Bài giải

\(N=\frac{9\cdot5^{20}\cdot27^9-3\cdot9^{15}\cdot25^9}{7\cdot3^{29}\cdot125^6-3\cdot3^9\cdot15^{19}}=\frac{3^2\cdot5^{20}\cdot\left(3^3\right)^9-3\cdot\left(3^2\right)^{15}\cdot\left(5^2\right)^9}{7\cdot3^{29}\cdot\left(5^3\right)^6-3\cdot3^9\cdot\left(3\cdot5\right)^{19}}=\frac{3^2\cdot5^{20}\cdot3^{27}-3\cdot3^{30}\cdot5^{18}}{7\cdot3^{29}\cdot5^{18}-3^{10}\cdot3^{19}\cdot5^{19}}\)

\(=\frac{3^{29}\cdot5^{20}-3^{31}\cdot5^{18}}{7\cdot3^{29}\cdot5^{18}-3^{29}\cdot5^{19}}=\frac{3^{29}\cdot5^{18}\left(5^2-3^2\right)}{3^{29}\cdot5^{18}\left(7-5\right)}=\frac{25-9}{7-5}=\frac{16}{2}=8\)

3².5²⁰.(3³)⁹-3.(3²)¹⁵.(5²)⁹

7.3²⁹.(5³)

\(=\dfrac{3^2\cdot5^{20}\cdot3^{27}-3\cdot3^{30}\cdot5^{18}}{7\cdot3^{29}\cdot5^{13}-3^{29}\cdot5^{19}}=\dfrac{3^{29}\cdot5^{20}-5^{18}\cdot3^{31}}{3^{29}\cdot5^{13}\cdot7-3^{29}\cdot5^{19}}\)

\(=\dfrac{3^{29}\cdot5^{18}\left(5^2-3^2\right)}{3^{29}\cdot5^{13}\left(7-5^6\right)}=5^5\cdot\dfrac{4^2}{7-5^6}\)

1. \(\dfrac{9.5^{20}.27^9-3.9^{15}:25^9}{7.3^{29}.125^6-3.3^9.15^{19}}\)

\(=\dfrac{3^2.5^{20}.3^{27}-3.3^{30}.5^{18}}{7.3^{29}.5^{18}-3^{10}.3^{19}.5^{19}}\)

\(=\dfrac{3^{29}.5^{20}-3^{31}.5^{18}}{7.3^{29}.5^{18}-3^{29}.5^{19}}\)

\(=\dfrac{3^{28}.5^{18}.\left(5^2-3^2\right)}{3^{29}.5^{18}.\left(7-5\right)}\)

\(=\dfrac{5^2-3^2}{7-5}=\dfrac{16}{2}=8\)

2.\(\left(\dfrac{1}{2}\right)^{15}.\left(\dfrac{1}{-4}\right)^{20}=\left(\dfrac{1}{2}\right)^{15}.\left(\dfrac{1}{4}\right)^{20}\)

\(=\left(\dfrac{1}{2}\right)^{15}.\left(\dfrac{1}{2}^2\right)^{20}=\left(\dfrac{1}{2}\right)^{15}.\left(\dfrac{1}{2}\right)^{40}\)

\(=\left(\dfrac{1}{2}\right)^{55}\)

3.\(\dfrac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}=\dfrac{2^{10}.3^8-2.2^9.3^9}{2^{10}.3^8+2^8.3^8.2^2.5}\)

\(=\dfrac{2^{10}.3^8-2^{10}.3^9}{2^{10}.3^8+2^{10}.3^8.5}=\dfrac{2^{10}.3^8\left(1-3\right)}{2^{10}.3^8\left(1+5\right)}\)

\(=\dfrac{1-3}{1+5}=\dfrac{-2}{6}=\dfrac{-1}{3}\)

Chúc học tốt!!