Ex 4

(16/25 * 5/8 + 4/9 - (10-8)/45) / ((9/4 * (1-1/2) -1)*8)

(2/5 + 4/9 - 2/45) / ((9/8 - 1) * 8)

(18+20)/45 - 2/45) / (1/8 * 8)

36/45

4/5 

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$\small \dfrac{\left(1-\dfrac{1}{5}\right)^2÷\left(2-\dfrac{2}{5}\right)+\left(\dfrac{2}{3}\right)^2-\left(\dfrac{2}{9}-\dfrac{8}{45}\right)}{\left[\left(1+\dfrac{1}{2}\right)^2×\left(1-\dfrac{\cancel4^1}{\cancel3_1}×\dfrac{\cancel3^1}{\cancel8_2}\right)-1\right]÷\left(\dfrac{1}{2}\right)^3}=$

$\small = \dfrac{\left(\dfrac{5-1}{5}\right)^2÷\left(\dfrac{10-2}{5}\right)+\dfrac{4}{9}-\left(\dfrac{10-8}{45}\right)}{\left[\left(\dfrac{2+1}{2}\right)^2×\left(1-\dfrac{1}{1}×\dfrac{1}{2}\right)-1\right]÷\dfrac{1}{8}}=$

$\small = \dfrac{\left(\dfrac{4}{5}\right)^2÷\dfrac{8}{5}+\dfrac{4}{9}-\dfrac{2}{45}}{\left[\left(\dfrac{3}{2}\right)^2×\left(1-\dfrac{1}{2}\right)-1\right]×8}=$

$\small = \dfrac{\dfrac{\cancel{16}^2}{\cancel{25}_5}×\dfrac{\cancel5^1}{\cancel8_1}+\dfrac{4}{9}-\dfrac{2}{45}}{\left[\dfrac{9}{4}×\left(\dfrac{2-1}{2}\right)-1\right]×8}=$

$\small = \dfrac{\dfrac{2}{5}×\dfrac{1}{1}+\dfrac{4}{9}-\dfrac{2}{45}}{\left[\dfrac{9}{4}×\dfrac{1}{2}-1\right]×8}=$

$\small = \dfrac{\dfrac{2}{5}+\dfrac{4}{9}-\dfrac{2}{45}}{\left[\dfrac{9}{8}-1\right]×8}=$

$\small = \dfrac{\dfrac{18+20-2}{45}}{\left[\dfrac{9-8}{8}\right]×8}=$

$\small = \dfrac{\dfrac{\cancel{36}^4}{\cancel{45}_5}}{\dfrac{1}{\cancel8_1}×\cancel8^1}=$

$\small = \dfrac{\dfrac{4}{5}}{\dfrac{1}{1}×1}=$

$\small = \dfrac{\dfrac{4}{5}}{1}=$

$\small = \dfrac{4}{5}$