Question

Giup minh với:

Tính : \(\frac{4.^72^8}{3.2^{15}-16^2-5.2^2.\left(2^{10}\right)^2}\)

^Thanks

Answer 1

\(=\frac{2^{14}.2^8}{3.2^{15}-2^8-5.2^2.2^{20}}=\frac{2^{22}}{3.2^{15}-2^8-5.2^{22}}=\frac{2^{22}}{2^{22}.\left(3.2^{-7}-2^{-14}-5\right)}=\frac{1}{3.2^{-7}-2^{-14}-5}\)

Answer 2

\(\frac{4^7\times2^8}{3\times2^{15}-16^2-5\times2^2\times\left(2^{10}\right)^2}=\frac{\left(2^2\right)^7\times2^8}{3\times2^{15}-\left(2^4\right)^2-5\times2^2\times2^{20}}=\frac{2^{14}\times2^8}{3\times2^{15}-2^8-5\times2^{22}}=\frac{2^{14}\times2^8}{2^8\times\left(3\times2^7-1-2^{14}\times5\right)}\)