Question

[Nguyễn Vũ Hoàng]

Cho : \(\text{A}=\dfrac{\left(-2\right)^0+1^{2017}+\left(\dfrac{-1}{3}\right)^8\cdot3^8}{2^{15}}\)

\(\text{B}=\dfrac{6^2}{2^{16}}\)

Tính \(\dfrac{\text{A}}{\text{B}}\).

Answer 1

[Kiêm Hùng]

\(A=\dfrac{\left(-2\right)^0+1^{2017}+\left(\dfrac{-1}{3}\right)^8.3^8}{2^{15}}=\dfrac{3}{2^{15}}\left(1\right)\)

\(B=\dfrac{6^2}{2^{16}}\left(2\right)\)

\(\left(1\right);\left(2\right)\Rightarrow\dfrac{A}{B}=\dfrac{\dfrac{3}{2^{15}}}{\dfrac{6^2}{2^{16}}}=\dfrac{1}{6}\)