Espressione

\[\frac{9}{13} \times \left\{0,7\overline{3} + [0,14 + 0,16 + (0,\overline{1} \times 9)^2 - 0,\overline{5}] - \frac{5}{12} \times 0,08\right\}\]

\[\frac{9}{13} \times \left\{\frac{11}{15} + \left[\frac{7}{50} + \frac{4}{25} + 1 - \frac{3}{5}\right] - \frac{5}{12} \times \frac{2}{25}\right\}\]

\[\frac{9}{13} \times \left\{\frac{11}{15} + \left[\frac{63}{450} + \frac{72}{450} + \frac{450}{450} - \frac{250}{450}\right] - \frac{5}{12} \times \frac{2}{25}\right\}\]

\[\frac{9}{13} \times \left\{\frac{11}{15} + \frac{67}{90} - \frac{1}{30}\right\}\]

\[\frac{9}{13} \times \left\{\frac{66}{90} + \frac{67}{90} - \frac{3}{90}\right\}\]

\[\frac{9}{13} \times \frac{13}{9} = 1\,.\]

180)

$\dfrac{9}{13}×\left\{0,7\overline3+\left[0,14+0,16+\left(0,\overline1×9\right)^2-0,\overline5\right]-\dfrac{5}{12}×0,08\right\}=$

$=\dfrac{9}{13}×\left\{\dfrac{73-7}{90}+\left[\dfrac{14}{100}+\dfrac{16}{100}+\left(\dfrac{1-0}{9}×9\right)^2-\dfrac{5-0}{9}\right]-\dfrac{5}{12}×\dfrac{8}{100}\right\}=$

$=\dfrac{9}{13}×\left\{\dfrac{66}{90}+\left[\dfrac{\cancel{14}^7}{\cancel{100}_{50}}+\dfrac{\cancel{16}^4}{\cancel{100}_{25}}+\left(\dfrac{1}{9}×9\right)^2-\dfrac{5}{9}\right]-\dfrac{5}{12}×\dfrac{\cancel8^2}{\cancel{100}_{25}}\right\}=$

$=\dfrac{9}{13}×\left\{\dfrac{11}{15}+\left[\dfrac{63+72+450-250}{450}\right]-\dfrac{1}{30}\right\}=$

$=\dfrac{9}{13}×\left\{\dfrac{11}{15}+\dfrac{\cancel{335}^{67}}{\cancel{450}_{90}}-\dfrac{1}{30}\right\}=$

$=\dfrac{9}{13}×\left\{\dfrac{11}{15}+\dfrac{67}{90}-\dfrac{1}{30}\right\}=$

$=\dfrac{9}{13}×\left\{\dfrac{66+67-3}{90}\right\}=$

$=\dfrac{9}{13}×\dfrac{\cancel{130}^{13}}{\cancel{90}_9}=$

$=\dfrac{\cancel9^1}{\cancel{13}_1}×\dfrac{\cancel{13}^1}{\cancel9_1}=$

$=1$