Question

\(\frac{2^{12}.3^5-4^6.81}{\left(2^2.6\right)^6+8^4.3^5}\)

Answer 1

                                                                 Bài giải

\(\frac{2^{12}\cdot3^5-4^6\cdot81}{\left(2^2\cdot6\right)^6+8^4\cdot3^5}=\frac{2^{12}\cdot3^5-\left(2^2\right)^6\cdot3^4}{2^{12}\cdot6^6+\left(2^3\right)^4\cdot3^5}=\frac{2^{12}\cdot3^5-2^{12}\cdot3^4}{2^{12}\cdot2^6\cdot3^6+2^{12}\cdot3^5}=\frac{2^{12}\cdot3^4\left(3-1\right)}{2^{12}\cdot3^4\left(2^6\cdot3^2+3\right)}\)

\(=\frac{2}{64\cdot9+3}=\frac{2}{576+3}=\frac{2}{579}\)