Question

Va roggg multtt!!!Urgent!!Sa fie si corecta.

Answer 1

[tex]\displaystyle \mathtt{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}=}\\ \\ \mathtt{= \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} } =}[/tex]
[tex]\displaystyle \mathtt{= \sqrt{2}+ \sqrt{3}- \sqrt{3}- \sqrt{5}+ \sqrt{5}- \sqrt{2} }=0[/tex]

$\displaystyle \mathtt{b)\left( \frac{2}{3 \sqrt{6} }- \frac{1}{2 \sqrt{6} }\right) \cdot \sqrt{6}+ \left( \sqrt{24}- \frac{5}{2 \sqrt{6} } \right) \cdot \sqrt{6}- \frac{29}{3} =}\\ \\ \mathtt{= \left(\frac{ 2\sqrt{6} }{3 \sqrt{6} }- \frac{ \sqrt{6} }{2 \sqrt{6} }\right)+\left(2 \sqrt{6} - \frac{5}{2 \sqrt{6} }\right) \cdot \sqrt{6} - \frac{29}{3}= }$

$\displaystyle \mathtt{=\left( \frac{2}{3} - \frac{1}{2} \right)+\left(12- \frac{5 \sqrt{6} }{2 \sqrt{6} }\right)- \frac{29}{3} = \frac{4-3}{6} +\left(12- \frac{5}{2} \right)- \frac{29}{3} = }\\ \\ \mathtt{= \frac{1}{6}+ \frac{24-5}{2}- \frac{29}{3}= \frac{1}{6}+ \frac{19}{2}- \frac{29}{3} = \frac{1+57-58}{6}=0 }$

$\displaystyle \mathtt{c)3 \sqrt{125} \cdot \left( \frac{2}{ \sqrt{5} }+ \frac{3}{2 \sqrt{5} }- \frac{1}{ \sqrt{5}}\right): \sqrt{5}+\left( \frac{14}{3 \sqrt{5} }- \frac{ \sqrt{45} }{15}+ \frac{12}{ \sqrt{5} }\right) \cdot \frac{5}{47}=}$

$\displaystyle \mathtt{=15 \sqrt{5}\cdot\left(\frac{2}{ \sqrt{5} }+ \frac{3}{2 \sqrt{5} }- \frac{1}{ \sqrt{5}}\right) \cdot \frac{1}{ \sqrt{5} } + \left( \frac{14 \sqrt{5} }{15}- \frac{3 \sqrt{5} }{15} + \frac{12 \sqrt{5} }{5} \right) \cdot \frac{5}{47} =}$

$\displaystyle \mathtt{=15 \sqrt{5}\cdot \left( \frac{2}{5} + \frac{3}{10}- \frac{1}{5}\right)+ \frac{14 \sqrt{5}-3 \sqrt{5}+36 \sqrt{5}}{15} \cdot \frac{5}{47}= }\\ \\ \mathtt{=15 \sqrt{5}\cdot \frac{4+3-2}{10}+ \frac{47 \sqrt{5} }{15} \cdot \frac{5}{47} =15 \sqrt{5}\cdot \frac{5}{10}+ \frac{ \sqrt{5} }{3}=}\\ \\ \mathtt{=15 \sqrt{5}\cdot \frac{1}{2}+ \frac{ \sqrt{5} }{3}= \frac{15 \sqrt{5} }{2}+ \frac{ \sqrt{5} }{3}= \frac{45 \sqrt{5}+2 \sqrt{5} }{6}= \frac{47 \sqrt{5} }{6} }$

$\displaystyle \mathtt{d)\left( \frac{5}{ \sqrt{28} }- \frac{2 \sqrt{7} }{7}+ \frac{3}{2 \sqrt{7} }\right)}: \frac{1}{ \sqrt{7}}+ \frac{ \sqrt{14} }{2} \cdot \left( \sqrt{14}- \frac{1}{ \sqrt{14} }+ \frac{ \sqrt{14} }{7} - \frac{3 \sqrt{14} }{14}\right)=}$

$\displaystyle \mathtt{=\left( \frac{5}{2 \sqrt{7} }- \frac{2 \sqrt{7} }{7}+ \frac{3}{2 \sqrt{7} }\right) \cdot \sqrt{7}+ \frac{ \sqrt{14} }{2}\cdot\left( \sqrt{14}- \frac{ \sqrt{14} }{14} + \frac{ \sqrt{14} }{7}- \frac{3 \sqrt{14} }{14}\right)=}$

$\displaystyle \mathtt{=\left( \frac{5 \sqrt{7} }{2 \sqrt{7} }- \frac{14}{7}+ \frac{3 \sqrt{7} }{2 \sqrt{7} } \right)+\left( \frac{14}{2}- \frac{14}{28} + \frac{14}{14}- \frac{42}{28}\right)=} \\ \\ \mathtt{= \left(\frac{5}{2}-2+ \frac{3}{2}\right)+ \frac{196-14+28-42}{28}= \frac{5-4+3}{2}+ \frac{168}{28}=} \\ \\ \mathtt{= \frac{4}{2}+6=2+6=8 }$