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Освободись от знака корня в знаменателе:

\(\dfrac {2}{\sqrt 5 -1}=\dfrac {2(\sqrt 5 +1)}{(\sqrt 5 -1) \cdot (\sqrt 5 +1)}=\dfrac {2(\sqrt 5 +1)}{5-1}=\dfrac {2(\sqrt 5 +1)}{4}=\dfrac {\sqrt 5 +1}{2}\) ;

\(\dfrac {6-2\sqrt 5}{\sqrt 5 -1}=\dfrac {5-2\sqrt 5 +1}{\sqrt 5 -1}=\dfrac {(\sqrt 5 -1)^2}{\sqrt 5 -1}=\sqrt 5 -1\) .

а) \(\dfrac {5}{\sqrt 6 -1}=\) [ ];

б) \(\dfrac {3}{\sqrt 2 +1}=\) [ ];

в) \(\dfrac {3}{\sqrt 5 +\sqrt 2}=\) [ ];

г) \(\dfrac {18}{2\sqrt 5 - \sqrt 2}=\) [ ];

д) \(\dfrac {4\sqrt 3 +7}{\sqrt 3 +2}=\) [ ];

е) \(\dfrac {14-6\sqrt 5}{3- \sqrt 5}=\) [ ].