Which one of the following is an irrational number?
$\sqrt[3]{-27}$
$\sqrt{2}(3\sqrt{2}+2\sqrt{8})$
$\dfrac{3\sqrt{18}}{2\sqrt{6}}$
$\sqrt{\dfrac{1}{2}}\cdot\sqrt{\dfrac{25}{2}}$
$\dfrac{2\sqrt{5}}{\sqrt{45}}$
Option A: $\sqrt [ 3 ]{ -27 } ={ (-3) }^{ 3\times \frac { 1 }{ 3 } }=-3$Option B: $\sqrt { 2 } (3\sqrt { 2 } +2\sqrt { 8 } )=\sqrt { 2 } (3\sqrt { 2 } +4\sqrt { 2 } )=\sqrt { 2 } (7\sqrt { 2 } )=14$Option C: $\dfrac { 3\sqrt { 18 } }{ 2\sqrt { 6 } } =\dfrac { 3\sqrt { 3 } }{ 2 } $Option D: $\sqrt { \dfrac { 1 }{ 2 } } \sqrt { \dfrac { 25 }{ 2 } } =\dfrac { 5 }{ 2 } $Option E: $\dfrac { 2\sqrt { 5 } }{ \sqrt { 45 } } =\dfrac { 2\sqrt { 5 } }{ 3\sqrt { 5 } } =\dfrac { 2 }{ 3 } $Therefore, all are rational except option $C$.