Es 63 e 64 grazieee
{2·k + 6 > 0
{2·k < 0
risolvo:
{k > -3
{k < 0
Quindi: [-3 < k < 0]
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(k + 1)·x^2 + y^2 = 2·k
x^2·(k + 1)/(2·k) + y^2/(2·k) = 1
{2·k/(k + 1) > 0
{2·k < 0
quindi:
{k < -1 ∨ k > 0
{k < 0
Risolvo: [k < -1]