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Babacu
 Post subject: Re: Foarte tare!
PostPosted: 09 Aug 2010, 09:26 
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Măiestrie

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Jaggguar
 Post subject: Re: Foarte tare!
PostPosted: 19 Aug 2010, 18:56 
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istv4n
 Post subject: Re: Foarte tare!
PostPosted: 25 Aug 2010, 09:01 
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Daca vi se pare ca stati mult in trafic: Link :)

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MainEvent
 Post subject: Re: Foarte tare!
PostPosted: 25 Aug 2010, 09:23 
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Da, oamenii din zonă au prosperat cu ocazia incidentului. S-au apucat să vândă d-ale gurii.
Chiar se gândeau să facă spectacole ca să îi destreseze pe şoferi :lol:

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KALI
 Post subject: Re: Foarte tare!
PostPosted: 02 Sep 2010, 09:19 
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Moto...

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andreiblitz
 Post subject: Re: Foarte tare!
PostPosted: 02 Sep 2010, 12:13 
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KALI Quote:
Moto...
Foarte tare!! :naughty:

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dobermann
 Post subject: Re: Foarte tare!
PostPosted: 02 Sep 2010, 17:29 
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interesant fenomenul... :run:

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mobyle
 Post subject: Re: Foarte tare!
PostPosted: 12 Sep 2010, 22:06 
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A=4
B=5
C=1
C = B-A / (B-A)
C(B-A) = (B-A)2
CB-CA = B2 – 2AB + A2 / -A2
CB-CA- A2 = B2 – 2AB / +AB
AB+ CB-CA- A2= B2 – AB / -CB
AB-CA- A2= B2 – CB-AB
A(B-C-A) = B(B-C-A) / 1:(B-C-A)
A = B
4 = 5

A2, respectiv B2 este A, resp. B la puterea 2

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TwinTwin
 Post subject: Re: Foarte tare!
PostPosted: 13 Sep 2010, 07:57 
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[quote="mobyle"]
C = B-A / (B-A)
C(B-A) = (B-A)2
:nono:

C=B-A/(B-A) <=> C(B-A)=B-A
sunt convins ca nu e greseala ta ci asa ai preluat-o!

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mobyle
 Post subject: Re: Foarte tare!
PostPosted: 13 Sep 2010, 08:46 
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TwinTwin Quote:
mobyle Quote:
C = B-A / (B-A)
C(B-A) = (B-A)2
:nono:

C=B-A/(B-A) <=> C(B-A)=B-A
sunt convins ca nu e greseala ta ci asa ai preluat-o!

Problema e ca nu ai inteles operatia matematica: "\ (B-A)" inseamna ca atat membrul stang cat si cel drept al ecuatiei se inmulteste cu (B-A)

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mox_is
 Post subject: Re: Foarte tare!
PostPosted: 13 Sep 2010, 11:55 
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TwinTwin
 Post subject: Re: Foarte tare!
PostPosted: 13 Sep 2010, 13:51 
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mobyle Quote:
TwinTwin Quote:
mobyle Quote:
C = B-A / (B-A)
C(B-A) = (B-A)2
:nono:

C=B-A/(B-A) <=> C(B-A)=B-A
sunt convins ca nu e greseala ta ci asa ai preluat-o!

Problema e ca nu ai inteles operatia matematica: "\ (B-A)" inseamna ca atat membrul stang cat si cel drept al ecuatiei se inmulteste cu (B-A)

Crede-ma ca am inteles-o dar cel care a facut operatia initiala a uitat de (B-A) de la numitor cu care se reduce un (B-A) de la numarator. E un fel de alba-neagra cu matematica care se bazeaza tot pe neatentia privitorului.

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mobyle
 Post subject: Re: Foarte tare!
PostPosted: 13 Sep 2010, 14:13 
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Este operatiunea de inmultire cu (B-A) si nu impartire, doar am scris / (B-A) si nu / 1:(B-A). Pe tine te-a derutat semnul "/" care inseamna ca se aplica aceeasi operatiune atat in partea dreapta cat si in cea stanga a ecuatiei. Matematica, numai alba-neagra nu e.

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ex: Alfa Romeo 159 1.9 jtd grigio vesuvio
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andreiblitz
 Post subject: Re: Foarte tare!
PostPosted: 13 Sep 2010, 14:17 
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Dar nici "operatiune" nu e. :lol:

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mobyle
 Post subject: Re: Foarte tare!
PostPosted: 13 Sep 2010, 14:27 
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andreiblitz Quote:
Dar nici "operatiune" nu e. :lol:
Vad ca matale te-ai blocat inainte sa faci primul rationament.

