aiuto

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Come da regolamento di sos matematica poni una domanda per volta, ti rispondo alla prima, le altre postale di nuovo una alla volta.

434)

$\left[\left(\dfrac{1}{24}+1-\dfrac{37}{48}\right)·\dfrac{8}{13}-\left(1-\dfrac{11}{12}\right)\right]^2 : \left(\dfrac{1}{3}\right)^2 - \left(1+\dfrac{1}{2}\right)·\dfrac{1}{24}=$

$=\left[\left(\dfrac{2+48-37}{48}\right)·\dfrac{8}{13}-\left(\dfrac{12-11}{12}\right)\right]^2 : \dfrac{1}{9} - \left(\dfrac{2+1}{2}\right)·\dfrac{1}{24}=$

$=\left[\dfrac{\cancel{13}^1}{\cancel{48}_6}·\dfrac{\cancel8^1}{\cancel{13}_1}-\dfrac{1}{12}\right]^2 : \dfrac{1}{9} - \dfrac{\cancel3^1}{2}· \dfrac{1}{\cancel{24}_8}=$

$=\left[\dfrac{1}{6}·\dfrac{1}{1}-\dfrac{1}{12}\right]^2 : \dfrac{1}{9} - \dfrac{1}{2}· \dfrac{1}{8}=$

$=\left[\dfrac{1}{6}-\dfrac{1}{12}\right]^2 : \dfrac{1}{9} - \dfrac{1}{16}=$

$=\left[\dfrac{2-1}{12}\right]^2 : \dfrac{1}{9} - \dfrac{1}{16}=$

$=\left[\dfrac{1}{12}\right]^2 : \dfrac{1}{9} - \dfrac{1}{16}=$

$=\dfrac{1}{\cancel{144}_{16}} · \dfrac{\cancel9^1}{1} - \dfrac{1}{16}=$

$=\dfrac{1}{16} · \dfrac{1}{1} - \dfrac{1}{16}=$

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

$= 0$