[Risolto] Algebra: espressione

Data l’espressione:

$$\left(2-\frac{3}{5}\right)^{4} \cdot\left(\frac{5}{21}\right)^{4}:\left[\left(1-\frac{1}{3}\right)^{2} \cdot\left(-1+\frac{5}{3}\right)-\frac{5}{27}\right]^{2}$$

si ha il seguente svolgimento, applicando le proprietà delle potenze:

$\left(2-\frac{3}{5}\right)^{4} \times\left(\frac{5}{21}\right)^{4}:\left[\left(1-\frac{1}{3}\right)^{2} \times\left(-1+\frac{5}{3}\right)-\frac{5}{27}\right]^{2}=$
$=\left(\frac{10-3}{5}\right)^{4} \times\left(\frac{5}{21}\right)^{4}:\left(\left(\frac{3-1}{3}\right)^{2} \times\left(\frac{-3+5}{3}\right)-\frac{5}{27}\right]^{2}=$
$=\left(\frac{7}{5}\right)^{4} \times\left(\frac{5}{21}\right)^{4}:\left[\left(\frac{2}{3}\right)^{2} \times\left(\frac{2}{3}\right)-\frac{5}{27}\right]^{2}=$
$\left.=\left(\frac{2401}{625}\right) \times\left(\frac{5}{21}\right)^{4}:\left(\frac{4}{9}\right) \times\left(\frac{2}{3}\right)-\frac{5}{27}\right]^{2}=$
$=\left(\frac{2401}{625}\right) \times\left(\frac{5}{21}\right)^{4}:\left(\left(\frac{8}{27}\right)-\frac{5}{27}\right]^{2}=$
$=\left(\frac{2401}{625}\right) \times\left(\frac{5}{21}\right)^{4}:\left[\frac{8}{27}-\frac{5}{27}\right]^{2}=$
$=\left(\frac{2401}{625}\right) \times\left(\frac{5}{21}\right)^{4}:\left[\frac{8-5}{27}\right]^{2}=$
$=\left(\frac{2401}{625}\right) \times\left(\frac{5}{21}\right)^{4}:\left[\frac{3}{27}\right]^{2}=$
$=\left(\frac{2401}{625}\right) \times\left(\frac{5}{21}\right)^{4}:\left[\frac{\not^{1}}{27^{9}}\right]^{2}=$
$\left.=\left(\frac{2401}{625}\right) \times\left(\frac{5}{21}\right)^{4}: \frac{1}{9}\right]^{2}=$
$=\left(\frac{2401}{625}\right) \times\left(\frac{625}{194481}\right):\left[\frac{1}{9}\right]^{2}=$
$\left.=\left(\frac{2401}{625}\right) \times\left(\frac{625}{194481}\right): \frac{1}{81}\right]=$
$\left.=\left(\frac{1}{81}\right): \frac{1}{81}\right]=$
$=\left(\frac{1}{81}\right) \times\left[\frac{81}{1}\right]=$
$=(1)=$
$=1$