Question

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Answer 1

[tex]\displaystyle 1d) \\ \\ \left( \frac{12}{ \sqrt{10} } + \frac{ \sqrt{40} }{5} + \frac{20}{2 \sqrt{10}} \right) - \left( \frac{ 6}{ \sqrt{90} } + \frac{5 \sqrt{10} }{2} - \frac{3}{2 \sqrt{10}}\right)= \\ \\ = \frac{12}{ \sqrt{10} } + \frac{ \sqrt{40} }{5} + \frac{20}{2 \sqrt{10}} - \frac{ 6}{ 3\sqrt{10} } - \frac{5 \sqrt{10} }{2} + \frac{3}{2 \sqrt{10}}= [/tex]


[tex]\displaystyle \\ = \frac{12}{ \sqrt{10} } + \frac{ 2\sqrt{10} }{5} + \frac{10}{\sqrt{10}} - \frac{ 2}{ \sqrt{10} } - \frac{5 \sqrt{10} }{2} + \frac{3}{2 \sqrt{10}}= \\ \\ = \frac{120}{ 10\sqrt{10} } + \frac{ 40}{10\sqrt{10} } + \frac{100}{10\sqrt{10} } - \frac{ 20}{ 10\sqrt{10} } - \frac{250 }{10\sqrt{10} } + \frac{6}{10 \sqrt{10}}= \\ \\ = \frac{120+40+100-20-250+6=}{ 10\sqrt{10} }= \frac{-4}{10\sqrt{10}} = \frac{-2} {5\sqrt{10}} = \boxed{-\frac{\sqrt{10}} {25}}[/tex]


[tex]\displaystyle 2)\\ a= \frac{1}{2} + \frac{3}{4} + \frac{5}{6}+ \hdots+\frac{2013}{2014}\\ \\ \\ b= \frac{ \frac{1}{1} + \frac{1}{2}+\frac{1}{3}+ \hdots +\frac{1}{1007} }{2}= \frac{1}{2} + \frac{1}{4}+\frac{1}{6}+ \hdots +\frac{1}{2014} \\ \\ \text{Ambele sume au:} \frac{2014-2}{2}+1 =\frac{2012}{2}+1 = 1006+1= 1007~termeni)[/tex]


[tex]\displaystyle \\ a+b =\left(\frac{1}{2} + \frac{3}{4} + \frac{5}{6}+ \hdots+\frac{2013}{2014}\right)+\left(\frac{1}{2} + \frac{1}{4}+\frac{1}{6}+ \hdots +\frac{1}{2014}\right)= \\ \\ \left(\frac{1}{2}+\frac{1}{2}\right) + \left(\frac{3}{4}+ \frac{1}{4}\right) + \left(\frac{5}{6}+\frac{1}{6}\right)+ \hdots+\left(\frac{2013}{2014}+\frac{1}{2014}\right)= \\ \\ = \frac{2}{2} + \frac{4}{4} + \frac{6}{6} + \hdots+ \frac{2014}{2014} = \\ \\ = 1+1+1+ ... de ~1007~ori ...+1+1 = \boxed{1007}[/tex]


[tex]\displaystyle \\ m_a = \frac{a+b}{2} = \frac{1007}{2} = \boxed{503,5} [/tex]