64÷4^2-(3×4-3^2)+(8-2^4+1)×(3^2-3-5)=
64÷4^2-(3×4-3^2)+(8-2^4+1)×(3^2-3-5)=
64÷4^2-(3×4-3^2)+(8-2^4+1)×(3^2-3-5)=
$64÷16-(12-9)+(8-16+1)×(9-3-5)=$
$4-3+(-7)×1=$
$4-3-7=-6$
64/4^2 - (3·4 - 3^2) + (8 - 2^4 + 1)·(3^2 - 3 - 5)=
=64/16 - (12 - 9) + (8 - 16 + 1)·(9 - 3 - 5)=
=4 - 3 + (-7)·1=4 - 3 - 7= -6
@lucianop io ti ringrazio, sei gentilissimo ma mi servono tutti i passaggi
64÷4^2-(3×4-3^2)+(8-2^4+1)×(3^2-3-5) =
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$64~\colon4^2-(3·4-3^2)+(8-2^4+1)·(3^2-3-5)$ =
= $64~\colon16-(12-9)+(8-16+1)·(9-3-5)$ =
= $4-3+(-7)·1$ =
= $1-7·1$ =
= $1-7$ =
= $-6$