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Egalitate
Posted: Sun Feb 08, 2009 3:25 pm
by Claudiu Mindrila
Fie \( a,b\in \mathbb{R} \) astfel incat \( a^2(a-2b)=b^2(b-2a) \). Aratati ca \( a=b \).
Claudiu Mindrila, G.M.-B. 10/2007
Posted: Sun Feb 08, 2009 9:50 pm
by Marius Mainea
Relatia este echivalenta cu \( (a-b)(a^2-ab+b^2)=0 \)
1) Daca a=0 atunci b=0 deci a=b.
2) daca \( a\neq 0 \) atunci a doua paranteza este nenula deci a-b=0
Posted: Thu Mar 05, 2009 12:24 pm
by BurnerD1
Dupa inmultire se ajunge la \( a^3 - 2a^{2}b = b^3 - 2ab^2 \ \ \ \)
deci \(
a^3 - b^3 = 2a^{2}b - 2ab^2 \)
dupa cum stim \( a^n - b^n = (a-b)\cdot (a^{n-1} + a^{n-2}b + a^{n-3}b^3 + ... + ab^{n-2} + b^{n-1}) \)
deci \( a^3 - b^3 =(a-b)(a^2 + ab + b^2) \)
vom obtine \( (a-b)(a^2 + ab + b^2)= 2a^2b - 2ab^2 \)
Imi explica si mie cineva de unde putem sti ca \( 2a^b - 2ab^2 = 0 \) nestiind ca \( a=b=0 \)
Posted: Thu Mar 05, 2009 12:56 pm
by BogdanCNFB
pai \( (a-b)(a^2ab+b^2)=2a^2b-2ab^2\Leftrightarrow (a-b)(a^2+ab+b^2)=2ab(a-b)\Leftrightarrow (a-b)(a^2+ab+b^2-2ab)=0\Leftrightarrow (a-b)(a^2-ab+b^2)=0... \)
Posted: Thu Mar 05, 2009 3:09 pm
by BurnerD1
corect...
