Question

ajutati-ma va rog dau coroana va rog toate exercitiile​

Answer 1

$\it e)\ \sqrt{3-2\sqrt2}=\sqrt{2+1-2\sqrt2}=\sqrt{(\sqrt2)^2-2\sqrt2+1^2}=\\ \\ \\ =\sqrt{(\sqrt2-1)^2}=|\underbrace{\sqrt2-1}_{ > 0}|=\sqrt2-1\\ \\ \\ m_a=\dfrac{\sqrt2-1+\sqrt2+1}{2}=\dfrac{2\sqrt2}{2}=\sqrt2\\ \\ \\ m_g=\sqrt{(\sqrt2-1)(\sqrt2+1)}=\sqrt{(\sqrt2)^2-1^2}=\sqrt{2-1}=\sqrt1=1$

$\it c)\ \ m_a=\dfrac{7-4\sqrt3+7+4\sqrt3}{2}=\dfrac{14}{2}=7\\ \\ \\ m_g=\sqrt{(7-4\sqrt3)(7+4\sqrt3)}=\sqrt{7^2-(4\sqrt3)^2}=\sqrt{49-48}=\sqrt1=1$

$\it f)\ \sqrt{5+2\sqrt6}=\sqrt{3+2+2\sqrt6}=\sqrt{(\sqrt3)^2+2\sqrt6+(\sqrt2)^2}=\\ \\ =\sqrt{(\sqrt3+\sqrt2)^2}=\sqrt3+\sqrt2\\ \\ \\ m_a=\dfrac{\sqrt3+\sqrt2+\sqrt3-\sqrt2}{2}=\dfrac{2\sqrt3}{2}=\sqrt3\\ \\ \\ m_g=\sqrt{(\sqrt3+\sqrt2)(\sqrt3-\sqrt2)}=\sqrt{3-2}=1$