1)
Deoarece x=y = 2√5 - 4 ⇒ cele trei medii sunt egale :
[tex]\it m_h=m_g=m_a = 2\sqrt5-4[/tex]
2)
[tex]\it x = (2\sqrt3+1)^2 = (2\sqrt3)^2 +2\cdot2\sqrt3\cdot1+1^2 =12+4\sqrt3+1=
\\\;\\
=13+4\sqrt3
\\\;\\
y=13-4\sqrt3[/tex]
[tex]\it m_a =\dfrac{x+y}{2} =\dfrac{13+4\sqrt3+13-4\sqrt3}{2} =\dfrac{26}{2} =13
\\\;\\ \\\;\\m_g = \sqrt{xy} =\sqrt{(13+4\sqrt3)(13-4\sqrt3)} =\sqrt{13^2-(4\sqrt3)^2} =
\\\;\\
=\sqrt{169-48} = \sqrt{121} =11[/tex]
[tex]\it m_h=\dfrac{m_g^2}{m_a} = \dfrac{11^2}{13} =\dfrac{121}{13}[/tex]
