Question

DAU 22 DE PUNCTE!!!!!!!
REPEDE !! VA ROG!
( $\frac{2}{ \sqrt{7}- \sqrt{5} } - \sqrt{5} ) : \frac{ \sqrt{14} }{6} - ( \frac{24}{2 \sqrt{48} } - \frac{12}{ \sqrt{108} } ) \sqrt{6}$

Answer 1

$=[ \frac{2( \sqrt{7}+ \sqrt{5}) }{( \sqrt{7}- \sqrt{5})( \sqrt{7}+ \sqrt{5}) } - \sqrt{5}]* \frac{6}{ \sqrt{14} } -( \frac{24 \sqrt{6} }{2 \sqrt{48} }- \frac{12 \sqrt{6} }{ \sqrt{108} })= \\ = [\frac{2( \sqrt{7}+ \sqrt{5}) }{7-5} - \sqrt{5}]* \frac{6 \sqrt{14} }{ 14 }-( \frac{24}{2 \sqrt{8} }- \frac{12}{ \sqrt{18} } })= \\ =( \sqrt{7}+ \sqrt{5}- \sqrt{5})* \frac{3 \sqrt{14} }{7}-( \frac{24}{4 \sqrt{2} }- \frac{12}{3 \sqrt{2} })=$
$= \sqrt{7}* \frac{3 \sqrt{14} }{7}-( \frac{6}{ \sqrt{2} }- \frac{4}{\sqrt{2}} )= \\ = \frac{3*7 \sqrt{2} }{7}- \frac{2}{ \sqrt{2} } = \\ = 3 \sqrt{2}- \sqrt{2}= \\= \boxed{2 \sqrt{2}}.$

Answer 2

in prima paranteza amplifici prima fractie cu conjugata
$(\frac{2(\sqrt{7}+\sqrt{5})}{(\sqrt{7}+\sqrt{5})(\sqrt{7}-\sqrt{5})}-\sqrt{5})=(\frac{2(\sqrt{7}+\sqrt{5})}{7-5}-\sqrt{5})=(\frac{2(\sqrt{7}+\sqrt{5})}{2}-\sqrt{5})$=$\sqrt{7}+\sqrt{5}-\sqrt{5}=\sqrt{7}$
in a doua paranteza transformi radicalii de la numitori:2*√48=2*4√3=8√3; √108=6√3
$\frac{24}{8\sqrt{3}}-\frac{12}{6 \sqrt{3}}=\frac{3}{\sqrt{3}}-\frac{2}{\sqrt{3}}=\frac{1}{\sqrt{3}}$
calculul initial devine deci:
$\sqrt{7} : \frac{ \sqrt{14} }{6} - \frac{1}{ \sqrt{3} }* \sqrt{6} = \sqrt{7} * \frac{6}{ \sqrt{14} } - \sqrt{2}= \frac{6}{ \sqrt{2} } - \sqrt{2}= \frac{6}{ \sqrt{2} }- \frac{2}{ \sqrt{2} }= \frac{4}{ \sqrt{2} }$ =$\frac{4 \sqrt{2} }{2} =2 \sqrt{2}$