Question

Dau 40 puncte + coroana!!
ex. c) si d) de sus + ex. 13 si 14 tot.

Answer 1

$\displaystyle c) \frac{3 \sqrt{2} }{4 \sqrt{3} }- \frac{5}{ \sqrt{6} } + \frac{2 \sqrt{3} }{3 \sqrt{2} } -\left( \sqrt{24} + \frac{12}{2 \sqrt{6} } \right)=\\ \\ = \frac{3 \sqrt{6} }{12} - \frac{5 \sqrt{6} }{6} + \frac{2 \sqrt{6} }{6} -\left(2 \sqrt{6} + \frac{12 \sqrt{6} }{12} \right)=$
$\displaystyle = \frac{3 \sqrt{6} }{12} - \frac{5 \sqrt{6} }{6} + \frac{2 \sqrt{6} }{6} -\left(2 \sqrt{6} + \sqrt{6} \right)= \frac{3 \sqrt{6} }{12} - \frac{5 \sqrt{6} }{6} + \frac{2 \sqrt{6} }{6}-3 \sqrt{6} = \\ \\ = \frac{3 \sqrt{6}-10 \sqrt{6}+4 \sqrt{6}- 36 \sqrt{6} }{12} =- \frac{39 \sqrt{6} }{12} =- \frac{13 \sqrt{6} }{4}$
$\displaystyle d) \frac{2 \sqrt{5} }{5 \sqrt{2} } - \frac{20}{ \sqrt{10} }+ \left( \frac{2 \sqrt{10} }{15} - \frac{30}{ \sqrt{40} } \right)=\\ \\ = \frac{2 \sqrt{10} }{10} - \frac{20 \sqrt{10} }{10} + \left( \frac{2 \sqrt{10} }{15} - \frac{30}{2 \sqrt{10} } \right)= \\ \\ =\frac{2 \sqrt{10} }{10} - \frac{20 \sqrt{10} }{10} + \left( \frac{2 \sqrt{10} }{15} - \frac{30 \sqrt{10} }{20} \right)=$
$\displaystyle = \frac{2 \sqrt{10} }{10} - \frac{20 \sqrt{10} }{10} + \frac{8 \sqrt{10}- 90 \sqrt{10} }{60} = \frac{2 \sqrt{10} }{10} - \frac{20 \sqrt{10} }{10} - \frac{82 \sqrt{10} }{60} = \\ \\ = \frac{12 \sqrt{10}-120 \sqrt{10}-82 \sqrt{10} }{60} =- \frac{190 \sqrt{10} }{60} =- \frac{19 \sqrt{10} }{6}$
$\displaystyle 13a)\left( \frac{1}{ \sqrt{3} } + \frac{1}{ \sqrt{2} } \right) \cdot \sqrt{6} -\left( \frac{1}{ \sqrt{5} } + \frac{1}{ \sqrt{3} } \right) \cdot \sqrt{15} + \left( \frac{1}{ \sqrt{2} } - \frac{1}{ \sqrt{5} } \right) \cdot \sqrt{10} =$
$\displaystyle = \frac{ \sqrt{6} }{ \sqrt{3} } + \frac{ \sqrt{6} }{ \sqrt{2} } - \frac{ \sqrt{15} }{ \sqrt{5} } - \frac{ \sqrt{15} }{ \sqrt{3} } + \frac{ \sqrt{10} }{ \sqrt{2} } - \frac{ \sqrt{10} }{ \sqrt{5} } = \\ \\= \sqrt{2} + \sqrt{3} - \sqrt{3} - \sqrt{5} + \sqrt{5}- \sqrt{2} =0$
$\displaystyle b)\left( \frac{2}{3 \sqrt{6} } - \frac{1}{2 \sqrt{6} } \right) \cdot \sqrt{6} + \left( \sqrt{24} - \frac{5}{2 \sqrt{6} } \right) \cdot \sqrt{6} - \frac{29}{3} =\\ \\= \left(\frac{2 \sqrt{6} }{3 \sqrt{6} } - \frac{ \sqrt{6} }{2 \sqrt{6} } \right)+ \left(2 \sqrt{6} - \frac{5}{2 \sqrt{6} } \right) \cdot \sqrt{6} - \frac{29}{3} =$
