3,(4)=(34-3)/9=31/9
1,(81)=(181-1)/99=180/99=20/12
5,(6)=(56-5)/9=51/9=17/3
0,(36)=(36-0)/99=36/99=4/11
10,(6)=(106-10)/9=96/9=32/3
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114)
$\small 3,\overline4= \dfrac{34-3}{9} = \dfrac{31}{9};$
$\small 1,\overline{81} = \dfrac{181-1}{99} = \dfrac{\cancel{180}^{20}}{\cancel{99}_{11}} = \dfrac{20}{11};$
$\small 5,\overline6= \dfrac{56-5}{9} = \dfrac{\cancel{51}^{17}}{\cancel9_3} = \dfrac{17}{3};$
$\small 0,\overline{36}= \dfrac{36-0}{99} = \dfrac{\cancel{36}^4}{\cancel{99}_{11}} = \dfrac{4}{11};$
$\small 10,\overline6= \dfrac{106-10}{9} = \dfrac{\cancel{96}^{32}}{\cancel9_3} = \dfrac{32}{3}.$
3,4 (con 4 periodico) = (34 - 3)/ 9 = 31/9;
1,81 (con 81 periodico) = (181 - 1) / 99 = 180/99 = 20/11; (semplificando per 9).
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