[tex]\it a=\sqrt{128}=\sqrt{64\cdot2}=8\sqrt2\\ \\ b=10\Big(\dfrac{1}{\sqrt{25\cdot2}}+\dfrac{\sqrt5}{10}-\dfrac{1}{\sqrt{4\cdot5}}\Big)=10\Big(\dfrac{^{\sqrt2)}1}{5\sqrt{2}}+\dfrac{\sqrt5}{10}-\dfrac{^{\sqrt5)}1}{2\sqrt{5}}\Big)=\\ \\ \\ =10\Big(\dfrac{\sqrt2}{10}+\dfrac{\sqrt5}{10}-\dfrac{\sqrt5}{10}\Big)=\sqrt2+\sqrt5-\sqrt5=\sqrt2[/tex]
[tex]\it m_a=\dfrac{a+b}{2}=\dfrac{8\sqrt2+\sqrt2}{2}=\dfrac{9\sqrt2}{2}=4,5\cdot\sqrt2\approx4,5\cdot1,41\approx6,4[/tex]
