Question

не поможете мне????​

Answer 1

Ответ:

D) $\dfrac{3}{4}$

Пошаговое объяснение:

$\lg \left( 6 \cdot 2^{\tfrac{1}{2x}} - 8 \right) - \dfrac{1}{x} \cdot \lg 2 = 0$

$\lg \left( 6 \cdot 2^{\tfrac{1}{2x}} - 8 \right) - \lg 2^{\tfrac{1}{x}} = 0$

$\lg \left( \dfrac{6 \cdot 2^{\tfrac{1}{2x}} - 8}{2^{\tfrac{1}{x}}} \right) = 0$

$\dfrac{6 \cdot 2^{\tfrac{1}{2x}} - 8}{2^{\tfrac{1}{x}}} = 1 \quad \mid \cdot 2^{\tfrac{1}{x}}$

$6 \cdot 2^{\tfrac{1}{2x}} - 8 = 2^{\tfrac{1}{x}}$

$2^{\tfrac{1}{x}} - 6 \cdot 2^{\tfrac{1}{2x}} + 8 = 0$

$t := 2^{\tfrac{1}{2x}}, \quad t^2 = 2^{\tfrac{1}{x}}$

$t^2 - 6t + 8 = 0$

$t_{1, 2} = \left \{ {{2} \atop {4}} \right.$

$x = \dfrac{1}{2 \cdot \log_2 t}$

$x_{1, 2} = \left \{ {{\tfrac{1}{2}} \atop {\tfrac{1}{4}}} \right.$

$x_1 + x_2 = \tfrac{1}{2} + \tfrac{1}{4} = \tfrac{3}{4}$