Задание ЕГЭ
Сократите дробь \frac{48^{n}}{4^{2n–1}\cdot 3^{n–3}}.Решение
Решение:
\frac{48^{n}}{4^{2n–1}\cdot 3^{n–3}}=\frac{(4\cdot 4\cdot 3)^{n}}{4^{2n–1}\cdot 3^{n–3}}=\frac{4^{n}\cdot 4^{n}\cdot 3^{n}}{4^{2n–1}\cdot 3^{n–3}}=\frac{4^{n+n}\cdot 3^{n}}{4^{2n–1}\cdot 3^{n–3}}=\frac{4^{2n}\cdot 3^{n}}{4^{2n–1}\cdot 3^{n–3}}=\frac{4^{2n}}{4^{2n–1}}\cdot \frac{3^{n}}{3^{n–3}}=4^{2n–(2n–1)}\cdot 3^{n–(n–3)}=4^{2n–2n+1}\cdot 3^{n–n+3}=4^{1}\cdot 3^{3}=4\cdot 27=108
Ответ: 108.