il primo che inizia con la quadra

1,6 (con 6 periodico) = (16 - 1) /9 = 15/9 = 5/3;

0,4 (con 4 periodico) = (4 - 0) / 9 = 4/9;

0,83 (con solo 3 periodico) = (83 - 8) /90 = 75/90 = 5/6; (semplificando per 15);

0,5 = 5/10 = 1/2;

1,5 = 15/10 = 3/2;

[5/3 + (2 + 4/9) : (5/6 + 1/2)] - 3/2 = 

= [5/3 + (18/9 + 4/9) : (5/6 + 3/6)] - 3/2 =

= [5/3 + 22/9 : 8/6 ] - 3/2 =

(semplifichiamo 8/6 = 4/3;  22/9 : 4/3 = 22/9 * 3/4 = 22 * 3 / (9 *4 = 11 /6);

= [5/3 + 22/9 * 3/4] - 3/2 =

= [5/3 + 11/6] - 3/2 =

= [10/6 + 11/6] - 3/2 =

= 21/6 - 3/2 =

= 21/6 - 9/6 =

= 12/6 = 2.

@nicole_dotta  ciao;  non so se ho svolto quella richiesta...

ho corretto in grassetto;   avevo scritto 5/6 - 1/2 .... invece è  5/6 + 1/2;

@nicole_dotta  ciao  di nuovo.    

===========================================================

$\small \left[1,\overline6+\left(2+0,\overline4\right) : \left(0,8\overline3+0,5\right)\right]-1,5=$

$\small = \left[\dfrac{16-1}{9}+\left(2+\dfrac{4-0}{9}\right):\left(\dfrac{83-8}{90}+\dfrac{5}{10}\right)\right] - \dfrac{15}{10}=$

$\small = \left[\dfrac{15}{9}+\left(2+\dfrac{4}{9}\right):\left(\dfrac{75}{90}+\dfrac{5}{10}\right)\right] - \dfrac{15}{10}=$

$\small = \left[\dfrac{\cancel{15}^5}{\cancel9_3}+\left(\dfrac{18+4}{9}\right):\left(\dfrac{\cancel{75}^5}{\cancel{90}_6}+\dfrac{\cancel5^1}{\cancel{10}_2}\right)\right] - \dfrac{\cancel{15}^3}{\cancel{10}_2}=$

$\small =\left[\dfrac{5}{3}+\dfrac{22}{9} : \left(\dfrac{5}{6}+\dfrac{1}{2}\right)\right] - \dfrac{3}{2}=$

$\small =\left[\dfrac{5}{3}+\dfrac{22}{9} : \left(\dfrac{5+3}{6}\right)\right] - \dfrac{3}{2}=$

$\small =\left[\dfrac{5}{3}+\dfrac{22}{9} : \dfrac{\cancel8^4}{\cancel6_3}\right] - \dfrac{3}{2}=$

$\small =\left[\dfrac{5}{3}+\dfrac{22}{9} : \dfrac{4}{3}\right] - \dfrac{3}{2}=$

$\small =\left[\dfrac{5}{3}+\dfrac{\cancel{22}^{11}}{\cancel9_3}·\dfrac{\cancel3^1}{\cancel4_2}\right] - \dfrac{3}{2}=$

$\small =\left[\dfrac{5}{3}+\dfrac{11}{3}·\dfrac{1}{2}\right] - \dfrac{3}{2}=$

$\small =\left[\dfrac{5}{3}+\dfrac{11}{6}\right] - \dfrac{3}{2}=$

$\small =\left[\dfrac{10+11}{6}\right] - \dfrac{3}{2}=$

$\small = \dfrac{\cancel{21}^7}{\cancel6_2} - \dfrac{3}{2}=$

$\small = \dfrac{7}{2} - \dfrac{3}{2}=$

$\small = \dfrac{\cancel4^2}{\cancel2_1} = 2$