Question

(2²+2¹+2²+2³)2⁰.2¹.2².2³

Answer 1

Ta có: \(\left(2^2+2^1+2^2+2^3\right)\cdot2^0\cdot2^1\cdot2^2\cdot2^3\)

\(=\left(4+2+4+8\right)\cdot2\cdot4\cdot8\)

\(=64\cdot18=1152\)