Monthly Archives: ខែសីហា 2009

តោះ!លេងហ្គេមម្តង!!

Posted on ខែសីហា 23, 2009 | មតិ 6


Pc game Caesar III Highly compressed | 88mb only.

Caesar III: Build a Better Rome | Real-Time Strategy (Seira online) game | 88 MB


Emperor Caesar has assigned you to one of Rome’s provinces and it’s up to you to create a safe, pleasant city that will attract immigrants. Watch your city grow and evolve, and maintain its economy to keep the favor of your citizens… as well as Caesar. Play your cards right and you could become Emperor of Rome itself!

Game Features:
* Build, rule and defend on one screen – no more switching between city, province and battle screens!
* Use the City Construction Kit to build the perfect city, or climb the ladder of Roman politics.
* Now your citizens can give you a glimpse into the life of the common man – talk to them for clues about how to improve your city.
* Appease 5 Gods with temples and festivals. Each God has its own sphere of influence: Please Ceres, and your crops will thrive but dishonour Neptune, and watch your trade ships sink into the sea…
* Various new structures and challenges arise throughout your Governor’s career, for tremendous depth of play and replayabiltiy.
* A truly intuitive, helpful interface lets you jump right in and start building!

System requirements
Windows 95/98/98SE/ME/xp/Vista
Pentium 933 or better
128MB of RAM
32MB graphics/video card

Download these parts:

បើចូលចិត្តប្រាប់ផង!!!

6 មតិ

បានផ្សាយ​ក្នុង ហ្គេម

Download CSI Sap2000 v14 full

Posted on ខែសីហា 20, 2009 | មតិ 14
CSI Sap2000 v14 full

The SAP name has been synonymous with state-of-the-art analytical methods since its introduction over 30 years ago. SAP2000 follows in the same tradition featuring a very sophisticated, intuitive and versatile user interface powered by an unmatched analysis engine and design tools for engineers working on transportation, industrial, public works, sports, and other facilities.

From its 3D object based graphical modeling environment to the wide variety of analysis and design options completely integrated across one powerful user interface, SAP2000 has proven to be the most integrated, productive and practical general purpose structural program on the market today.

This intuitive interface allows you to create structural models rapidly and intuitively without long learning curve delays. Now you can harness the power of SAP2000 for all of your analysis and design tasks, including small day-to-day problems. Complex Models can be generated and meshed with powerful Templates built into the interface.

The Advanced Analytical Techniques allow for Step-by-Step Large Deformation Analysis, Multiple P-Delta, Eigen and Ritz Analyses, Cable Analysis, Tension or Compression Only Analysis, Buckling Analysis, Blast Analysis, Fast Nonlinear Analysis for Dampers, Base Isolators and Support Plasticity, Energy Methods for Drift Control and Segmental Construction Analysis.

Bridge Designers can use SAP2000 Bridge Templates for generating Bridge Models, Automated Bridge Live Load Analysis and Design, Bridge Base Isolation, Bridge Construction Sequence Analysis, Large Deformation Cable Supported Bridge Analysis and Pushover Analysis.

SAP2000 is for everyone! SAP2000 is for every project! From a simple small 2D static frame analysis to a large complex 3D nonlinear dynamic analysis, SAP2000 is the answer to all structural analysis and design needs.

Download all these parts:

14 មតិ

បានផ្សាយ​ក្នុង New Software

Download BricsCad Pro 9.3.10 Portable

Posted on ខែសីហា 13, 2009 | មតិ 9
BricsCad Pro 9.3.10 Portable

BricsCad v9 is the smart choice for architects, engineers,designers-virtually any professional who creates or uses CAD drawings.BricsCad provides unrivaled compatibility with Autodesk AutoCAD and is fully programmable with hundreds of third-party solutions. AutoCAD command line Most CAD professionals spend years mastering AutoCAD commands until they’re second nature.Since these commands help create the majority of CAD work,BricsCad transparently maps them to BricsCad equivalent commands.

