn/</link>
		<description>coo.mn</description>
		</image>
<webMaster>admin@coo.mn (Webmaster)</webMaster>
<item><title>zurh</title><link>https://mat-2011.coo.mn/44311/zurh.html</link><guid>https://mat-2011.coo.mn/44311/zurh.html</guid><description><![CDATA[<br /> <a href="http://www.slide.com/pivot?cy=lt&amp;at=un&amp;id=1585267068861737716&amp;map=1" target="_blank"><img border="0" alt="" src="http://widget-f4.slide.com/p1/1585267068861737716/lt_t062_v000_s0un_f00/images/xslide1.gif" ismap="ismap" /></a> <a href="http://www.slide.com/pivot?cy=lt&amp;at=un&amp;id=1585267068861737716&amp;map=2" target="_blank"><img border="0" alt="" src="http://widget-f4.slide.com/p2/1585267068861737716/lt_t062_v000_s0un_f00/images/xslide2.gif" ismap="ismap" /></a> <a href="http://www.slide.com/pivot?cy=lt&amp;at=un&amp;id=1585267068861737716&amp;map=F" target="_blank"><img border="0" alt="" src="http://widget-f4.slide.com/p4/1585267068861737716/lt_t062_v000_s0un_f00/images/xslide42.gif" ismap="ismap" /></a>]]></description><comments>https://mat-2011.coo.mn/set_bichih.php?w=mat-2011&amp;amp;e_id=44311</comments><pubDate>Mon, 21 Feb 2011 17:48:00 +0800</pubDate><author> no_email@coo.mn (ealtanbayar)</author></item>
<item><title></title><link>https://mat-2011.coo.mn/44304/</link><guid>https://mat-2011.coo.mn/44304/</guid><description><![CDATA[<br /> <a target="_blank" href="http://www.slide.com/pivot?cy=lt&amp;at=un&amp;id=1585267068861737716&amp;map=1"><img border="0" ismap="ismap" src="http://widget-f4.slide.com/p1/1585267068861737716/lt_t062_v000_s0un_f00/images/xslide1.gif" alt="" /></a> <a target="_blank" href="http://www.slide.com/pivot?cy=lt&amp;at=un&amp;id=1585267068861737716&amp;map=2"><img border="0" ismap="ismap" src="http://widget-f4.slide.com/p2/1585267068861737716/lt_t062_v000_s0un_f00/images/xslide2.gif" alt="" /></a> <a target="_blank" href="http://www.slide.com/pivot?cy=lt&amp;at=un&amp;id=1585267068861737716&amp;map=F"><img border="0" ismap="ismap" src="http://widget-f4.slide.com/p4/1585267068861737716/lt_t062_v000_s0un_f00/images/xslide42.gif" alt="" /></a>]]></description><comments>https://mat-2011.coo.mn/set_bichih.php?w=mat-2011&amp;amp;e_id=44304</comments><pubDate>Mon, 21 Feb 2011 17:40:00 +0800</pubDate><author> no_email@coo.mn (ealtanbayar)</author></item>
<item><title></title><link>https://mat-2011.coo.mn/44292/</link><guid>https://mat-2011.coo.mn/44292/</guid><description><![CDATA[<br /> <a href="http://www.slide.com/pivot?cy=lt&amp;at=un&amp;id=1585267068861737716&amp;map=1" target="_blank"><img border="0" alt="" src="http://widget-f4.slide.com/p1/1585267068861737716/lt_t062_v000_s0un_f00/images/xslide1.gif" ismap="ismap" /></a> <a href="http://www.slide.com/pivot?cy=lt&amp;at=un&amp;id=1585267068861737716&amp;map=2" target="_blank"><img