[tex]Inseamna~a~transforma~fractia~intr-o~fractie~cu~numitorul \\ \\ un~numar~rational. \\ \\ De~exemplu:~ \frac{1}{\sqrt{2}}~are~numitorul~ \sqrt{2},~numar~irational.~Amplificand \\ \\ fractia~cu~\sqrt{2},~obtinem: \frac{1}{\sqrt{2}}= \frac{\sqrt{2}}{2}.~Astfel~am~rationalizat~fractia. [/tex]
[tex]Sa~luam~un~alt~exemplu:~ \frac{1}{\sqrt{5}- \sqrt{2}} .~In~acest~caz~vom~folosi~ \\ \\ identitatea~(formula)~a^2-b^2=(a+b)(a-b).~Alegem~a= \sqrt{5} \\ \\ si~b= \sqrt{2}.~Vom~amplifica~fractia~cu~(a+b)=(\sqrt{5}+ \sqrt{2}). \\ \\ \frac{1}{\sqrt{5}-\sqrt{2}}= \frac{\sqrt{5}+\sqrt{2}}{(\sqrt{5}-\sqrt{2})(\sqrt{5}+\sqrt{2})}= \frac{\sqrt{5}+\sqrt{2}}{5-2}= \frac{\sqrt{5}+\sqrt{2}}{3}. [/tex]
[tex]O~fractie~de~tipul~ \frac{a}{\sqrt{b}}~se~va~amplifica~cu~\sqrt{b},~iar~o~fractie \\ \\ ~tipul~ \frac{a}{\sqrt{b}- \sqrt{c}}~se~va~rationaliza~cu ~\sqrt{b}+ \sqrt{c}. \\ \\ Spunem~ca ~\sqrt{b}-\sqrt{c}~si~\sqrt{b}+\sqrt{c}~sunt~conjugate.[/tex]
