Question

DOAR CE E INCERCUIT
Si cu asta
$a) \: \: ( \frac{2}{ \sqrt{3} } + \frac{3}{ \sqrt{2} } ) \times \sqrt{6} - ( \frac{2}{ \sqrt{5} } + \frac{3}{ \sqrt{3} } ) \times \sqrt{15} + ( \frac{3}{ \sqrt{2} } - \frac{4}{ \sqrt{5} }) \times \sqrt{10} \\ b) \: \: \sqrt{6} ( \frac{2}{3 \sqrt{6 } } - \frac{1}{2 \sqrt{6} } ) - ( \sqrt{54} - \frac{5}{6 \sqrt{6} } ) \times \sqrt{6} - \frac{25}{6} \\ c) \: \: 3 \sqrt{27} ( \frac{2}{ \sqrt{3} } + \frac{4}{3 \sqrt{3} } - \frac{1}{ \sqrt{3} } ) \div \sqrt{3} + ( \frac{8}{3 \sqrt{3} } - \frac{ \sqrt{108} }{9} + \frac{4}{ \sqrt{3} } ) \times \frac{3}{7} \\ d) \: \: ( \frac{6}{ \sqrt{20} } - \frac{2 \sqrt{5} }{5} + \frac{3}{2 \sqrt{5} } ) \div \frac{1}{ \sqrt{5} } + \frac{ \sqrt{10} }{2} ( \sqrt{10} - \frac{1}{ \sqrt{10} } + \frac{ \sqrt{10} }{5} - \frac{3 \sqrt{10} }{10} )$
URGENT!!!!!!!!

Answer 1

$a)\frac{12}{5 \sqrt{3}}= \frac{12 \sqrt{3} }{5\cdot3} = \frac{4 \sqrt{3}}{5} \\ \\ \frac{16}{7 \sqrt{2}}= \frac{16 \sqrt{2} }{7\cdot2} = \frac{8 \sqrt{2}}{7} \\ \\ \frac{8}{3\sqrt{2}}= \frac{8\sqrt{2} }{3\cdot2} = \frac{4 \sqrt{2}}{3} \\ \\ \frac{6}{\sqrt{3}}= \frac{6 \sqrt{3} }{3} = 2\sqrt{3} \\ \\ \frac{24}{5 \sqrt{6}}= \frac{24 \sqrt{6} }{5\cdot6} = \frac{4 \sqrt{6}}{5} \\ \\ \frac{28}{3\sqrt{7}}= \frac{28 \sqrt{7} }{3\cdot7}= \frac{4 \sqrt{7}}{3}$

$b))\frac{15}{2\sqrt{3}}= \frac{15\sqrt{3} }{2\cdot3} = \frac{5\sqrt{3}}{2} \\ \\ \frac{4}{3\sqrt{2}}= \frac{4\sqrt{2} }{3\cdot2} = \frac{2\sqrt{2}}{3} \\ \\ \frac{9}{\sqrt{3}}= \frac{9\sqrt{3} }{3} =3\sqrt{3}} \\ \\ \frac{12}{5\sqrt{2}}= \frac{12\sqrt{2}}{5\cdot2} = \frac{6}{5} \sqrt{2} \\ \\ \frac{10}{3\sqrt{5}}= \frac{10\sqrt{5} }{3\cdot5} = \frac{2 \sqrt{5}}{3} \\ \\ \frac{18}{7\sqrt{6}}= \frac{18 \sqrt{6} }{6\cdot7}= \frac{3 \sqrt{6}}{7}$

$\frac{36}{5\sqrt{6}}= \frac{36\sqrt{6} }{6\cdot5}= \frac{6\sqrt{6}}{5}$

4)
$(\frac{18}{5 \sqrt{3}} - \frac{12}{4 \sqrt{3}} - \frac{6}{2\sqrt{3}}+ \frac{8}{2 \sqrt{3}})-[tex] (\frac{4}{3\sqrt{2}}+ \frac{3}{2 \sqrt{2}} - \frac{40}{12\sqrt{2}})=$
  $(\frac{18}{5 \sqrt{3}} - \frac{12}{4 \sqrt{3}} - \frac{6}{2\sqrt{3}}+ \frac{8}{2 \sqrt{3}})= \frac{18\cdot4 \sqrt{3} -12\cdot5 \sqrt{3}-6\cdot5\cdot2 \sqrt{3} +8\cdo10 \sqrt{3}}{5\cdot2\cdot2\cdot3} = \frac{8 \sqrt{3}}{5}$

$( \frac{16 \sqrt{2}+18 \sqrt{2}-40 \sqrt{2}}{12\cdot2} )= \frac{-6 \sqrt{2}}{24}={-(\frac{\sqrt{2}}{4})}\\ \\ \frac{8 \sqrt{3}}{5}-(\frac{- \sqrt{2}}{4})= \frac{32 \sqrt{3}+5 \sqrt{3}}{20}$