Question

3supra2radical din 3+7supra radical din3)•15 supra radical din 48

Answer 1

Gasesti atasat rezolvarea

Answer 2

   
$\displaystyle \bf\\ \left(\ \frac{3}{2 \sqrt{3}} + \frac{7}{\sqrt{3}} \right)\cdot \frac{15}{\sqrt{48}} =\\\\\\ =\left(\ \frac{3}{2 \sqrt{3}} + \frac{2\cdot 7}{2\sqrt{3}} \right)\cdot \frac{15}{\sqrt{16\cdot 3}} =\\\\\\ =\left(\ \frac{3}{2 \sqrt{3}} + \frac{14}{2\sqrt{3}} \right)\cdot \frac{15}{4\sqrt{3}} =\\\\\\ =\frac{17}{2\sqrt{3}} \cdot \frac{15}{4\sqrt{3}} = \frac{17\times 15}{2\sqrt{3}\times 4\sqrt{3}} =\frac{255}{8\times 3} = \frac{255}{24}= \boxed{\bf \frac{85}{8}}$