Đáp án:
$\min B =6045\Leftrightarrow x =\dfrac12$
Giải thích các bước giải:
$B = \dfrac{2015x}{1 - x} +\dfrac{2015}{x}$
$\to B = 2015\cdot\left(\dfrac{x}{1 - x} +\dfrac{1}{x}\right)$
$\to B = 2015\cdot\left(\dfrac{x}{1 - x} +\dfrac{1}{x}- 1 + 1\right)$
$\to B = 2015\cdot\left(\dfrac{x}{1 - x} +\dfrac{1-x}{x} + 1\right)$
$\to B \geq 2015\cdot\left(2\sqrt{\dfrac{x}{1 - x}\cdot\dfrac{1-x}{x}} + 1\right)$
$\to B \geq 2015\cdot(2 + 1)$
$\to B \geq 6045$
Dấu $=$ xảy ra $\Leftrightarrow x^2 = (1-x)^2 \Leftrightarrow x =\dfrac12$
Vậy $\min B =6045\Leftrightarrow x =\dfrac12$
