Odpowiedź:
[tex]a)\ \ \dfrac{1}{\sqrt{5}}=\dfrac{1\cdot\sqrt{5}}{\sqrt{5}\cdot\sqrt{5}}=\dfrac{\sqrt{5}}{5}\\\\\\b)\ \ \dfrac{4}{\sqrt{11}}=\dfrac{4\cdot\sqrt{11}}{\sqrt{11}\cdot\sqrt{11}}=\dfrac{4\sqrt{11}}{11}\\\\\\c)\ \ \dfrac{4}{\sqrt{2}}=\dfrac{4\cdot\sqrt{2}}{\sqrt{2}\cdot\sqrt{2}}=\dfrac{\not4^2\sqrt{2}}{\not2_{1}}=2\sqrt{2}\\\\\\d)\ \ \dfrac{6}{\sqrt{3}}=\dfrac{6\cdot\sqrt{3}}{\sqrt{3}\cdot\sqrt{3}}=\dfrac{\not6^2\sqrt{3}}{\not3_{1}}=2\sqrt{3}[/tex]
[tex]e)\ \ \dfrac{3}{2\sqrt{7}}=\dfrac{3\cdot\sqrt{7}}{2\sqrt{7}\cdot\sqrt{7}}=\dfrac{3\sqrt{7} }{2\cdot7}=\dfrac{3\sqrt{7}}{14}\\\\\\f)\ \ \dfrac{9}{2\sqrt{3}}=\dfrac{9\cdot\sqrt{3}}{2\sqrt{3}\cdot\sqrt{3}}=\dfrac{9\sqrt{3}}{2\cdot3}=\dfrac{\not9^3\sqrt{3}}{\not6_{2}}=\dfrac{3\sqrt{3}}{2}[/tex]
