[tex]\it x=\Big(\dfrac{8}{\sqrt{18}}+\dfrac{6}{\sqrt2}\Big)\cdot\dfrac{\sqrt2}{13}=\dfrac{8}{\sqrt{18}}\cdot\dfrac{\sqrt2}{13}+\dfrac{6}{\sqrt2}\cdot\dfrac{\sqrt2}{13}=\dfrac{8}{3}\cdot\dfrac{1}{13}+\dfrac{6}{1}\cdot\dfrac{1}{13}=\\ \\ \\ =\dfrac{8}{39}+\dfrac{^{3)}6}{\ 13}=\dfrac{8+18}{39}=\dfrac{\ 26^{(13}}{39}=\dfrac{2}{3}[/tex]
