Proporzioni

0,3 periodico = 3/9 = 1/3;

0,6 periodico = 6/9 = 2/3;

x : (1/3 + 1/5) = radicequadrata[11/24 : (9/6 + 2/6)] : (2 - 2/3);

x : (5/15 + 3/15) = radice[11/24 * 6/11] : (6/3 - 2/3);

x : 8/15 = radice[1/4] : 4/3;

x : 8/15 = 1/2 : 4/3;    (il prodotto dei medi è uguale al prodotto degli estremi).

x = (8/15 * 1/2) : 4/3;

x = 4/15 * 3/4;

x = 1/5.

Ciao @lolorena

x/(1/3 + 1/5) = √(11/24/(3/2 + 1/3))/(2 - 2/3)

x/(8/15) = √(11/24/(11/6))/(2 - 2/3)

x/(8/15) = √(11/24/(11/6))/(4/3)

x/(8/15) = √(1/4)/(4/3)

x/(8/15) = 1/2/(4/3)

x/(8/15) = 3/8

8·x = 3·8/15

8·x = 8/5

x = 1/5

x : (1/3+1/5) = √(11/24*6/11) : (2-2/3) 

x : 8/15 = 1/2 : 4/3 

4x/3 = 8/30

x = 8/30*3/4 = 2/10 = 1/5  

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$\small x : \left(0,\overline3+\dfrac{1}{5}\right) = \sqrt{\dfrac{11}{24} : \left(\dfrac{3}{2}+\dfrac{1}{3}\right)} : \left(2-0,\overline6\right)$

$\small x : \left(\dfrac{3-0}{9}+\dfrac{1}{5}\right) = \sqrt{\dfrac{11}{24} : \left(\dfrac{9+2}{6}\right)} : \left(2-\dfrac{6-0}{9}\right)$

$\small x : \left(\dfrac{\cancel3^1}{\cancel9_3}+\dfrac{1}{5}\right) = \sqrt{\dfrac{11}{24} : \dfrac{11}{6}} : \left(2-\dfrac{\cancel6^2}{\cancel9_3}\right)$

$\small x : \left(\dfrac{1}{3}+\dfrac{1}{5}\right) = \sqrt{\dfrac{\cancel{11}^1}{\cancel{24}_4} · \dfrac{\cancel6^1}{\cancel{11}_1}} : \left(2-\dfrac{2}{3}\right)$

$\small x : \left(\dfrac{5+3}{15}\right) = \sqrt{\dfrac{1}{4} · \dfrac{1}{1}} : \left(\dfrac{6-2}{3}\right)$

$\small x : \dfrac{8}{15} = \sqrt{\dfrac{1}{4} } : \dfrac{4}{3}$

$\small x : \dfrac{8}{15} = \dfrac{1}{2} : \dfrac{4}{3}$

$\small x = \dfrac{\cancel8^4}{15}·\dfrac{1}{\cancel2_1} : \dfrac{4}{3}$

$\small x = \dfrac{4}{15}·\dfrac{1}{1} · \dfrac{3}{4}$

$\small x = \dfrac{\cancel4^1}{\cancel{15}_5}· \dfrac{\cancel3^1}{\cancel4_1}$

$\small x = \dfrac{1}{5}· \dfrac{1}{1}$

$\small x = \dfrac{1}{5}$