Question

пожалуйста, решите примеры

Answer 1

Ответ:

а)

$2x + 3 = {4}^{3} \\ 2x + 3 = 64 \\ 2x = 61 \\ x = 30.5$

б)

$log_{2}((x - 2)(x - 3)) = 1 \\ {x}^{2} - 3x - 2x + 6 = {2}^{1} \\ {x}^{2} - 5x + 6 - 2 = 0 \\ {x}^{2} - 5x + 4 = 0 \\ x1 = 1 \\ x2 = 4$

По ОДЗ: x>3, значит х1=1 не подходит.

Ответ: 4.

в)

$\frac{4 + 2x}{x - 5} = {4}^{2} \\ 4 + 2x = 16(x - 5) \\ 4 + 2x = 16x - 80 \\ - 12x = - 84 \\ x = 7$

д)

${x}^{2} - 5x + 6 = { \frac{1}{2} }^{ - 1} \\ {x}^{2} - 5x + 6 = 2 \\ {x}^{2} - 5x + 4 = 0 \\ \\ x1 = 1 \\ x2 = 4$

е)

$log_{3}(4) - log_{3}(x - 1) = log_{3}(3) + log_{3}(5) \\ log_{3}( \frac{4}{x - 1} ) = log_{3}(15) \\ \frac{4}{x - 1} = 15 \\ 4 = 15(x - 1) \\ 15x - 15 = 4 \\ 15x = 19 \\ x = \frac{19}{15}$