Soluzioni sulle conclusioni sui numeri razionali

 

(1) \cfrac{\left(\cfrac{3}{4}-0,0\overline{3}\right)\cdot\cfrac{3}{43}+\left(2,\overline{4}-1,2\right)\cdot\cfrac{9}{7}-\cfrac{8}{5}}{\left[\cfrac{2,3-2,\overline{15}}{0,2+1,\overline{3}}\cdot\left(5+\cfrac{8}{49}\right)+\cfrac{3}{4}\right]:\left(1+\cfrac{1}{4}\right)}+\cfrac{4}{0,\overline{2}+1,2}

quindi

\cfrac[l]{\left(\cfrac{3}{4}-\cfrac{3}{90}\right)\cdot\cfrac{3}{43}+\left(\cfrac[l]{24-2}{9}-\cfrac[l]{12}{10}\right)\cdot\cfrac{9}{7}-\cfrac[l]{8}{5}}{\left[\cfrac{\cfrac{23}{10}-\cfrac{215-2}{99}}{\cfrac{2}{10}+\cfrac{13-1}{9}}\cdot\left(5+\cfrac{8}{49}\right)+\cfrac{3}{4}\right]:\left(\cfrac{5}{4}\right)}+\cfrac{4}{\cfrac{2}{9}+\cfrac{12}{10}}

quindi

\cfrac[l]{\left(\cfrac{3}{4}-\cfrac{1}{30}\right)\cdot\cfrac{3}{43}+\left(\cfrac[l]{22}{9}-\cfrac[l]{6}{5}\right)\cdot\cfrac{9}{7}-\cfrac[l]{8}{5}}{\left[\cfrac{\cfrac{23}{10}-\cfrac{213}{99}}{\cfrac{1}{5}+\cfrac{12}{9}}\cdot\left(\cfrac{253}{49}\right)+\cfrac{3}{4}\right]\cdot\left(\cfrac{4}{5}\right)}+\cfrac{4}{\cfrac{20+108}{90}}

passaggio successivo

\cfrac[l]{\left(\cfrac{45-2}{60}\right)\cdot\cfrac{3}{43}+\left(\cfrac[l]{110-54}{45}\right)\cdot\cfrac{9}{7}-\cfrac[l]{8}{5}}{\left[\cfrac{\cfrac{23}{10}-\cfrac{71}{33}}{\cfrac{1}{5}+\cfrac{12}{9}}\cdot\left(\cfrac{253}{49}\right)+\cfrac{3}{4}\right]\cdot\left(\cfrac{4}{5}\right)}+\cfrac{4}{\cfrac{128}{90}}

e quindi

\cfrac[l]{\left(\cfrac{43}{60}\right)\cdot\cfrac{3}{43}+\left(\cfrac[l]{56}{45}\right)\cdot\cfrac{9}{7}-\cfrac[l]{8}{5}}{\left[\cfrac{\cfrac{759-710}{330}}{\cfrac{9+60}{45}}\cdot\left(\cfrac{253}{49}\right)+\cfrac{3}{4}\right]\cdot\left(\cfrac{4}{5}\right)}+4\cdot\cfrac{90}{128}

e

\cfrac[l]{\cfrac{1}{20}+\cfrac{8}{5}-\cfrac[l]{8}{5}}{\left[\cfrac{\cfrac{49}{330}}{\cfrac{69}{45}}\cdot\left(\cfrac{253}{49}\right)+\cfrac{3}{4}\right]\cdot\left(\cfrac{4}{5}\right)}+\cfrac{90}{32}

e

\cfrac[l]{\cfrac{1}{20}}{\left[\cfrac{49}{330}\cdot\cfrac{45}{69}\cdot\left(\cfrac{253}{49}\right)+\cfrac{3}{4}\right]\cdot\left(\cfrac{4}{5}\right)}+\cfrac{90}{32}

e

\cfrac[l]{\cfrac{1}{20}}{\left[\cfrac{1}{2}+\cfrac{3}{4}\right]\cdot\left(\cfrac{4}{5}\right)}+\cfrac{90}{32}

e

\cfrac[l]{\cfrac{1}{20}}{\left[\cfrac{5}{4}\right]\cdot\left(\cfrac{4}{5}\right)}+\cfrac{90}{32}

e

\cfrac{1}{20}+\cfrac{90}{32}=\cfrac{32+1800}{640}

ed infine

\cfrac{1832}{640}=\cfrac{229}{80}

(2) \cfrac{\left[\left(2+\cfrac{1}{2}\right)^{2}\cdot\left(3-\cfrac{4}{3}\right)^{2}\cdot\left(-1-\cfrac{1}{5}\right)^{2}\right]^{-1}:\left(-1+\cfrac{4}{5}\right)^{2}}{\cfrac{1}{2}\cdot\left(-\cfrac{1}{3}\right)^{-2}+\left(\cfrac{1}{4}\right)^{2}\cdot\left(1-\cfrac{3}{8}\right)^{-2}:\left(1-\cfrac{3}{5}\right)^{2}-\cfrac{5}{2}}

Risultato \cfrac[l]{1}{3}

\cfrac{\left[\left(\cfrac{5}{2}\right)^{2}\cdot\left(\cfrac{5}{3}\right)^{2}\cdot\left(-\cfrac{6}{5}\right)^{2}\right]^{-1}:\left(-\cfrac{1}{5}\right)^{2}}{\cfrac{1}{2}\cdot9+\cfrac{1}{4^{2}}\cdot\left(\cfrac{5}{8}\right)^{-2}:\left(\cfrac{2}{5}\right)^{2}-\cfrac{5}{2}}

e

\cfrac{\left[\cfrac{5^{2}}{2^{2}}\cdot\cfrac{5^{2}}{3^{2}}\cdot\cfrac{6^{2}}{5^{2}}\right]^{-1}\cdot5^{2}}{\cfrac{9}{2}+\cfrac{1}{4^{2}}\cdot\cfrac{8^{2}}{5^{2}}\cdot\cfrac{5^{2}}{2^{2}}-\cfrac{5}{2}}

ed infine

\cfrac{5^{-2}\cdot5^{2}}{\cfrac{9}{2}+1-\cfrac{5}{2}}=\cfrac{1}{\cfrac{9+2-5}{2}}=\cfrac{1}{\cfrac{12}{6}}=\cfrac{1}{3}

Come conclusione

About Francesco Bragadin

Insegno informatica e telecomunicazioni al liceo scienze applicate ed all'indirizzo informatica e telecomunicazioni. Ho terminato gli studi in ingegneria elettronica e telecomunicazioni lavorando per molti anni come libero professionista nell'ambito della gestione storage e disaster recovery su mainframe.
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