a) $\dfrac{3}{\sqrt5} + \sqrt{\dfrac{4}{5}} - \sqrt{125}$
$= \dfrac{3}{\sqrt5} + \dfrac{2}{\sqrt5} - 5\sqrt5$
$= \dfrac{5}{\sqrt5} - 5\sqrt5$
$= \sqrt5 - 5\sqrt5$
$= -4\sqrt5$
b) $(\sqrt5 - \sqrt3)\sqrt{8 + 2\sqrt{15}}$
$= (\sqrt5 - \sqrt3)\sqrt{5 + 2\sqrt{5}.\sqrt3 + 3}$
$= (\sqrt5 - \sqrt3)\sqrt{(\sqrt5 + \sqrt3)^2}$
$= (\sqrt5 - \sqrt3)(\sqrt5 + \sqrt3)$
$= 5 - 3 = 2$
c) $3\sqrt{x -1} + 2\sqrt{4x - 4} -\dfrac{1}{3}\sqrt{9x - 9} - 12 = 0\qquad (*)$
$ĐK:\, x \geq 1$
$(*)\Leftrightarrow 3\sqrt{x -1} + 2\sqrt{4(x - 1)} -\dfrac{1}{3}\sqrt{9(x - 1)} =12$
$\Leftrightarrow 3\sqrt{x -1} + 4\sqrt{x -1} - \sqrt{x -1} = 12$
$\Leftrightarrow 6\sqrt{x -1} = 12$
$\Leftrightarrow \sqrt{x -1} = 2$
$\Rightarrow x -1 = 4$
$\Leftrightarrow x = 5$ (nhận)
Vậy phương trình có nghiệm $x = 5$
