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.