Preposizioni sottoforma di espressioni

(1+1/4-1/6):×=(9/2:5/3+6/5):(3/2-3/8)

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$\small \left(1+\dfrac{1}{4}-\dfrac{1}{6}\right) : x = \left(\dfrac{9}{2} : \dfrac{5}{3}+\dfrac{6}{5}\right) : \left(\dfrac{3}{2}-\dfrac{3}{8}\right)$

$\small \left(\dfrac{12+3-2}{12}\right) : x = \left(\dfrac{9}{2}·\dfrac{3}{5}+\dfrac{6}{5}\right) : \left(\dfrac{12-3}{8}\right)$

$\small \dfrac{13}{12} : x = \left(\dfrac{27}{10}+\dfrac{6}{5}\right) : \dfrac{9}{8}$

$\small \dfrac{13}{12} : x = \left(\dfrac{27+12}{10}\right) : \dfrac{9}{8}$

$\small \dfrac{13}{12} : x = \dfrac{39}{10} : \dfrac{9}{8}$

$\small x = \dfrac{13}{\cancel{12}_4} · \dfrac{\cancel9^3}{8} : \dfrac{39}{10}$

$\small x = \dfrac{13}{4} · \dfrac{3}{8}·\dfrac{10}{39}$

$\small x = \dfrac{13}{4} · \dfrac{\cancel3^1}{\cancel8_4}·\dfrac{\cancel{10}^5}{\cancel{39}_{13}}$

$\small x = \dfrac{\cancel{13}^1}{4} · \dfrac{1}{4}·\dfrac{5}{\cancel{13}_1}$

$\small x = \dfrac{1}{4} · \dfrac{1}{4}·\dfrac{5}{1} = \dfrac{5}{16}$ $\;(= 0,3125).$