$\displaystyle =\left( \frac{2}{3} - \frac{1}{2} \right)+ \left(12- \frac{5 \sqrt{6} }{2 \sqrt{6} } \right)- \frac{29}{3} = \frac{4-3}{6} +\left(12- \frac{5}{2} \right)- \frac{29}{3} = \\ \\= \frac{1}{6} + \frac{24-5}{2} - \frac{29}{3} = \frac{1}{6} + \frac{19}{2} - \frac{29}{3} = \frac{1+57-58}{6} =0$
$\displaystyle c)3 \sqrt{125} \cdot \left( \frac{2}{ \sqrt{5} } + \frac{3}{2 \sqrt{5} }- \frac{1}{ \sqrt{5} } \right): \sqrt{5} + \left( \frac{14}{3 \sqrt{5} }- \frac{ \sqrt{45} }{15} + \frac{12}{ \sqrt{5} } \right)\cdot \frac{5}{47} =$
$\displaystyle =15 \sqrt{5} \cdot \left( \frac{2}{ \sqrt{5} } + \frac{3}{2 \sqrt{5} }- \frac{1}{ \sqrt{5} } \right) \cdot \frac{1}{ \sqrt{5} } +\left( \frac{14 \sqrt{5} }{15} - \frac{3 \sqrt{5} }{15} + \frac{12 \sqrt{5} }{5} \right) \cdot \frac{5}{47} =\\ \\ =15 \sqrt{5} \cdot \left( \frac{2}{5} + \frac{3}{10} - \frac{1}{5} \right)+ \frac{14 \sqrt{5} -3 \sqrt{5} +36 \sqrt{5} }{15} \cdot \frac{5}{47} =$
$\displaystyle =15 \sqrt{5} \cdot \frac{4+3-2}{10} + \frac{47 \sqrt{5} }{15} \cdot \frac{5}{47} =15 \sqrt{5} \cdot \frac{5}{10} + \frac{ \sqrt{5} }{3} = \\ \\ =15 \sqrt{5} \cdot \frac{1}{2} + \frac{ \sqrt{5} }{3} = \frac{15 \sqrt{5} }{2} + \frac{ \sqrt{5} }{3} = \frac{45 \sqrt{5}+2 \sqrt{5} }{6} = \frac{47 \sqrt{5} }{6}$
$\displaystyle d)\left( \frac{5}{ \sqrt{28} } - \frac{2 \sqrt{7} }{7} + \frac{3}{2 \sqrt{7} } \right): \frac{1}{ \sqrt{7} } + \frac{ \sqrt{14} }{2} \cdot \left( \sqrt{14} - \frac{1}{ \sqrt{14} } + \frac{ \sqrt{14} }{7} - \frac{3 \sqrt{14} }{14} \right)=$
$\displaystyle =\left( \frac{5}{2 \sqrt{7} } - \frac{2 \sqrt{7} }{7} + \frac{3}{2 \sqrt{7} } \right) \cdot \sqrt{7} + \frac{ \sqrt{14} }{2} \cdot \left( \sqrt{14} - \frac{ \sqrt{14} }{14} + \frac{ \sqrt{14} }{7} - \frac{3 \sqrt{14} }{14} \right)=$
$\displaystyle =\left( \frac{5 \sqrt{7} }{2 \sqrt{7} } - \frac{14}{7} + \frac{3 \sqrt{7} }{2 \sqrt{7} } \right)+\left( \frac{14}{2} - \frac{14}{28} + \frac{14}{14} - \frac{42}{28} \right)= \\ \\ =\left( \frac{5}{2} -2+ \frac{3}{2} \right)+ \frac{196-14+28-42}{28} = \frac{5-4+3}{2} + \frac{168}{28} = \\ \\ = \frac{4}{2} +6=2+6=8$