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aqu4

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Post Posted: 08-11-2008, 15:34:50 | Translate post to: ... (Click for more languages)

Nu sunt bun la mate..care stie sa faca asta are un 10=))
a+b=10
(a+b)²=70

Calculati a³+b³=? o rezolvare completa pls mult -

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Levier
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Post Posted: 08-11-2008, 18:24:37 | Translate post to: ... (Click for more languages)

pai ba, a+b totu la patrat se rezolva prin a patrat plus 2 ori a ori b plus b patrat.

si a + b totu la cub inseamna a cub + 3 ori a patrat ori b + 3 ori a ori b patrat + b cub.

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ion

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Post Posted: 08-11-2008, 18:34:52 | Translate post to: ... (Click for more languages)

a+b=10 => (a+b)(a+b)=100 in contradictie cu (a+b)patrat=70 => problema stupida


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Malpraxxiss

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Post Posted: 08-11-2008, 23:01:10 | Translate post to: ... (Click for more languages)

ion.mzk wrote:
a+b=10 => (a+b)(a+b)=100 in contradictie cu (a+b)patrat=70 => problema stupida


Corect.

Cred ca problema era asa:

a+b=10

a^2 + b^2 = 70

a^3 + b^3 = ?


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Neo

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Post Posted: 09-11-2008, 00:05:48 | Translate post to: ... (Click for more languages)

nerds -


joking -

/extrem: +1 warnin pentru post inutil.

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Levier
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Post Posted: 09-11-2008, 04:56:29 | Translate post to: ... (Click for more languages)

am gresit formula.
@!Star#&!@#( mama cui a inchinat-o prima oara...

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Andiku

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Post Posted: 09-11-2008, 11:44:36 | Translate post to: ... (Click for more languages)

daca a + b = 10
(a+b)^2 = a^2 + 2ab + b^2 = 70

=> a = 10 - b
-> (10-b)^2 + 2b(10-b) + b^2 = 70
100 - 20b + b^2 + 20b - 2b^2 + b^2 = 70 ... daca esti atent se simplifica tot si ramane 100 = 70 ( ceea ce nu este adevarat , deci ori e gresita problema ori ai scris-o tu gresit )

cat despre formulele ridicarii la anumite puteri..ia si citeste de aci , poate asa le retii -
(a+b)^2 = a^2 + 2ab + b^2
(a-b)^2 = a^2 - 2ab + b^2
(a+b)(a-b) = (a-b)(a+b) = a^2 - b^2
(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3
(a-b)^3 = a^3 - 3a^2b + 3ab^2 - b^3
a^3 + b^3 = (a+b)(a^2 - ab + b^2)
a^3 - b^3 = (a-b)(a^2 + ab + b^2)
(a+b+c)^2 = a^2 + b^2 + c^2 + 2ab + 2ac + 2bc
(a+b-c)^2 = a^2 + b^2 + c^2 + 2ab - 2ac - 2bc
(a-b-c)^2 = a^2 + b^2 + c^2 - 2ab - 2ac + 2bc



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aqu4

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Post Posted: 12-11-2008, 19:16:02 | Translate post to: ... (Click for more languages)

Malpraxxiss wrote:
ion.mzk wrote:
a+b=10 => (a+b)(a+b)=100 in contradictie cu (a+b)patrat=70 => problema stupida


Corect.

Cred ca problema era asa:

a+b=10

a^2 + b^2 = 70

a^3 + b^3 = ?



ms moolt

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^^aMaT0Rr <3 ^^

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Post Posted: 01-12-2008, 17:52:02 | Translate post to: ... (Click for more languages)

aqu4 wrote:
Nu sunt bun la mate..care stie sa faca asta are un 10=))
a+b=10
(a+b)²=70

Calculati a³+b³=? o rezolvare completa pls mult -






(a+b)2=70
asta inseamna ca a+b da un numar care la puterea a doua da +70

si a3+b3 poti sa faci mai multe variane :exemplu + schimbi ( a = 1 ) iar (b + 2) si calculezi 1 la a doua da 1 + 8 ( 2la a 3-a) si poti sa dai mai multe exemple .... astea sunt exerciti de a 5-a -

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^^aMaT0Rr <3 ^^

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Post Posted: 01-12-2008, 17:53:02 | Translate post to: ... (Click for more languages)

aqu4 wrote:
Nu sunt bun la mate..care stie sa faca asta are un 10=))
a+b=10
(a+b)²=70

Calculati a³+b³=? o rezolvare completa pls mult -






(a+b)2=70
asta inseamna ca a+b da un numar care la puterea a doua da +70

si a3+b3 poti sa faci mai multe variane :exemplu = schimbi ( a = 1 ) iar (b + 2) si calculezi 1 la a doua da 1 + 8 ( 2 la a 3-a) si poti sa dai mai multe exemple .... astea sunt exerciti de a 5-a -

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