Question

Știe careva s-o facă?

Answer 1

$\left\{\begin{array}{ll} x+y+z=a\Rightarrow z = a-(x+y) ~~(1)\\ 2xy-z^2=9~~(2) \end{array} \right \\ \\ \text{Inlocuim (1) in (2):} \\ \\ \Rightarrow 2xy - \Big[a-(x+y)\Big]^2 =9\Rightarrow \\ \\ \Rightarrow 2xy - \Big[a^2-2a(x+y)+(x+y)^2\Big] =9 \\ \\ \Rightarrow 2xy - a^2+2ax+2ay -x^2-2xy-y^2 = 9$

$\Rightarrow -x^2-y^2+2ax+2ay-a^2-9 = 0 \\ \Rightarrow x^2+y^2-2ax-2ay+a^2+9 = 0 \\ \Rightarrow (x^2-2ax +a^2) +y^2-2ay+9 = 0 \\ \Rightarrow \underset{\geq 0}{\underbrace{(x-a)^2}}+(y^2-2ay+9) = 0 \\ \\ \Rightarrow y^2-2ay+9 = 0\\ \\ \Delta = 0 \Rightarrow 4a^2-36 \Rightarrow a^2-9 = 0 \Rightarrow a^2 = 9 \Rightarrow a = \pm 3 \Rightarrow \\ \Rightarrow a_1 = -3,~a_2 = 3\\ \\ \Rightarrow \sum\limits_{a\in A} a = a_1+a_2 = -3+3 = \boxed{0}$

$\Rightarrow \boxed{\text{d) }S = 0} \rightarrow \text{r\u{a}spuns corect}$