Es. N 653

Questo è il procedimento migliore de seguire. Il risultato è 0

653)

$\big[\frac{1}{2}+\big(\frac{2}{3}\big)^2·\big(\frac{3}{4}\big)^2\big]·\big[\big(\frac{8}{15}·\frac{3}{4}+\frac{8}{3}\big) : \frac{23}{5}+\big(1-\frac{5}{3}\big)^2\big]+\frac{1}{6}-1$ =

= $\big[\frac{1}{2}+\frac{4}{9}·\frac{9}{16}\big]·\big[\big(\frac{2}{5}·\frac{1}{1}+\frac{8}{3}\big) · \frac{5}{23}+\big(\frac{3-5}{3}\big)^2\big]+\frac{1}{6}-1$ =

= $\big[\frac{1}{2}+\frac{1}{1}·\frac{1}{4}\big]·\big[\big(\frac{2}{5}+\frac{8}{3}\big) · \frac{5}{23}+\big(-\frac{2}{3}\big)^2\big]+\frac{1}{6}-1$ =

= $\big[\frac{1}{2}+\frac{1}{4}\big]·\big[\big(\frac{6+40}{15}\big) · \frac{5}{23}+\frac{4}{9}\big]+\frac{1}{6}-1$ =

= $\big[\frac{2+1}{4}\big]·\big[\big(\frac{46}{15}\big) · \frac{5}{23}+\frac{4}{9}\big]+\frac{1}{6}-1$ =

= $\frac{3}{4}·\big[\frac{2}{3}·\frac{1}{1}+\frac{4}{9}\big]+\frac{1}{6}-1$ =

= $\frac{3}{4}·\big[\frac{2}{3}+\frac{4}{9}\big]+\frac{1}{6}-1$ =

= $\frac{3}{4}·\big[\frac{6+4}{9}\big]+\frac{1}{6}-1$ =

= $\frac{3}{4}·\frac{10}{9}+\frac{1}{6}-1$ =

= $\frac{1}{2}·\frac{5}{3}+\frac{1}{6}-1$ =

= $\frac{5}{6}+\frac{1}{6}-1$ =

= $\frac{5+6-6}{6} = \frac{0}{6}= 0$.