Stabiliti intervalul caruia apartine fiecare din numerele reale \( x,y,z \) daca \( x^2+4y^2+z^2+13=4\left(3y+2z-x\right) \)
Sacota Florian, Zalau
Concursul "Teodor Topan" - problema 2
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Marius Dragoi
- Thales
- Posts: 126
- Joined: Thu Jan 31, 2008 5:57 pm
- Location: Bucharest
\( x^2+4y^2+z^2+13=4(3y+2z-x) \Rightarrow {(x+2)}^2 + {(2y-3)}^2 + {(z-4)}^2 = 16 \)
Atunci avem: \( |x+2| \leq 4 \) , \( |2y-3| \leq 4 \) , \( |z-4| \leq 4 \) \( \Rightarrow x \in [-6,2] , y \in [-\frac {1}{2} , \frac {7}{2}] , z \in [0,8] \)
Atunci avem: \( |x+2| \leq 4 \) , \( |2y-3| \leq 4 \) , \( |z-4| \leq 4 \) \( \Rightarrow x \in [-6,2] , y \in [-\frac {1}{2} , \frac {7}{2}] , z \in [0,8] \)
Politehnica University of Bucharest
The Faculty of Automatic Control and Computers
The Faculty of Automatic Control and Computers