A=\(\dfrac{2^{10}\left(13+65\right)}{2^8.104}\)

=\(\dfrac{2.78}{104}\)=\(\dfrac{78}{52}\)=\(\dfrac{39}{26}\)

Sửa đề: \(C=1+3^1+3^2+...+3^{100}\)

b) Ta có: \(C=1+3^1+3^2+...+3^{100}\)

\(\Leftrightarrow3\cdot C=3+3^2+...+3^{101}\)

\(\Leftrightarrow C-3\cdot C=1+3+3^2+...+3^{100}-3-3^2-...-3^{100}-3^{101}\)

\(\Leftrightarrow-2\cdot C=1-3^{101}\)

hay \(C=\dfrac{3^{101}-1}{2}\)

b) Ta có: C=1+31+32+...+3100C=1+31+32+...+3100

⇔3⋅C=3+32+...+3101⇔3⋅C=3+32+...+3101

⇔C−3⋅C=1+3+32+...+3100−3−32−...−3100−3101⇔C−3⋅C=1+3+32+...+3100−3−32−...−3100−3101

⇔−2⋅C=1−3101

\(\frac{2^{10}.13+2^{10}.65}{2^8.104}\)

Xét : 210 . 13 + 210. 65 = 210 . ( 13 + 65 ) = 210 . 78 

         2⁸  . 104 = 2⁸ . ( 2² . 26 ) = 2⁸. 2². 26 = 2¹⁰.26

có : \(\frac{2^{10}.13+2^{10}.65}{2^8.104}\)= \(\frac{2^{10}.78}{2^{10}.26}\)= \(\frac{78}{26}\)= 3

\(\frac{2^{10}.13+2^{10}.65}{2^8.104}=\frac{2^{10}.\left(13+65\right)}{2^8.104}=\frac{2^{10}.78}{2^8.104}\)

\(=\frac{2^2.39}{52}=\frac{2^2.39}{2^2.13}\)

\(=\frac{39}{13}\)

\(=3\)

\(\frac{2^{10}.13+2^{10}.65}{2^8.104}\)

\(=\frac{2^{10}.\left(13+65\right)}{2^8.104}\)

\(=\frac{2^{10}.78}{2^8.104}\)

\(=\frac{2^8.2^2.78}{2^8.104}\)

\(=\frac{2^8.4.78}{2^8.104}\)

\(=\frac{2^8.312}{2^8.104}\)

\(=\frac{2^8.3.104}{2^8.104}=3\)

\(\frac{2^{10}.13+2^{10}.65}{2^8.104}\)

\(=\frac{2^{10}.\left(13+65\right)}{2^8.104}\)

\(=\frac{2^{10}.78}{2^8.104}\)

\(=\frac{2^{10}.2.39}{2^8.2^3.13}\)

\(=\frac{2^{11}.39}{2^{11}.13}=3\)

\(\frac{2^{10}\cdot13+2^{10}\cdot65}{2^8\cdot104}=\frac{2^{10}\cdot\left(13+65\right)}{2^8\cdot104}=\frac{2^{10}\cdot78}{2^8\cdot104}=\frac{2^{10}\cdot2\cdot3\cdot13}{2^8\cdot2^2\cdot13}=\frac{2^{11}\cdot3\cdot13}{2^{10}\cdot13}=2\cdot3=6\)

2¹⁰.13+2¹⁰.65 = 2¹⁰(13 + 65) = 2¹⁰ . 78 = 1024 . 78 = 79872

2^8.104 = 2⁸³²

2¹⁰.13+2¹⁰.65= (13 + 65 ) . 2¹⁰

                    =  78 . 2¹⁰

                           = 78 . 1024

                     =  79872

 2^8.104= 2¹¹²