Răspuns :
[tex]\it a=\sqrt{12}+\sqrt3=\sqrt{4\cdot3}+\sqrt3=2\sqrt3+\sqrt3=3\sqrt3\\ \\ \\ b=\dfrac{^{\sqrt3)}15}{\ \sqrt3}=\dfrac{15\sqrt3}{3}=5\sqrt3\\ \\ \\ m_a=\dfrac{a+b}{2}=\dfrac{3\sqrt3+5\sqrt3}{2}=\dfrac{8\sqrt3}{2}=4\sqrt3\ .[/tex]
[tex]\it a=\sqrt{12}+\sqrt3=\sqrt{4\cdot3}+\sqrt3=2\sqrt3+\sqrt3=3\sqrt3\\ \\ \\ b=\dfrac{^{\sqrt3)}15}{\ \sqrt3}=\dfrac{15\sqrt3}{3}=5\sqrt3\\ \\ \\ m_a=\dfrac{a+b}{2}=\dfrac{3\sqrt3+5\sqrt3}{2}=\dfrac{8\sqrt3}{2}=4\sqrt3\ .[/tex]