Soluzione del esercizio 133 di Matematica.blu: Algebra, Geometria, Statistica, Vol. 1

$\{[(2809-2025):7+3\cdot16]:[(441-56):7-51]^2+2\cdot25\}^2:(4+8+16+32)+10\cdot4$ $=$ $\{[(784):7+48]:[(385):7-51]^2+50\}^2:(60)+40$ $=$ $\{[112+48]:[55-51]^2+50\}^2:60+40$ $=$ $\{[160]:[4]^2+50\}^2:60+40$ $=$ $\{160:16+50\}^2:60+40$ $=$ $\{10+50\}^2:60+40$ $=$ $\{60\}^2:60+40$ $=$ $3600:60+40$ $=$ $60+40$ $=$ $100$