Question

Poate sa ma ajute cineva cu acest exerciu?
trebuie sa imi dea 2/9 .. 
Multumesc.

Answer 1

sper sa intelegi ce am scris...bafta

Answer 2

$( \frac{1}{ \sqrt{3} } + \frac{1}{3} )( \frac{ \sqrt{5}- \sqrt{3} }{ \sqrt{15} } +\frac{ \sqrt{4}- \sqrt{5} }{ \sqrt{20} }+\frac{ \sqrt{9}- \sqrt{4} }{ \sqrt{36} } )$

$\frac{1}{ \sqrt{3} } + \frac{1}{3} = \frac{3+ \sqrt{3} }{3 \sqrt{3} } = \frac{ \sqrt{3} (3+ \sqrt{3} )}{3*3} = \frac{3 \sqrt{3}+3 }{9} = \frac{3( \sqrt{3}+1 )}{9} = \frac{\sqrt{3}+1}{3}$

$\frac{ \sqrt{5} - \sqrt{3} }{ \sqrt{15} } = \frac{ \sqrt{15}( \sqrt{5} - \sqrt{3})}{ \sqrt{15*15} } = \frac{ 5\sqrt{3} -3 \sqrt{5} }{15}$

$\frac{ \sqrt{4} - \sqrt{5} }{ \sqrt{20} } = \frac{2- \sqrt{5} }{2 \sqrt{5} } = \frac{ \sqrt{5}( 2 - \sqrt{5})}{ 2\sqrt{5*5} } = \frac{ 2\sqrt{5} -5 }{10}$

$\frac{ \sqrt{9}- \sqrt{4} }{ \sqrt{36} } = \frac{3-2}{6} = \frac{1}{6}$

$( \frac{ \sqrt{5}- \sqrt{3} }{ \sqrt{15} } +\frac{ \sqrt{4}- \sqrt{5} }{ \sqrt{20} } +\frac{ \sqrt{9}- \sqrt{4} }{ \sqrt{36} } )=\frac{ 5\sqrt{3} -3 \sqrt{5} }{15} +\frac{ 2\sqrt{5} -5 }{10}+\frac{1}{6}=$$= \frac{2(5 \sqrt{3}-3 \sqrt{5} )+3(2 \sqrt{5}-5)+5 }{30} =\frac{10 \sqrt{3}-6 \sqrt{5}+6 \sqrt{5}-15+5 }{30} = \frac{10 \sqrt{3}-10 }{30} = \frac{10( \sqrt{3} -1)}{30}$$=\frac{ \sqrt{3} -1}{3}$

$( \frac{1}{ \sqrt{3} } + \frac{1}{3} )( \frac{ \sqrt{5}- \sqrt{3} }{ \sqrt{15} } +\frac{ \sqrt{4}- \sqrt{5} }{ \sqrt{20} }+\frac{ \sqrt{9}- \sqrt{4} }{ \sqrt{36} } ) =\frac{\sqrt{3}+1}{3} *\frac{ \sqrt{3} -1}{3} = \frac{( \sqrt{3} )^{2}-1 }{9} = \frac{3-1}{9}$$= \frac{2}{9}$