Sa se determine \(
x,y \in \mathbb{Z}
\) care verifica egalitatea: \(
\sqrt {10 - x} + \sqrt {y + 3} = 1 + \left| {x - y} \right|
\)
O.G.: 489, G.M. 12/2006, V. Chiriac)
O.G.: 489, G.M. 12/2006, V. Chiriac
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Claudiu Mindrila
- Fermat
- Posts: 520
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- Location: Targoviste
- Contact:
O.G.: 489, G.M. 12/2006, V. Chiriac
elev, clasa a X-a, C. N. "C-tin Carabella", Targoviste
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Marius Dragoi
- Thales
- Posts: 126
- Joined: Thu Jan 31, 2008 5:57 pm
- Location: Bucharest
Observam ca: \( x=10-a^2 \) si \( y=b^2-3 \) , unde \( a,b \in Z \).
\( \Rightarrow |a|+|b|=1+|13-(a^2+b^2)| \geq 1+|a^2+b^2|-13 = a^2+b^2-12 \) \( \Rightarrow |a|+|b|+12 \geq a^2+b^2 \) \( \Rightarrow a,b\in [-4,...,4] \)
Se trateaza relativ usor toate cazurile si obtinem perechile: \( x,y\in {(-6,-3)},{(1,-2)},{(1,-3)},{(9,6)},{(10,13)} \).
\( \Rightarrow |a|+|b|=1+|13-(a^2+b^2)| \geq 1+|a^2+b^2|-13 = a^2+b^2-12 \) \( \Rightarrow |a|+|b|+12 \geq a^2+b^2 \) \( \Rightarrow a,b\in [-4,...,4] \)
Se trateaza relativ usor toate cazurile si obtinem perechile: \( x,y\in {(-6,-3)},{(1,-2)},{(1,-3)},{(9,6)},{(10,13)} \).
Politehnica University of Bucharest
The Faculty of Automatic Control and Computers
The Faculty of Automatic Control and Computers