Question

[Mk tên là Chi]

Tính A, biết:

\(A=\frac{4^6\cdot9^5+6^9\cdot120}{8^4\cdot3^{12}-6^{11}}\)

Answer 1

[Hoàng Minh Chi]

\(A=\frac{4^6\cdot9^5+6^9\cdot120}{8^4\cdot3^{12}-6^{11}}=\frac{\left(2^2\right)^6\cdot\left(3^2\right)^5+\left(2\cdot3\right)^9\cdot2^3\cdot3\cdot5}{\left(2^3\right)^4\cdot3^{12}-\left(2\cdot3\right)^{11}}=\frac{2^{12}\cdot3^{10}+2^9\cdot3^9\cdot2^3\cdot3\cdot5}{2^{12}\cdot3^{12}-2^{11}\cdot3^{11}}=\frac{2^{12}\cdot3^{10}+2^{12}\cdot3^{10}\cdot5}{2^{12}\cdot3^{12}-2^{11}\cdot3^{11}}=\frac{2^{12}\cdot3^{10}\left(1+5\right)}{2^{11}\cdot3^{11}\left(2\cdot3-1\right)}=\frac{2\left(1+5\right)}{3\left(6-1\right)}=\frac{2\cdot6}{3\cdot5}=\frac{2\cdot2}{5}=\frac{4}{5}\)

                                                             \(\approx GOOD\)\(LUCK\approx\)