Question

Rationalizati numitorii: √2+√3 supra √6

Answer 1

$\frac{ \sqrt{2} }{1} + \frac{ \sqrt{3}}{ \sqrt{6} } } (amplificam.a.doua.fractie.cu. \sqrt{6}) \\ \sqrt{2} + \frac{ \sqrt{3}* \sqrt{6} }{ \sqrt{6}* \sqrt{6} } \\ \sqrt{2} + \frac{ \sqrt{18} }{ \sqrt{36} }(scoatem.de.sub.radical) \\ \sqrt{2} + \frac{3 \sqrt{2} }{6}(simplificam.prin.3.pe.3.si.pe.6) \\ \frac{ \sqrt{2} }{1} + \frac{ 1\sqrt{2} }{2} (amplificam.prima.fractie.cu.2.sa.avem.acelasi.num.) \\ \frac{2 \sqrt{2} }{2} + \frac{1 \sqrt{2} }{2} \\ =\ \textgreater \ \frac{3 \sqrt{2} }{2}.Sper.ca.te-am.ajutat$