[tex]\it \dfrac{a}{b} =3-\sqrt5
\\\;\\\;
\dfrac{a^2+b^2}{ab} =\dfrac{a^2}{ab} + \dfrac{b^2}{ab} = \dfrac{a}{b} + \dfrac{b}{a} = 3-\sqrt5 +\dfrac{1}{3-\sqrt5} =
\\\;\\ \\\;\\
=3-\sqrt5+\dfrac{3+\sqrt5}{4} = \dfrac{12-4\sqrt5+3+\sqrt5}{4} =\dfrac{15-3\sqrt5}{4}
[/tex]
