A expressao 2^3+x - 2^×-3 ÷ 2^× + 2^x-3 =

\[{a^x} \cdot {a^y} = {a^{x + y}}\]

\[{{{a^x}} \over {{a^y}}} = {a^{x - y}}\]

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Resolvendo a equação temos:

\[\eqalign{ & {{{2^{3 + x}} - {2^{x - 3}}} \over {{2^x} + {2^{x - 3}}}} = \cr & {{{2^3} \cdot {2^x} - {{{2^x}} \over {{2^3}}}} \over {{2^x} + {{{2^x}} \over {{2^3}}}}} = \cr & {{8 \cdot {2^x} - {{{2^x}} \over 8}} \over {{2^x} + {{{2^x}} \over 8}}} = \cr & {{{{64 \cdot {2^x} - {2^x}} \over 8}} \over {{{8 \cdot {2^x} + {2^x}} \over 8}}} = \cr & {{64 \cdot {2^x} - {2^x}} \over {8 \cdot {2^x} + {2^x}}} = \cr & {{63 \cdot {2^x}} \over {9 \cdot {2^x}}} = \cr & {{63} \over 9} = \cr & 7 }\]

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Concluímos então que \(\boxed{{{{2^{3 + x}} - {2^{x - 3}}} \over {{2^x} + {2^{x - 3}}}} = 7}\)