Question

розвязати рівняння чере 2^x=t

Answer 1

${2}^{x} = t. \: \: t > 0 \\ \frac{4}{t + 2} - \frac{1}{t - 3} = 2 \\ \frac{4(t - 3) - (t + 2) - 2(t + 2)(t - 3)}{(t + 2)(t - 3)} = 0 \\ \frac{4t - 12 - t - 2 - 2 {t}^{2} - 4t + 6t + 12 }{(t + 2)(t - 3)} = 0 \\ \frac{ - 2 {t}^{2} + 5t - 2 }{(t + 2)(t - 3)} = 0 \\ \frac{2 {t}^{2} - 5t + 2}{(t + 2)(t - 3)} = 0$
ОДЗ: t + 2 ≠ 0, t ≠ -2
t - 3 ≠ 0, t ≠ 3

$2 {t}^{2} - 5t + 2 = 0 \\ d = {b}^{2} - 4ac = 25 - 4 \times 2 \times 2 = 25 - 16 = 9 \\ t1 = \frac{5 + 3}{2 \times 2} = \frac{8}{4} = 2 \\ t2 = \frac{5 - 3}{2 \times 2} = \frac{2}{4} = \frac{1}{2}$
${2}^{x} = 2 \\ x = 1 \\ \\ {2}^{x} = \frac{1}{2} \\ {2}^{x} = {2}^{ - 1} \\ x = - 1$
Ответ: -1; 1.