border="0" alt="" src="http://widget-f4.slide.com/p2/1585267068861737716/lt_t062_v000_s0un_f00/images/xslide2.gif" ismap="ismap" /></a> <a href="http://www.slide.com/pivot?cy=lt&amp;at=un&amp;id=1585267068861737716&amp;map=F" target="_blank"><img border="0" alt="" src="http://widget-f4.slide.com/p4/1585267068861737716/lt_t062_v000_s0un_f00/images/xslide42.gif" ismap="ismap" /></a>]]></description><comments>https://mat-2011.coo.mn/set_bichih.php?w=mat-2011&amp;amp;e_id=44292</comments><pubDate>Mon, 21 Feb 2011 17:32:00 +0800</pubDate><author> no_email@coo.mn (ealtanbayar)</author></item>
<item><title></title><link>https://mat-2011.coo.mn/44288/</link><guid>https://mat-2011.coo.mn/44288/</guid><description><![CDATA[<br /> <a href="http://www.slide.com/pivot?cy=lt&amp;at=un&amp;id=1585267068861737716&amp;map=1" target="_blank"><img border="0" alt="" src="http://widget-f4.slide.com/p1/1585267068861737716/lt_t062_v000_s0un_f00/images/xslide1.gif" ismap="ismap" /></a> <a href="http://www.slide.com/pivot?cy=lt&amp;at=un&amp;id=1585267068861737716&amp;map=2" target="_blank"><img border="0" alt="" src="http://widget-f4.slide.com/p2/1585267068861737716/lt_t062_v000_s0un_f00/images/xslide2.gif" ismap="ismap" /></a> <a href="http://www.slide.com/pivot?cy=lt&amp;at=un&amp;id=1585267068861737716&amp;map=F" target="_blank"><img border="0" alt="" src="http://widget-f4.slide.com/p4/1585267068861737716/lt_t062_v000_s0un_f00/images/xslide42.gif" ismap="ismap" /></a>]]></description><comments>https://mat-2011.coo.mn/set_bichih.php?w=mat-2011&amp;amp;e_id=44288</comments><pubDate>Mon, 21 Feb 2011 17:30:00 +0800</pubDate><author> no_email@coo.mn (ealtanbayar)</author></item>
<item><title>zurag</title><link>https://mat-2011.coo.mn/44282/zurag.html</link><guid>https://mat-2011.coo.mn/44282/zurag.html</guid><description><![CDATA[<br />
<a target="_blank" href="http://www.slide.com/r/gb2s304G5D91xFgGuP0p3oIH8NfeCrJz?previous_view=mscd_embedded_url&amp;view=original">http://www.slide.com/r/gb2s304G5D91xFgGuP0p3oIH8NfeCrJz?previous_view=mscd_embedded_url&amp;view=original<br />
</a>]]></description><comments>https://mat-2011.coo.mn/set_bichih.php?w=mat-2011&amp;amp;e_id=44282</comments><pubDate>Mon, 21 Feb 2011 17:25:00 +0800</pubDate><author> no_email@coo.mn (ealtanbayar)</author></item>
<item><title>монгол улс</title><link>https://mat-2011.coo.mn/44178/mongol-uls.html</link><guid>https://mat-2011.coo.mn/44178/mongol-uls.html</guid><description><![CDATA[<br />
<a title="Altanbayar" href="http://www.slideshare.net/zob_bayar/altanbayar">Altanbayar</a>



View more <a href="http://www.slideshare.net/">presentations</a> from <a href="http://www.slideshare.net/zob_bayar">zob_bayar</a>.
