Question

[Fudo]

Bài 22 : a, Cho A = 4 + 2² + 2³ + 2⁴ + ... + 2²⁰

              Hỏi A có chia hết cho 128 không ?

             b, Tính giá trị biểu thức

       \(\frac{2^{12}\cdot13+2^{12}\cdot65}{2^{10}\cdot104}+\frac{3^{10}\cdot11+3^{10}\cdot5}{3^9\cdot2^4}\)

Answer 1

[TuiTenQuynh]

b)

\(\frac{2^{12}.13+2^{12}.65}{2^{10}.104}+\frac{3^{10}.11+3^{10}.5}{3^9.2^4}\)

\(=\frac{2^{10}\left(4.13+4.65\right)}{2^{10}.104}+\frac{3^9\left(11.3+5.3\right)}{3^9.16}\)

\(=\frac{312}{104}+\frac{48}{16}=3+3=6\)

Answer 2

[Trần Tiến Pro ✓]

a) \(A=4+2^2+2^3+2^4+....+2^{20}\)

\(\Rightarrow2A=2^3+2^3+2^4+.....+2^{21}\)

\(\Rightarrow2A-A=\left(2^3+2^3+2^4+....+2^{21}\right)-\left(2^2+2^3+2^4+...+2^{20}\right)\)

\(\Rightarrow A=2^3+2^{21}-\left(2^2+2^2\right)\)

\(\Rightarrow A=2^{21}\)

\(\text{Vì }2^{21}⋮2^7\Rightarrow A⋮128\)

b) \(\frac{2^{12}.13+2^{12}.65}{2^{10}.104}+\frac{3^{10}.11+3^{10}.5}{3^9.2^4}\)

\(=\frac{2^{12}\left(13+65\right)}{2^{10}.2^3.13}+\frac{3^{10}\left(11+5\right)}{3^9.2^4}\)

\(=\frac{2^{12}.78}{2^{13}.13}+\frac{3^{10}.16}{3^9.16}=\frac{6}{2}+\frac{3^{10}}{3^9}\)

\(=3+3=6\)