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Sa se efectueze:[(3^2+3×4) (2^3-2^0)+4×3+2]×2-2^5. Repede va rog.Dau coroniță

Răspuns :

[(3²+3*4)(2³-2⁰)+4*3+2]*2-2⁵=
=[(9+12)(8-1)+12+2]*2-32
=[21*7+12+2]*2-32
=[147+12+2]*2-32
=161*2-32
=322-32
=290