a)Cho A=4+\(^{2^2+2^3+2^4}\)+...+\(2^{20}\).Hỏi A có chia hết cho 128 không?
b)Tính giá trị biểu thức:\(\frac{2^{12}x13+2^{12}x65}{2^{10}104}\)+\(\frac{3^{10}x11+3^{10}x5}{3^9x2^4}\)
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b)
\(\frac{2^{12}.13+2^{12}.65}{2^{10}.104}+\frac{3^{10}.11+3^{10}.5}{3^9.2^4}\)
\(=\frac{2^{10}\left(4.13+4.65\right)}{2^{10}.104}+\frac{3^9\left(11.3+5.3\right)}{3^9.16}\)
\(=\frac{312}{104}+\frac{48}{16}=3+3=6\)
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) \(\frac{2^{12}.13+2^{12}.65}{2^{10}.104}+\frac{3^{10}.11+3^{10}.5}{3^9.2^4}\)
\(=\frac{2^{12}\left(13+65\right)}{2^{10}.2^3.13}+\frac{3^{10}\left(11+5\right)}{3^9.2^4}\)
\(=\frac{2^{12}.78}{2^{13}.13}+\frac{3^{10}.16}{3^9.16}=\frac{6}{2}+\frac{3^{10}}{3^9}\)
\(=3+3=6\)
a)\(A=\frac{5.2^{13}.2^{22}-2^{36}}{\left(3.2^{17}\right)^2}\)
\(A=\frac{5.2^{35}-2^{36}}{3^2.2^{34}}\)
\(A=\frac{2^{35}\left(5-2\right)}{3^2.2^{34}}\)
\(A=\frac{2.3}{3^2}=\frac{2}{3}\)
b) \(B=\frac{2^{19}.27^3+15.4^9.9^4}{6^9.2^{10}+12^{10}}\)
\(B=\frac{2^{19}.3^9+3.5.2^{18}.3^8}{2^9.3^9.2^{10}+2^{20}.3^{10}}\)
\(B=\frac{2^{18}.3^9\left(2+5\right)}{2^{19}.3^9\left(2+3\right)}\)
\(B=\frac{7}{2.5}=\frac{7}{10}\)