Exercitiu mate

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AlexGooDz
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20 Oct 2013, 20:40

Salut, am un exercitiu care imi da de cap care nu am mai fac de un ( cls 8 )

-2cos30+(3tg60-2tg45)=
-2√3+(3√3-2)
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2

Daca mi-l puteti face voi de aici.
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aNNakin
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20 Oct 2013, 20:50

Ce trebuie sa afli? in orice caz cos30 e ^3 pe 2, 2 cu -2 se simplifica si ramana -^3 + 3*^3 - 2, rezultatul final fiind 2 (^3 - 1) daca dai factor comun
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X2060
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20 Oct 2013, 21:01

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BrZ;x
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20 Oct 2013, 22:16

cos30 = √3/2
tg60 = √3
tg45 = 1

-2cos30+(3tg60-2tg45)=-2√3+(3√3-2)
-2√3/2+(3√3-2)=-2√3+(3√3-2) care este = -√3+2√3+3√3-2-3√3+2 = √3

@aNNakin nu inteleg ce ai facut acolo.
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super maN
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20 Oct 2013, 23:14

aNNakin wrote:Ce trebuie sa afli? in orice caz cos30 e ^3 pe 2, 2 cu -2 se simplifica si ramana -^3 + 3*^3 - 2, rezultatul final fiind 2 (^3 - 1) daca dai factor comun
Aprob.
cos 30=√3/2
tg 60=√3
tg 45=1
= -2√3/2+3√3-2
= 2√3-2
= 2(√3-1)
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aNNakin
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21 Oct 2013, 14:55

BrZ;x wrote:cos30 = √3/2
tg60 = √3
tg45 = 1

-2cos30+(3tg60-2tg45)=-2√3+(3√3-2)
-2√3/2+(3√3-2)=-2√3+(3√3-2) care este = -√3+2√3+3√3-2-3√3+2 = √3

@aNNakin nu inteleg ce ai facut acolo.
exercitiu e de forma -2 ori cos30, adica -2 ori ^3/2, -2 cu 2 se simplifica si ramana -^3...ce nu e de inteles?
Last edited by BrZ;x on 21 Oct 2013, 22:01, edited 1 time in total.
Reason: El a postat acel exercitiu in forma asta : -2cos30+(3tg60-2tg45)= -2√3+(3√3-2) ,cel putin asa am inteles eu si m-a indus in eroare.
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