0 \\ t + 2 \times \frac{1}{t} = 3 \\ {t}^{2} + 2 = 3t \\ {t}^{2} - 3t + 2 = 0 \\ d = 9 - 8 = 1 > 0 \\ x1 = \frac{3 - 1}{2} = 1 \\ x2 = 2 \\ {2}^{x} = 1 \\ x = 0 \\ {2}^{x} = 2 \\ x = 1" alt="{2}^{x} + 2 \times {2} ^{ - x} = 3 \\ {2}^{x} = t > 0 \\ t + 2 \times \frac{1}{t} = 3 \\ {t}^{2} + 2 = 3t \\ {t}^{2} - 3t + 2 = 0 \\ d = 9 - 8 = 1 > 0 \\ x1 = \frac{3 - 1}{2} = 1 \\ x2 = 2 \\ {2}^{x} = 1 \\ x = 0 \\ {2}^{x} = 2 \\ x = 1" align="absmiddle" class="latex-formula">
Ответ: 0; 1