√(x²-2x+5) + √(y²+4y+13) = 5
Doar doua numere naturale adunate pot da un numar natural.
Avem doua cazuri:
(1) √(x²-2x+5) = 2 sau (2) √(x²-2x+5) = 3
√(y²+4y+13) = 3 √(y²+4y+13) = 2
(1) √(x²-2x+5) = 2 => x²-2x+5 = 4 => x²-2x+1=0 => (x-1)² = 0 => x = 1
√(y²+4y+13) = 3 => y²+4y+13 = 9 => y²+4y+4 = 0 => (y+2)² = 0 => y=-2
(2) √(x²-2x+5) = 3 => x²-2x+5 = 9 => x²-2x-4 = 0;
[tex]\Delta = 4+16 = 20 \Rightarrow x_{1,2} = \frac{2\pm 2\sqrt{5} }{2} \Rightarrow x_{1,2} = 1\pm \sqrt{5[/tex]
√(y²+4y+13) = 2 => y²+4y+13 = 4 => y²+4y+9 = 0
[tex]\Delta = 16-4\cdot 9 = 16-36 = -20 \ \textless \ 0 \Rightarrow y \in \O[/tex]
[tex]\Rightarrow (x,y) = \Big\{(1,-2)\Big\} [/tex]
