Ta có:
$\dfrac{52}{9} =5 + \dfrac{1}{a + \dfrac{1}{ b + \dfrac{1}{c}}}$
$⇔ 5 + \dfrac{7}{9} = 5 + \dfrac{1}{a + \dfrac{1}{ b + \dfrac{1}{c}}}$
$⇔ \dfrac{7}{9} = \dfrac{1}{a + \dfrac{1}{ b + \dfrac{1}{c}}}$
$⇔ \dfrac{1}{\dfrac{9}{7}} = \dfrac{1}{a+ \dfrac{1}{ b + \dfrac{1}{c}}}$
$⇔ \dfrac{9}{7} = a + \dfrac{1}{b + \dfrac{1}{c}}$
Vì : $\dfrac{1}{b + \dfrac{1}{c}}$ $<1$
$⇒ 1 + \dfrac{2}{7} = 1 + \dfrac{1}{b + \dfrac{1}{c}}$ $⇒ a=1$
$⇔ \dfrac{2}{7} = \dfrac{1}{b + \dfrac{1}{c}}$
$⇔ \dfrac{1}{\dfrac{7}{2}} = \dfrac{1}{b + \dfrac{1}{c}}$
$⇔ \dfrac{7}{2} = b + \dfrac{1}{c}$
Vì : $\dfrac{1}{c}< 1$
$⇒ 3 +\dfrac{1}{2} = b + \dfrac{1}{c}$
$⇒ b=3;c=2$
Vậy `(a;b;c)=(1;3;2)`
