Vasiliu Costel Andrei at 10:39, 13 January 2025

2025-01-13T10:39:47Z

← Older revision		Revision as of 10:39, 13 January 2025
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			'''[[E:15346]] (Mihai Vijdeluc și Vasile Ienuțaș, Baia Mare)'''

			''a) Calculați <math>1^3 + 1^3 + 5^3 + 6^3 - 7^3</math>.

			b) Arătați că <math>5^{2018}+6^{2018}<7^{2018}</math>''

	'''Soluție.'''		'''Soluție.'''

Vasiliu Costel Andrei: Created page with "'''Soluție.''' a) $1^3 + 1^3 + 5^3 + 6^3 - 7^3 = 0$ . b) Din punctul a) putem scrie $1^3 + 1^3 + 5^3 + 6^3 = 7^3$ sau $\left(\frac{1}{7}\right)^3 + \left(\frac{1}{7}\right)^3 + \left(\frac{5}{7}\right)^3+\left(\frac{6}{7}\right)^3 = 1$ . Acum $\left(\frac{1}{7}\right)^{2018}<\left(\frac{1}{7}\right)^3 , \left(\frac{5}{7}\right)^{2018}<\left(\frac{5}{7}\right)^3, \left(\frac{6}{7}\right)^{2018}<\left(\frac{6}{7}\right)^3$ , de..."

2025-01-13T10:38:09Z

Created page with "'''Soluție.''' a) <math>1^3 + 1^3 + 5^3 + 6^3 - 7^3 = 0</math>. b) Din punctul a) putem scrie <math>1^3 + 1^3 + 5^3 + 6^3 = 7^3</math> sau <math>\left(\frac{1}{7}\right)^3 + \left(\frac{1}{7}\right)^3 + \left(\frac{5}{7}\right)^3+\left(\frac{6}{7}\right)^3 = 1</math>. Acum <math>\left(\frac{1}{7}\right)^{2018}<\left(\frac{1}{7}\right)^3 , \left(\frac{5}{7}\right)^{2018}<\left(\frac{5}{7}\right)^3, \left(\frac{6}{7}\right)^{2018}<\left(\frac{6}{7}\right)^3</math>, de..."

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'''Soluție.'''

a) <math>1^3 + 1^3 + 5^3 + 6^3 - 7^3 = 0</math>.

b) Din punctul a) putem scrie <math>1^3 + 1^3 + 5^3 + 6^3 = 7^3</math> sau <math>\left(\frac{1}{7}\right)^3 + \left(\frac{1}{7}\right)^3 + \left(\frac{5}{7}\right)^3+\left(\frac{6}{7}\right)^3 = 1</math>.

Acum <math>\left(\frac{1}{7}\right)^{2018}<\left(\frac{1}{7}\right)^3
, \left(\frac{5}{7}\right)^{2018}<\left(\frac{5}{7}\right)^3, \left(\frac{6}{7}\right)^{2018}<\left(\frac{6}{7}\right)^3</math>, de unde <math>\left(\frac{1}{7}\right)^{2018}+\left(\frac{1}{7}\right)^{2018}+\left(\frac{5}{7}\right)^{2018}+\left(\frac{6}{7}\right)^{2018} < \left(\frac{1}{7}\right)^3 + \left(\frac{1}{7}\right)^3 + \left(\frac{5}{7}\right)^3 + \left(\frac{6}{7}\right)^3.</math>

Obținem astfel <math>\left(\frac{1}{7}\right)^{2018}+\left(\frac{1}{7}\right)^{2018}+\left(\frac{5}{7}\right)^{2018}+\left(\frac{6}{7}\right)^{2018}<1</math> sau <math>1^{2018}+1^{2018}+5^{2018}+6^{2018}<7^{2018}</math>.

Deoarece <math>5^{2018}+6^{2018} < 1^{2018}+1^{2018}+5^{2018}+6^{2018} < 7^{2018} </math>, inegalitatea este demonstrată.

E:15346 - Revision history

Vasiliu Costel Andrei at 10:39, 13 January 2025