4) (3¹⁵+3¹²) 2⁹

4) $$ \frac{(3^{15} + 3^{12}) \cdot 2^9}{(3^{14} + 3^{12}) \cdot 1024} = \frac{(3^{15} + 3^{12}) \cdot 2^9}{(3^{14} + 3^{12}) \cdot 2^{10}} = \frac{3^{12}(3^3+1) \cdot 2^9}{3^{12}(3^2+1) \cdot 2^{10}} = \frac{3^{12}(27+1) \cdot 2^9}{3^{12}(9+1) \cdot 2^{10}} = \frac{28 \cdot 2^9}{10 \cdot 2^{10}} = \frac{28}{10 \cdot 2} = \frac{28}{20} = \frac{7}{5} $$

Ответ: $$\frac{7}{5}$$