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\(A=\frac{4^5.13+32^2.65}{4^4.104}=\frac{2^{10}.13+2^{10}.13.5}{2^8.2^3.13}=\frac{2^{10}.13.\left(1+5\right)}{2^{11}.13}=\frac{6}{2}=3\)
Vậy A = 3
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Ư(a)={-13;-5;-1;1;5;13}
Ư(b)={-81;-27;-9;-3;-1;1;3;9;27;81}
Ư(b)={-24;-12;-8;-6;-4;-3;-2;-1;1;2;3;4;6;8;12;24}
a = 5 x 13
a = 65
Ư ( 65) = { 1; 5; 1; 65 }
b = 34
b = 81
Ư ( 81 ) = { 1; 3; 9; 27 ; 81 }
c = 23 x 3
c = 24
Ư ( 24 ) = { 1; 2; 3; 4; 6; 8; 12; 24 }
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\(S=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+...+\frac{1}{2^n}=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^n}\)
=>\(\frac{S}{2}=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{n+1}}\)
=> \(\frac{S}{2}-S=\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+..+\frac{1}{2^{n+1}}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+..+\frac{1}{2^n}\right)\)
=> \(-\frac{S}{2}=\frac{1}{2^{n+1}}-1\)
=> S= \(2-\frac{1}{2^n}\)
\(A=\dfrac{2^{10}\cdot13+2^{10}\cdot65}{2^8\cdot2^3\cdot13}=\dfrac{2^{11}\cdot3\cdot13}{2^{11}\cdot13}=3\)