Giá trị $x$ thỏa mãn $\left( {x:\dfrac{2}{5} + \dfrac{1}{6}} \right)\left( {\dfrac{{14}}{{15}} + \dfrac{1}{5}.x} \right) = 0$ là:

Ta có: $\left( {x:\dfrac{2}{5} + \dfrac{1}{6}} \right)\left( {\dfrac{{14}}{{15}} + \dfrac{1}{5}.x} \right) = 0\,$

Suy ra $x:\dfrac{2}{5} + \dfrac{1}{6} = 0\,$ hoặc $\dfrac{{14}}{{15}} + \dfrac{1}{5}.x = 0\,$

TH1: $x:\dfrac{2}{5} + \dfrac{1}{6} = 0\,$

$x:\dfrac{2}{5} = \dfrac{{ - 1}}{6}$

$x = \dfrac{{ - 1}}{6}.\dfrac{2}{5}$

$x = \dfrac{{ - 1.2}}{{6.5}}$

$x = \dfrac{{ - 1.2}}{{2.3.5}}$

$x = \dfrac{{ - 1}}{{15}}$

TH2: $\dfrac{{14}}{{15}} + \dfrac{1}{5}.x = 0\,$

$\dfrac{1}{5}.x = \dfrac{{ - 14}}{{15}}$

$x = \dfrac{{ - 14}}{{15}}:\dfrac{1}{5}$

$x = \dfrac{{ - 14}}{{15}}.\dfrac{5}{1}$

$x = \dfrac{{ - 14.5}}{{3.5}}$

$x = \dfrac{{ - 14}}{3}$

Do đó có hai giá trị của $x$ thỏa mãn $\left( {x:\dfrac{2}{5} + \dfrac{1}{6}} \right)\left( {\dfrac{{14}}{{15}} + \dfrac{1}{5}.x} \right) = 0\,$ là $x = \dfrac{{ - 1}}{{15}}$; $x = \dfrac{{ - 14}}{3}$.