Secondo me hai svolto bene, risulta 3/8 non 1/4, infatti:
$\small \left[\dfrac{4}{3}-\left(\dfrac{3}{7}-\dfrac{2}{5}\right)·\dfrac{20}{1}-\dfrac{\cancel9^3}{\cancel{21}_7}\right] : \left(\dfrac{1}{3} : \dfrac{3}{1} -\dfrac{17}{36}+\dfrac{5}{4}\right)=$
$\small =\left[\dfrac{4}{3}-\left(\dfrac{15-14}{35}\right)·20-\dfrac{3}{7}\right] : \left(\dfrac{1}{3} · \dfrac{1}{3} -\dfrac{17}{36}+\dfrac{5}{4}\right)=$
$\small =\left[\dfrac{4}{3}-\dfrac{1}{\cancel{35}_7}·\cancel{20}^4-\dfrac{3}{7}\right] : \left(\dfrac{1}{9} -\dfrac{17}{36}+\dfrac{5}{4}\right)=$
$\small =\left[\dfrac{4}{3}-\dfrac{1}{7}·4-\dfrac{3}{7}\right] : \left(\dfrac{4-17+45}{36}\right)=$
$\small =\left[\dfrac{4}{3}-\dfrac{4}{7}-\dfrac{3}{7}\right] : \dfrac{\cancel{32}^8}{\cancel{36}_9} =$
$\small =\left[\dfrac{28-12-9}{21}\right] : \dfrac{8}{9} =$
$\small = \dfrac{\cancel7^1}{\cancel{21}_3} · \dfrac{9}{8} =$
$\small = \dfrac{1}{\cancel3_1} · \dfrac{\cancel9^3}{8} =$
$\small = \dfrac{1}{1} · \dfrac{3}{8} = $
$\small = \dfrac{3}{8} $