Aritmetica

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$\left(\dfrac{8}{9}-\dfrac{11}{15}\right)+\left\{\left[\left(\dfrac{3}{5}+\dfrac{5}{6}-1+\dfrac{1}{15}\right) : \left(2-\dfrac{2}{3}·\dfrac{9}{4}·\dfrac{5}{18}+\dfrac{1}{2}\right)-\dfrac{1}{5}\right] : \dfrac{3}{5}\right\}-\dfrac{2}{9}=$

$=\left(\dfrac{40-33}{45}\right)+\left\{\left[\left(\dfrac{18+25-30+2}{30}\right) : \left(2-\dfrac{\cancel2^1}{3}·\dfrac{\cancel9^1}{\cancel4_2}·\dfrac{5}{\cancel{18}_2}+\dfrac{1}{2}\right)-\dfrac{1}{5}\right]·\dfrac{5}{3}\right\}-\dfrac{2}{9}=$

$=\dfrac{7}{45}+\left\{\left[\dfrac{15}{30} : \left(2-\dfrac{1}{3}·\dfrac{1}{2}·\dfrac{5}{2}+\dfrac{1}{2}\right)-\dfrac{1}{5}\right]·\dfrac{5}{3}\right\}-\dfrac{2}{9}=$

$=\dfrac{7}{45}+\left\{\left[\dfrac{\cancel{15}^1}{\cancel{30}_2} : \left(2-\dfrac{5}{12}+\dfrac{1}{2}\right)-\dfrac{1}{5}\right]·\dfrac{5}{3}\right\}-\dfrac{2}{9}=$

$=\dfrac{7}{45}+\left\{\left[\dfrac{1}{2} : \left(\dfrac{24-5+6}{12}\right)-\dfrac{1}{5}\right]·\dfrac{5}{3}\right\}-\dfrac{2}{9}=$

$=\dfrac{7}{45}+\left\{\left[\dfrac{1}{2} : \dfrac{25}{12}-\dfrac{1}{5}\right]·\dfrac{5}{3}\right\}-\dfrac{2}{9}=$

$=\dfrac{7}{45}+\left\{\left[\dfrac{1}{2}·\dfrac{12}{25}-\dfrac{1}{5}\right]·\dfrac{5}{3}\right\}-\dfrac{2}{9}=$

$=\dfrac{7}{45}+\left\{\left[\dfrac{1}{\cancel2_1}·\dfrac{\cancel{12}^6}{25}-\dfrac{1}{5}\right]·\dfrac{5}{3}\right\}-\dfrac{2}{9}=$

$=\dfrac{7}{45}+\left\{\left[\dfrac{1}{1}·\dfrac{6}{25}-\dfrac{1}{5}\right]·\dfrac{5}{3}\right\}-\dfrac{2}{9}=$

$=\dfrac{7}{45}+\left\{\left[\dfrac{6}{25}-\dfrac{1}{5}\right]·\dfrac{5}{3}\right\}-\dfrac{2}{9}=$

$=\dfrac{7}{45}+\left\{\left[\dfrac{6-5}{25}\right]·\dfrac{5}{3}\right\}-\dfrac{2}{9}=$

$=\dfrac{7}{45}+\left\{\dfrac{1}{25}·\dfrac{5}{3}\right\}-\dfrac{2}{9}=$

$=\dfrac{7}{45}+\left\{\dfrac{1}{\cancel{25}_5}·\dfrac{\cancel5^1}{3}\right\}-\dfrac{2}{9}=$

$=\dfrac{7}{45}+\left\{\dfrac{1}{5}·\dfrac{1}{3}\right\}-\dfrac{2}{9}=$

$=\dfrac{7}{45}+\dfrac{1}{15}-\dfrac{2}{9}=$

$=\dfrac{7+3-10}{45}=$

$=\dfrac{0}{45}=$

$= 0$