Non riesco a risolvere 532

GIA' RISOLTA!

https://www.sosmatematica.it/forum/domande/aiuto-la-532-non-riesco-a-risolverla/#post-282939

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$\small \left[\left(-\dfrac{5}{6}\right)^3·\dfrac{25}{36}÷\left(\dfrac{6}{5}\right)^{-3}\right]^{-6}·\left[\left(-\dfrac{32}{15}\right)^ {-10}·\left(\dfrac{9}{16}\right)^{-10}\right]·\dfrac{(-6)^3·6^{-1}}{15^{-2}·(-36)^2} =$

$\small = \left[\left(-\dfrac{5}{6}\right)^3·\left(\dfrac{5}{6}\right)^2÷\left(\dfrac{5}{6}\right)^{3}\right]^{-6}·\left[\left(-\dfrac{15}{32}\right)^{10}·\left(\dfrac{16}{9}\right)^{10}\right]·\dfrac{-\cancel{216}^{36}·\dfrac{1}{\cancel6_1}}{\left(\dfrac{1}{15}\right)^2·1296} =$

$\small = \left[\left(-\dfrac{5}{6}\right)^{3+2-3}\right]^{-6}·\left[\left(-\dfrac{15·16}{32·9}\right)^{10}\right]·\dfrac{-36·1}{\dfrac{1}{\cancel{225}_{25}}·\cancel{1296}^{144}} =$

$\small = \left[\left(-\dfrac{5}{6}\right)^2\right]^{-6}·\left[\left(-\dfrac{\cancel{15}^5·\cancel{16}^1}{\cancel{32}_2·\cancel9_3}\right)^{10}\right]·\dfrac{-\cancel{36}^1}{\dfrac{1}{25}·\cancel{144}_4} =$

$\small = \left(-\dfrac{5}{6}\right)^{2·(-6)}·\left[\left(-\dfrac{5·1}{2·3}\right)^{10}\right]·\dfrac{-1}{\dfrac{1}{25}·4} =$

$\small = \left(-\dfrac{5}{6}\right)^{-12}·\left(-\dfrac{5}{6}\right)^{10}·(-1)·\dfrac{25}{4} =$

$\small = \left(-\dfrac{5}{6}\right)^{-12+10}·-\left(\dfrac{25}{4}\right) =$

$\small = \left(-\dfrac{5}{6}\right)^{-2}·-\left(\dfrac{5}{2}\right)^2 =$

$\small = \left(-\dfrac{\cancel6^3}{\cancel5_1}\right)^2·-\left(\dfrac{\cancel5^1}{\cancel2_1}\right)^2 =$

$\small = (-3)^2·-\left(1\right)^2 =$

$\small = 9·-\left(1\right) =$

$\small = -9$