DWG COMPATIBILITY
– Native DWG 2007 file format
– Support for AutoCAD versions 2.5 to 2008
– CUI menus and toolbars
– Extended raster image support
– User data file manager
– New LISP engine
– Full SDS support
– New highly compatible COM API
– Support for Object ARX

PRODUCTIVITY
– Powerful Drawing Explorer
– Dockable Properties Bar
– Consolidated Settings Manager
– Customize dialog and editor
– Full ACIS (3D solids) modeling
– Visual Basic for Applications (VBA)

Download

Enjoy!!!

9 មតិ

បានផ្សាយ​ក្នុង New Software

ប្រយ័ត្នច្រឡំ!!!

Posted on ខែសីហា 11, 2009 | មតិ 10

គណនា\mathop {\lim }\limits_{x \to \infty } {{\sin x} \over x}

សំរាយ

\forall x \in \Re _ +យើងមាន- 1 \le \sin x \le 1

ដូចនេះ- {1 \over x} \le {{\sin x} \over x} \le {1 \over x}

\Rightarrow \mathop {\lim }\limits_{x \to \infty } \left( { - {1 \over x}} \right) \le \mathop {\lim }\limits_{x \to \infty } {{\sin x} \over x} \le \mathop {\lim }\limits_{x \to \infty } {1 \over x}

តែ\mathop {\lim }\limits_{x \to \infty } \left( { - {1 \over x}} \right) = \mathop {\lim }\limits_{x \to \infty } {1 \over x} = 0

ដូច្នេះ\mathop {\lim }\limits_{x \to \infty } {{\sin x} \over x} = 0

ខ្មាសគេមិនស្ទើរដោយសារលំហាត់នឹង​ ទៅជាស្រាយថាបាន១អីណា៎ យាប់ម៉ង :mrgreen:

10 មតិ

បានផ្សាយ​ក្នុង គណិតវិទ្យា, ពិជគណិត

ធរណីមាត្រម្តង!!!

Posted on ខែសីហា 5, 2009 | មតិ 36

គេអោយចតុកោណកែងABCDមួយដែលBC=3AB។PនិងQជាពីរចំនុចនៃជ្រុង[BC]ដែលBP=PQ=QC។ចូរ​បង្ហាញថា\angle DQC = \angle DBC + \angle DPC។

ដំណោះស្រាយ

បង្ហាញថា\angle DQC = \angle DBC + \angle DPC

1

តាង\angle DBC= \alpha ; \angle DPC = \beta ; \angle DQC = \lambda

យើងមាន\tan \alpha = {1 \over 3} , \tan \beta = {1 \over 2} , \tan \lambda= 1

ដោយ\tan \left( {\alpha + \beta } \right) = {{\tan \alpha + \tan \beta } \over {1 - \tan \alpha \tan \beta }} = {{{1 \over 2} + {1 \over 3}} \over {1 - {1 \over 6}}} = 1 = \tan \lambda

ដូចនេះ\alpha +\beta = \lambdaឬ\angle DQC = \angle DBC + \angle DPC។

36 មតិ

បានផ្សាយ​ក្នុង គណិតវិទ្យា, ធរណីមាត្រ

លំហាត់កំសាន្ត!!!

Posted on ខែសីហា 3, 2009 | បញ្ចេញមតិ

១.គេអោយបីចំនួនពិតវិជ្ជមានa,bនិតc។​ចូរបង្ហាញថាa^2 + b^2 + c^2 \ge ab + bc + ca។

ដំណោះស្រាយ

បង្ហាញថាa^2 + b^2 + c^2 \ge ab + bc + ca

តាមវិសមភាពAM-GMគេមាន៖

{{a^2 + b^2 } \over 2} \ge ab​ (1)

{{b^2 + c^2 } \over 2} \ge bc (2)

{{c^2 + a^2 } \over 2} \ge ca (3)

បូកវិសមភាព (1), (2) និង ​(3) អង្គនឹងអង្គគេបាន៖

{{a^2 + b^2 + b^2 + c^2 + c^2 + a^2 } \over 2} \ge ab + bc + ca

ដូចនេះa^2 + b^2 + c^2 \ge ab + bc + ca។

២.​ គេអោយចំនួនពិតវិជ្ជមានaនិងb។ ចូរបង្ហាញថា\left( {1 + a} \right)\left( {1 + b} \right) \ge \left( {1 + \sqrt {ab} } \right)^2

