$B = \dfrac{x - 25}{\sqrt{x} - 5}$
$ĐK: \, x \geq 0; \, x \ne 25$
$B =\dfrac{(\sqrt{x} - 5)(\sqrt{x} +5)}{\sqrt{x} - 5}$
$= \sqrt{x} + 5$
Do $\sqrt{x} \geq 0, \forall x \geq0; \, x \ne 5$
nên $\sqrt{x} + 5 \geq 5$
Vậy $B_{min} = 5 \Leftrightarrow x = 0$