Espressione che non torna

(((4/9 - 1/3)^2 - (3/5 - 1/3)^2·(5/3)^2 + 1/2)/(- 17/27) + 3/4)^2=

=(((1/9)^2 - (4/15)^2·(5/3)^2 + 1/2)/(- 17/27) + 3/4)^2=

=((1/81 - (4/15)^2·(5/3)^2 + 1/2)/(- 17/27) + 3/4)^2=

=((1/81 - 16/81 + 1/2)/(- 17/27) + 3/4)^2=

=(17/54/(- 17/27) + 3/4)^2=

=(- 1/2 + 3/4)^2=

=(1/4)^2= 1/16

 Intanto controlla il passaggio (la sesta riga) dove, per una svista, hai messo il minimo comun denominatore 81 che invece doveva essere 162, secondo me l'errore è lì. Controllo meglio il resto e semmai ti mando un esempio. Saluti.

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$\small \left\{\left[\left(\dfrac{4}{9}-\dfrac{1}{3}\right)^2-\left(\dfrac{3}{5}-\dfrac{1}{3}\right)^2·\left(\dfrac{5}{3}\right)^2+\dfrac{1}{2}\right]÷\left(-\dfrac{17}{27}\right)+\dfrac{3}{4}\right\}^2 =$

$\small = \left\{\left[\left(\dfrac{4-3}{9}\right)^2-\left(\dfrac{9-5}{15}\right)^2·\dfrac{25}{9}+\dfrac{1}{2}\right]·\left(-\dfrac{27}{17}\right)+\dfrac{3}{4}\right\}^2 =$

$\small = \left\{\left[\left(\dfrac{1}{9}\right)^2-\left(\dfrac{4}{15}\right)^2·\dfrac{25}{9}+\dfrac{1}{2}\right]·\left(-\dfrac{27}{17}\right)+\dfrac{3}{4}\right\}^2 =$

$\small = \left\{\left[\dfrac{1}{81}-\dfrac{16}{\cancel{225}_9}·\dfrac{\cancel{25}^1}{9}+\dfrac{1}{2}\right]·\left(-\dfrac{27}{17}\right)+\dfrac{3}{4}\right\}^2 =$

$\small = \left\{\left[\dfrac{1}{81}-\dfrac{16}{9}·\dfrac{1}{9}+\dfrac{1}{2}\right]·\left(-\dfrac{27}{17}\right)+\dfrac{3}{4}\right\}^2 =$

$\small = \left\{\left[\dfrac{1}{81}-\dfrac{16}{81}+\dfrac{1}{2}\right]·\left(-\dfrac{27}{17}\right)+\dfrac{3}{4}\right\}^2 =$

$\small = \left\{\left[\dfrac{2-32+81}{162}\right]·\left(-\dfrac{27}{17}\right)+\dfrac{3}{4}\right\}^2 =$

$\small = \left\{\dfrac{\cancel{51}^3}{\cancel{162}_6}·\left(-\dfrac{\cancel{27}^1}{\cancel{17}_1}\right)+\dfrac{3}{4}\right\}^2 =$

$\small = \left\{\dfrac{\cancel3^1}{\cancel6_2}·\left(-\dfrac{1}{1}\right)+\dfrac{3}{4}\right\}^2 =$

$\small = \left\{\dfrac{1}{2}·\left(-1\right)+\dfrac{3}{4}\right\}^2 =$

$\small = \left\{-\dfrac{1}{2}+\dfrac{3}{4}\right\}^2 =$

$\small = \left\{\dfrac{-2+3}{4}\right\}^2 =$

$\small = \dfrac{1}{16} $