Question

exercitiul 4 doar a si b

Answer 1

   
[tex]\displaystyle\\ \text{Rationalizam numitorii amplificand fractia cu radicalul de la numitor.}\\\\ a)~~~\frac{\sqrt{2}+\sqrt{3}}{\sqrt{6}}=\frac{\sqrt{6} (\sqrt{2}+ \sqrt{3} ) }{ \sqrt{6} \times \sqrt{6} }= \frac{ \sqrt{6}\times \sqrt{2}+\sqrt{6}\times \sqrt{3} }{ \sqrt{6} \times \sqrt{6} }=\\\\ = \frac{ \sqrt{6\times 2}+\sqrt{6\times 3}}{ \sqrt{6\times 6} }= \frac{ \sqrt{12}+\sqrt{18}}{ \sqrt{36} }=\frac{ \sqrt{4\times 3}+\sqrt{9\times2}}{ \sqrt{36} }=\boxed{\frac{ 2\sqrt{3}+3\sqrt{2}}{6}}[/tex]


[tex]\displaystyle\\ b)~~~\frac{\sqrt{5}-\sqrt{2}}{\sqrt{10}}=\frac{\sqrt{10} (\sqrt{5}- \sqrt{2} ) }{ \sqrt{10} \times \sqrt{10} }= \frac{ \sqrt{10}\times \sqrt{5}-\sqrt{10}\times \sqrt{2} }{ \sqrt{10} \times \sqrt{10} }=\\\\ = \frac{ \sqrt{10\times 5}-\sqrt{10\times 2}}{ \sqrt{10\times 10} }= \frac{ \sqrt{50}-\sqrt{20}}{ \sqrt{100} }=\\\\ =\frac{ \sqrt{25\times 2}-\sqrt{4\times5}}{ \sqrt{100} }=\boxed{\frac{ 5\sqrt{2}-2\sqrt{5}}{10}}[/tex]