Atunci cand avem o fractie cu un radical sau mai multi radicali la numitor, vrem sa scapam de radicalii de la numitor.
Operatia se numeste rationalizarea numitorului.
Cum se procedeaza ?
Se amplifica fractia cu un numar convenabil ales, astfel incat radicalii sa dispara.
Rezolvare:
[tex]a)\,\, \frac{4}{ \sqrt{6}-2 }\,\,se\,\,amplifica\,\,cu\,\,(\sqrt{6}+2)\\ \\\frac{4*(\sqrt{6}+2)}{ (\sqrt{6}-2)(\sqrt{6}+2)}= \frac{4 \sqrt{6}+8 }{(\sqrt{6})^{2}- 2^{2}}=\frac{4 \sqrt{6}+8 }{6- 4}=\frac{4 \sqrt{6}+8 }{2}=2 \sqrt{6}+4[/tex]
[tex]b)\,\, \frac{ \sqrt{6}+ \sqrt{5} }{\sqrt{6}- \sqrt{5}} \,\,se\,\,amplifica\,\,cu\,\,(\sqrt{6}+ \sqrt{5}) \\ \\ \frac{ (\sqrt{6}+ \sqrt{5})(\sqrt{6}+ \sqrt{5}) }{(\sqrt{6}- \sqrt{5})(\sqrt{6}+ \sqrt{5})}= \frac{6+2\sqrt{6} \sqrt{5}+5 }{6-5}=11+2\sqrt{30}
[tex] c)\,\,\frac{1}{3\sqrt{2}-2 \sqrt{5}} \,\,se\,\,amplifica\,\,cu\,\,(3\sqrt{2}+ 2\sqrt{5}) \\ \\ \frac{3\sqrt{2}+ 2\sqrt{5}}{(3\sqrt{2}-2 \sqrt{5})(3\sqrt{2}+ 2\sqrt{5})}=\frac{3\sqrt{2}+ 2\sqrt{5}}{9*2-4 *5}=\frac{3\sqrt{2}+ 2\sqrt{5}}{-2}[/tex]
