Выбери правильные ответы

Освободись от иррациональности в знаменателях дробей:

а) \(\cfrac{1}{\sqrt{6}-2}=\) [ \(\frac{\sqrt{6}\,+\,2\,}{2}\) | \(\frac{\sqrt{6}\,-\,2}{2}\) | \(\frac{\sqrt{6}}{2}\) | \(\sqrt{6}+2\) ];

б) \(\cfrac{10}{\sqrt{7}+\sqrt{2}}=\) [ \(\frac{\sqrt{7}\,+\,\sqrt{2}}{10}\) | \(\frac{\sqrt{7}\,-\,\sqrt{2}}{2}\) | \(2\cdot (\sqrt{7}-\sqrt{2})\) ];

в) \(\cfrac{12}{\sqrt{x}+\sqrt{y}}=\) [ \(\frac{\sqrt{x}\,-\,\sqrt{y}}{12}\) | \(\frac{12\,\cdot\, \sqrt{(x\,-\,y)}}{x\,-\,y}\) | \(\frac{12\,\cdot\, (\sqrt{x}\,-\,\sqrt{y})}{(x\,-\,y)}\) ];

г) \(\cfrac{36(a-b)}{\sqrt{a}-\sqrt{b}}=\) [ \(36\,\cdot\, (\sqrt{a}\,+\,\sqrt{b})\) | \(\frac{36\,\cdot\, (\sqrt{a}\,+\,\sqrt{b})}{(a\,-\,b)}\) | \(\frac{36\,\cdot\, \sqrt{(a\,+\,b)}}{(a\,-\,b)}\) | \(\frac{36\,\cdot\, \sqrt{a\,-\,b}}{(a\,+\,b)}\) ].