]]></description><comments>https://mat-2011.coo.mn/set_bichih.php?w=mat-2011&amp;amp;e_id=44178</comments><pubDate>Mon, 21 Feb 2011 16:24:00 +0800</pubDate><author> no_email@coo.mn (ealtanbayar)</author></item>
<item><title>Тригонометр функц-3</title><link>https://mat-2011.coo.mn/44154/trigonomyetr-funkts-3.html</link><guid>https://mat-2011.coo.mn/44154/trigonomyetr-funkts-3.html</guid><description><![CDATA[<p class="title">Тригнометр Функц (3-р хэсэг)</p>       <p class="theorem">Туслах Теорем<br />         <br />         Хэрвээ h функц нь 2-р эрэмбийн уламжлалтай (бvх бодит тооны хувьд), ба<br />         <br />         h'' + h = 0,<br />         h(0) = 0,<br />         h'(0) = 0.<br />         <br />         бол h = 0. (h нь налуу функц байна.)</p>       <p class="body">Баталгаа</p>       <p class="body">1-р тэнцэтгэлийг нь h' - ээр vржvvлвэл</p>       <p align="center" class="body">h'h'' + h'h = 0</p>       <p class="body">гарна. Тэгэхлээр</p>       <p align="center" class="body">[(h')2 + h2]' = 0 = 2(h'h'' + h'h)</p>       <p class="body">байна. Өөрөөр хэлбэл (h')2 + h2 чинь          налуу функц байна. (Хэрвээ уламжлал нь бvх бодит тооны хувьд 0 бол функц          нь налуу байдаг.) h(0) = 0, h'(0) = 0 гэсэн болохлээр</p>       <p align="center" class="body">[h'(x)]2 + [h(x)]2          = 0 (бvх бодит х - ийн хувьд)</p>       <p class="body">Энэ нь h(x)=0 налуу функц болохыг хэлж байна.<img width="5" height="1...   <br><br><a href="https://mat-2011.coo.mn/44154/trigonomyetr-funkts-3.html">[дэлгэрэнгүй]</a>]]></description><comments>https://mat-2011.coo.mn/set_bichih.php?w=mat-2011&amp;amp;e_id=44154</comments><pubDate>Mon, 21 Feb 2011 16:05:00 +0800</pubDate><author> no_email@coo.mn (ealtanbayar)</author></item>
<item><title>Тригонометр функц-2</title><link>https://mat-2011.coo.mn/44150/trigonomyetr-funkts-2.html</link><guid>https://mat-2011.coo.mn/44150/trigonomyetr-funkts-2.html</guid><description><![CDATA[<p class="title">Тригнометрийн Функц (2-р хэсэг)</p>       <p class="body">Бид sin(x), cos(x) функцvvдийн графикийг уламжлалыг нь ашиглаж          байгаад хялбархан зурж болно. Тодорхойлогдох муж нь бvх бодит тоо, утгын          муж нь [-1,1] билээ.</p>       <p align="center" class="body"><img border="1" width="461" height="123" src="http://www.asuult.net/achbold/images/graphs/trig/sin.gif" alt="" /></p>       <p align="center" class="body"><img border="1" width="461" height="123" src="http://www.asuult.net/achbold/images/graphs/trig/cos.gif" alt="" /></p>       <p class="defenition">Бvх х - ийн хувьд f(x+a) = f(x) бол f функцийг vет          функц гэж нэрэлдэг. a - г f функцийн vе гэнэ.</p>       <p class="body">Бидний vзсэн тригнометрийн хоёр функц vет функц болохыг          харж болно. Vе нь 2<img align="absbottom" width="12" height="12" src="http://www.asuult.net/achbold/images/eq/trig/pi.gif" alt="" />          болно.</p>       <p class="body">sin(x), cos(x) хоёрыг л мэдэж байхад бусад тригнометрийн    ...   <br><br><a href="https://mat-2011.coo.mn/44150/trigonomyetr-funkts-2.html">[дэлгэрэнгүй]</a>]]></description><comments>https://mat-2011.coo.mn/set_bichih.php?w=mat-2011&amp;amp;e_id=44150</comments><pubDate>Mon, 21 Feb 2011 16:02:00 +0800</pubDate><author> no_email@coo.mn (ealtanbayar)</author></item>
