1) [tex] \frac{3}{ \sqrt{5}- \sqrt{2} } = \frac{3( \sqrt{5}+ \sqrt{2}) }{( \sqrt{5}- \sqrt{2})( \sqrt{5}+ \sqrt{2}) } = [/tex][tex] \frac{3( \sqrt{5} + \sqrt{2}) }{ (\sqrt{5})^2- (\sqrt{2})^2 } = \frac{3( \sqrt{5}+ \sqrt{2}) }{5-2} = \frac{3( \sqrt{5} +\sqrt{2}) }{3} = \sqrt{5}+ \sqrt{2} [/tex]
2) Diagonala cubului este: d = a√3
d = 6√3
