[Risolto] Algebra

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$\left\{-\left({11\over5}+{3\over4}-{7\over2}\right)÷\left[\left(-{3\over2}\right)^2·\left(-{3\over2}\right)+{11\over8}+{3\over10}+{1\over60}+{7\over12}\right]\right\}^4=$

$=\left\{-\left({44+15-70\over20}\right)÷\left[{9\over4}·\left(-{3\over2}\right)+{11\over8}+{3\over10}+{1\over60}+{7\over12}\right]\right\}^4=$

$=\left\{-\left(-{11\over20}\right)÷\left[-{27\over8}+{11\over8}+{3\over10}+{1\over60}+{7\over12}\right]\right\}^4=$

$=\left\{{11\over20}÷\left[-{\cancel{16}^2\over\cancel8_1}+{3\over10}+{1\over60}+{7\over12}\right]\right\}^4=$

$=\left\{{11\over20}÷\left[-{2\over1}+{3\over10}+{1\over60}+{7\over12}\right]\right\}^4=$

$=\left\{{11\over20}÷\left[{-240+36+2+70\over120}\right]\right\}^4=$

$=\left\{{11\over20}÷\left[-{\cancel{132}^{11}\over\cancel{120}_{10}}\right]\right\}^4=$

$=\left\{{11\over20}÷\left[-{11\over10}\right]\right\}^4=$

$=\left\{{\cancel{11}^1\over\cancel{20}_2}·\left[-{\cancel{10}^1\over\cancel{11}_1}\right]\right\}^4=$

$=\left\{{1\over2}·\left[-{1\over1}\right]\right\}^4=$

$=\left\{-{1\over2}\right\}^4=$

$={1\over16}$