អនុវត្តន៍ រកតំលៃតូចបំផុតនៃអនុគមន៍៖

f\left( x \right) = \left( {1 + 4^{\sin ^2 x} } \right)\left( {1 + 4^{\cos ^2 x} } \right)ដែលx \in \Re។

ដំណោះស្រាយ

បង្ហាញថា\left( {1 + a} \right)\left( {1 + b} \right) \ge \left( {1 + \sqrt {ab} } \right)^2

យើងមាន\left( {1 + a} \right)\left( {1 + b} \right) = 1 + \left( {a + b} \right) + ab(1)

តាមវិសមភាពAM-GMគេមានa + b \ge 2\sqrt {ab}

តាម(1)គេទាញបាន\left( {1 + a} \right)\left( {1 + b} \right) = 1 + 2\sqrt {ab} + ab

ដោយ1 + 2\sqrt {ab} + ab = \left( {1 + \sqrt {ab} } \right)^2

ដូចនេះ\left( {1 + a} \right)\left( {1 + b} \right) \ge \left( {1 + \sqrt {ab} } \right)^2

ដោយប្រើវិសមភាពខាងលើយើងទាញបាន៖

f\left( x \right) = \left( {1 + 4^{\sin ^2 x} } \right)\left( {1 + 4^{\cos ^2 x} } \right) \ge \left( {1 + \sqrt {4^{\sin ^2 x + \cos ^2 x} } } \right)^2 = 9

ដូចនេះតំលៃអប្បបរិមានៃអនុគមន៍នេះស្មើនឹង9។

បញ្ចេញមតិ

បានផ្សាយ​ក្នុង គណិតវិទ្យា, ពិជគណិត

លំហាត់ពិសេស!!!

Posted on ខែសីហា 3, 2009 | បញ្ចេញមតិ

េគអោយពហុធាដឹក្រេទី៤៖

p\left( x \right) = 2009x^4 + 2004x^3 + 2007x^2 + 2003x + 2005ស្រាយបញ្ជាក់ថាp\left( x \right) > 0គ្រប់x \in \mathbb{R}។

ដំណោះស្រាយ

-ស្រាយបញ្ជាក់ថាp\left( x \right) > 0គ្រប់x \in \mathbb{R}

+ករណីx \geqslant 0នោះយើងឃើញថាp\left( x \right) \ge 2005 \ge 0ពិត។

+ករណីx < 0

យើងមាន

p\left( x \right) = 2009x^4 + 2004x^3 + 2007x^2 + 2003x + 2005

p\left( x \right) = 5x^4+2004\left( {x^4 + x^3 + x^2 + x + 1} \right) +3x^2-x+1

p\left( x \right) = 5x^4+3x^2 + \left( { - x} \right)+1+2004\left( {x^4+x^3+x^2+x+1} \right)

ចំពោះកន្សោមx^4+ x^3+ x^2+ x+1

= \left( {x^4+ x^3+{{x^2 } \over 4}} \right) + \left( {{{x^2 } \over 4} + x + 1} \right) + {{x^2 } \over 2}

= x^2 \left( {x + {1 \over 2}} \right)^2+ \left( {{x \over 2} + 1} \right)^2+ {{x^2 } \over 2} > 0

នាំអោយp\left( x \right) > 0ពិត

ដូចនេះp\left( x \right) > 0គ្រប់x \in \mathbb{R}។

បញ្ចេញមតិ

បានផ្សាយ​ក្នុង គណិតវិទ្យា, ពិជគណិត

លំហាត់ជុំវិញពិភពលោក(៤)

Posted on ខែសីហា 3, 2009 | បញ្ចេញមតិ

គេអោយnចំនួនពិតវិជ្ជមានa_1 ;a_2 ;a_3 ; \ldots ;a_nដែលផល​គុណa_1 a_2 a_3 \ldots a_n = 1។