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ex: Alfa Romeo 159 1.9 jtd grigio vesuvio
ex: Alfa Romeo 166 Super 2.0 tb grigio chiaro metallizzato
ex: Alfa Romeo 156 1.8 ts rosso proteo
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TwinTwin
 Post subject: Re: Foarte tare!
PostPosted: 13 Sep 2010, 16:40 
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mobyle Quote:
Este operatiunea de inmultire cu (B-A) si nu impartire, doar am scris / (B-A) si nu / 1:(B-A). Pe tine te-a derutat semnul "/" care inseamna ca se aplica aceeasi operatiune atat in partea dreapta cat si in cea stanga a ecuatiei. Matematica, numai alba-neagra nu e.
Ai dreptate, am interpretat gresit "/". Rest my case!

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Alexandrus
 Post subject: Re: Foarte tare!
PostPosted: 14 Sep 2010, 06:30 
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Fascinant, intrigant, nemaipomenit. Acum puteti, va rog, inceta ?
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Soul_assassin
 Post subject: Re: Foarte tare!
PostPosted: 14 Sep 2010, 12:38 
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Am facut curatenie, rezolvati-va problemele pe privat va rog.

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gabrieldima
 Post subject: Re: Foarte tare!
PostPosted: 15 Sep 2010, 19:17 
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probabil dupa ore in sir de filmare si cateva bodykituri sfaramate , ramane insa o realizare de exceptie zic eu a montajului , cat despre sofer... foarte tare frate :D

http://www.youtube.com/watch?v=4TshFWSs ... r_embedded" onclick="window.open(this.href);return false;

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silencer
 Post subject: Re: Foarte tare!
PostPosted: 15 Sep 2010, 19:52 
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Speechless!
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Cristi84
 Post subject: Re: Foarte tare!
PostPosted: 15 Sep 2010, 19:54 
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Asta chiar se potriveste la "Foarte tare". Mi se par foarte interesante cele 2 faze in care se roteste cu masina aproape pe loc :blink: Asta da sofer, cred ca mi-ar desfiinta masina in 5 minute :lol: . Bafta!

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ALXege
 Post subject: Re: Foarte tare!
PostPosted: 15 Sep 2010, 21:00 
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Nu puteai bre sa zici ca-i vorba de Ken Block la telefon? Ca stiam la ce te referi atunci :lol: Foarte tare, intr-adevar!

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istv4n
 Post subject: Re: Foarte tare!
PostPosted: 16 Sep 2010, 09:18 
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Asta da "shitty job" :)

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------
When the fast car picks you up, you will weep and smile,
And see Heaven in the headlights,
Mile after mile, after mile, after mile..
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Dany
 Post subject: Re: Foarte tare!
PostPosted: 16 Sep 2010, 10:00 
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mobyle Quote:
A=4
B=5
C=1
C = B-A / (B-A)
C(B-A) = (B-A)2
CB-CA = B2 – 2AB + A2 / -A2
CB-CA- A2 = B2 – 2AB / +AB
AB+ CB-CA- A2= B2 – AB / -CB
AB-CA- A2= B2 – CB-AB
A(B-C-A) = B(B-C-A) / 1:(B-C-A)
A = B
4 = 5

A2, respectiv B2 este A, resp. B la puterea 2
A(B-C-A) = B(B-C-A) / 1:(B-C-A) operatia asta nu e posibila - impartire cu 0, de fapt rezultatul fiind 0=0.

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mobyle
 Post subject: Re: Foarte tare!
PostPosted: 16 Sep 2010, 19:38 
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Dany Quote:
mobyle Quote:
A=4
B=5
C=1
C = B-A / (B-A)
C(B-A) = (B-A)2
CB-CA = B2 – 2AB + A2 / -A2
CB-CA- A2 = B2 – 2AB / +AB
AB+ CB-CA- A2= B2 – AB / -CB
AB-CA- A2= B2 – CB-AB
A(B-C-A) = B(B-C-A) / 1:(B-C-A)
A = B
4 = 5

A2, respectiv B2 este A, resp. B la puterea 2
A(B-C-A) = B(B-C-A) / 1:(B-C-A) operatia asta nu e posibila - impartire cu 0, de fapt rezultatul fiind 0=0.
corect: nederminare, impartire cu zero :clapping:

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ex: Alfa Romeo 159 1.9 jtd grigio vesuvio
ex: Alfa Romeo 166 Super 2.0 tb grigio chiaro metallizzato
ex: Alfa Romeo 156 1.8 ts rosso proteo
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