Bài 1:
a) Ta có: \(\frac{2^8\cdot4\cdot13+2^7\cdot8\cdot65}{2^9\cdot39}\)
\(=\frac{2^8\cdot4\cdot13+2^8\cdot4\cdot13\cdot5}{2^9\cdot39}\)
\(=\frac{2^{10}\cdot13\left(1+5\right)}{2^9\cdot13\cdot3}=\frac{6}{3}=2\)
b) Đặt \(A=4+2^2+2^3+2^4+...+2^{20}\)
Ta có: \(A=4+2^2+2^3+2^4+...+2^{20}\)
\(\Rightarrow2A=2^3+2^3+2^4+...+2^{21}\)
Ta có: \(2A-A=2^3+2^{21}-2^2-2^2=8+2^{21}-8=2^{21}\)
hay \(A=2^{21}\)
Vậy: \(4+2^2+2^3+2^4+...+2^{20}=2^{21}\)