ចូរស្រាយបញ្ជាក់ថា\left( {1 + a_1 } \right)\left( {1 + a_2 } \right) \ldots\left( {1 + a_n } \right) \geqslant 2^n

ដំណោះស្រាយ

ស្រាយបញ្ជាក់ថា\left( {1 + a_1 } \right)\left( {1 + a_2 } \right) \ldots\left( {1 + a_n } \right) \geqslant 2^n

តាមវិសមភាពAM-GM

1 + a_1 \geqslant 2\sqrt {a_1 }

1 + a_2 \geqslant 2\sqrt {a_2 }

1 + a_3 \geqslant 2\sqrt {a_3 }

------------

1 + a_n \geqslant 2\sqrt {a_n }

គេបាន\left( {1 + a_1 } \right)\left( {1 + a_2 } \right) \ldots \left( {1 + a_n} \right) \ge 2^n \sqrt {a_1 a_2 a_3 \ldots a_n }

ដោយa_1 a_2 a_3 \ldots a_n = 1

ដូចនេះ\left( {1 + a_1 } \right)\left( {1 + a_2 } \right) \ldots \left( {1 + a_n } \right) \geqslant 2^n។

បញ្ចេញមតិ

បានផ្សាយ​ក្នុង គណិតវិទ្យា, ពិជគណិត

លំហាត់ជុំវិញពិភពលោក(២)

Posted on ខែសីហា 3, 2009 | បញ្ចេញមតិ

ចូរបង្ហាញថា\tan 3a - \tan 2a - \tan a = \tan 3a\tan 2a\tan aដែល​a \ne {{k\pi } \over 2}គ្រប់ចំនួនគត់រ៉ឺឡាទីបk។

ដំណោះស្រាយ

បង្ហាញថា\tan 3a - \tan 2a - \tan a = \tan 3a\tan 2a\tan a

គេមាន\tan 3a = \tan \left( {2a + a} \right)

\tan 3a = {{\tan 2a + \tan a} \over {1 - \tan 2a\tan a}}

\tan 3a\left( {1 - \tan 2a\tan a} \right) = \tan 2a + \tan a

\tan 3a - \tan 3a\tan 2a\tan a = \tan 2a + \tan a

ដូចនេះ\tan 3a - \tan 2a - \tan a = \tan 3a\tan 2a\tan a។

បញ្ចេញមតិ

បានផ្សាយ​ក្នុង គណិតវិទ្យា, ពិជគណិត

លំហាត់ជុំវិញពិភពលោក(1)

Posted on ខែសីហា 3, 2009 | បញ្ចេញមតិ

ពីរចំនួនវិជ្ជមានxនិងyផ្ទៀងផ្ទាត់ទំនាក់ទំនង4x+3y=11ចូរ​​កំនត់​រក​តំលៃ​​អតិបរិមា​នៃ​អនុគមន៍៖

f\left( {x,y} \right) = \left( {x + 6} \right)\left( {y + 7} \right)\left( {3x + 2y} \right)

ដំណោះស្រាយ

កំនត់រកតំលៃអតិបរមានៃអនុគមន៍៖

f\left( {x,y} \right) = \left( {x + 6} \right)\left( {y + 7} \right)\left( {3x + 2y} \right)

តាមវិសមភាពAM-GM (cauchy)យើងបាន៖

{{\left( {x + 6} \right) + \left( {y + 7} \right) + \left( {3x + 2y} \right)} \over 3} \ge \root 3 \of {\left( {x + 6} \right)\left( {y + 7} \right)\left( {3x + 2y} \right)}

{{4x + 3y + 13} \over 3} \ge \root 3 \of {f\left( {x,y} \right)}

គេទាញf\left( {x,y} \right) \le \left( {{{4x + 3y + 13} \over 3}} \right)^3ដោយ4x+3y=11

គេបានf\left( {x,y} \right) \le \left( {{{11 + 13} \over 3}} \right)^3=512។

ដូចនេះតំលៃអតិបរមានៃអនុគមន៍f(x,y)ស្មើនឹង512។

បញ្ចេញមតិ

បានផ្សាយ​ក្នុង គណិតវិទ្យា, ពិជគណិត