Odpowiedź:
1
[tex] - {x}^{2} + 2x - 1 < 0 \\ ∆ = {2}^{2} - 4 \times ( - 1) \times ( - 1) = 4 - 4 = 0 \\ ∆ = 0 \\ x_{0} = \frac{ - 2}{ - 2} = 1[/tex]
x e R \ {1}
2.
[tex] \frac{2}{3} {x}^{2} + 6 \leqslant 0 \\∆ = 0 - 4 \times \frac{2}{3} \times 6 = - 16 \\ ∆ < 0[/tex]
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