<item><title>Тригонометр-1</title><link>https://mat-2011.coo.mn/42847/trigonomyetr-1.html</link><guid>https://mat-2011.coo.mn/42847/trigonomyetr-1.html</guid><description><![CDATA[<p class="title">Тригнометрийн Функц</p>       <p class="body">Энэ хичээл дээрээ бид тригнометрийн функцvvдийг  талбайгаар          тодорхойлно. (Ихэнх сурах бичгvvд тойргийн уртыг ашигладаг.) Бас  энэ хичээлийг          ойлгоход <a href="http://www.asuult.net/achbold/radians.htm">радиан  хэмжигдхvvн, нэгж тойргийн</a> мэдлэг          шаардагдана.</p>       <p class="body">Тойргийн талбай A=<img align="absbottom" width="12" height="12" src="http://www.asuult.net/achbold/images/eq/trig/pi.gif" alt="" />r2,          тойргийн урт C=2<img align="absbottom" width="12" height="12" src="http://www.asuult.net/achbold/images/eq/trig/pi.gif" alt="" />r          (r нь радиус) байдгийг санавал нэгж тойргийн урт 2<img align="absbottom" width="12" height="12" src="http://www.asuult.net/achbold/images/eq/trig/pi.gif" alt="" />,          талбай нь <img align="absbottom" width="12" height="12" src="http://www.asuult.net/achbold/images/eq/trig/pi.gif" alt="" />          байна.</p>       <p class="body">Хэрвээ х тойргийн урттай...   <br><br><a href="https://mat-2011.coo.mn/42847/trigonomyetr-1.html">[дэлгэрэнгүй]</a>]]></description><comments>https://mat-2011.coo.mn/set_bichih.php?w=mat-2011&amp;amp;e_id=42847</comments><pubDate>Wed, 16 Feb 2011 17:34:00 +0800</pubDate><author> no_email@coo.mn (ealtanbayar)</author></item>
<item><title>Ньютон-Лейбницийн томъёо</title><link>https://mat-2011.coo.mn/42845/niyuton-lyeibnitsiin-tomyoo.html</link><guid>https://mat-2011.coo.mn/42845/niyuton-lyeibnitsiin-tomyoo.html</guid><description><![CDATA[<p class="title">Ньютон - Лейбницийн Теорем</p>       <p class="body">Хоорондоо ямар ч холбоогvй мэт харагдах уламжлал  интеграл          хоёрыг холбодог энэ теорем нь гайхалтай гэмээр энгийн. Баталгаа  нь ч хялбар.</p>       <p class="body">Бид гол зорьлогоо дахин нэг хэлнэ: функцийг дээд,  доод нийлбэрvvдийг          нь олно гэж ядаргаатахгvйгээр интегралчихал. (f(x)=x2  функцийн          интегралыг тэгэж олох ямар хэцvv байсныг санаж байна уу?) Бид  өмнө нь          F(x) функцийг тодорхойлж байсан. Хэрвээ энэ функцийн хялбар  томъёог олоодохвол          интегралыг хялбарханаар олох гээд байна. Гэхдээ яг F(x) функцрvv  дайраад          юу ч олж долоохгvй. Харин F - ийн уламжлалруу дайрвал яах бол?</p>       <p class="theorem">Ньютон - Лейбницийн Теорем (1-р хэсэг)<br />         <br />         [a, b] завсар дээр интегралтай f функц байг. Бас<img align="absmiddle" width="96" height="44" src="http://www.asuult.net/achbold/images/eq/t_f_t_o_c/int_f.gif" alt="" />          гэе. Хэрвээ f нь [a, b] завсарын дурын ...   <br><br><a href="https://mat-2011.coo.mn/42845/niyuton-lyeibnitsiin-tomyoo.html">[дэлгэрэнгүй]</a>]]></description><comments>https://mat-2011.coo.mn/set_bichih.php?w=mat-2011&amp;amp;e_id=42845</comments><pubDate>Wed, 16 Feb 2011 17:33:00 +0800</pubDate><author> no_email@coo.mn (ealtanbayar)</author></item>
</channel></rss>