Odpowiedź:
[tex]a)\ \ \dfrac{4}{\sqrt{10}}=\dfrac{4}{\sqrt{10}}*\dfrac{\sqrt{10}}{\sqrt{10}}=\dfrac{\not4^2\sqrt{10}}{\not10_{5}}=\dfrac{2\sqrt{10}}{5}\\\\\\\\b)\ \ \dfrac{1}{\sqrt{48}+\sqrt{27}}=\dfrac{1}{\sqrt{16*3}+\sqrt{9*3}}=\dfrac{1}{4\sqrt{3}+3\sqrt{3}}=\dfrac{1}{7\sqrt{3}}*\dfrac{\sqrt{3}}{\sqrt{3}}=\dfrac{\sqrt{3}}{7*3}=\dfrac{\sqrt{3}}{21}[/tex]
