\(A=2^{100}-\left(2^{99}+2^{98}+...+2+1\right)\)

Đặt \(B=2^{99}+2^{98}+...+2+1\)

\(\Rightarrow2B=2^{100}+2^{99}+...+2^2+2\)

\(\Rightarrow2B-B=2^{100}-1\Leftrightarrow B=2^{100}-1\)

\(\Rightarrow A=2^{100}-\left(2^{100}-1\right)=1\)

\(A=2^{100}-2^{99}+2^{98}-2^{97}+....-2^3+2^2-2+1\\ A=\left(2^{100}+2^{98}+...+2\right)-\left(2^{99}+2^{97}+...+1\right)\)

Gọi \(\left(2^{100}+2^{98}+...+2\right)\)là B

\(B=\left(2^{100}+2^{98}+...+2\right)\\ 2B=2^{102}+2^{100}+.....+2^2\\ 2B-B=\left(2^{102}+2^{100}+.....+2^2\right)-\left(2^{100}+2^{98}+...+2\right)\\ B=2^{102}-2\)

Gọi \(\left(2^{99}+2^{97}+...+1\right)\) là C

\(C=\left(2^{99}+2^{97}+...+1\right)\\ 2C=2^{101}+2^{99}+....+2\\ 2C-C=\left(2^{101}+2^{99}+9^{97}+...+2\right)-\left(2^{99}+9^{97}+...+1\right)\\ C=2^{101}-1\)

\(A=B+C\\ =>A=2^{102}-2+2^{101}-1\\ A=2^{101}\left(2+1\right)-3\\ A=2^{101}\cdot3-3\\ A=3\cdot\left(2^{101}-1\right)\)

\(\dfrac{1}{2}A=2^{99}-2^{98}+...-1+\dfrac{1}{2}\\ \Rightarrow A-\dfrac{1}{2}A=2^{100}-\dfrac{1}{2}\\ \Rightarrow A=2^{101}-1\)

\(2^{100}-2^{99}+2^{98}-2^{97}+2^{96}-2^{95}+...+2^4-2^3+2^2\)

\(=\left(2^{100}-2^{99}+2^{98}\right)-\left(2^{97}-2^{96}+2^{95}\right)+...+\left(2^4-2^3+2^2\right)\)

\(=2^{96}\left(2^4-2^3+2^2\right)-2^{93}\left(2^4-2^3+2^2\right)+...+\left(2^4-2^3+2^2\right)\)

\(=12\left(2^{96}-2^{93}+...+1\right)⋮12\)

a) \(A=1+2+2^2+...+2^{50}\)

\(\Rightarrow2A=2+2^2+...+2^{51}\)

\(\Rightarrow A=2A-A=2+2^2+...+2^{51}-1-2-2^2-...-2^{50}=2^{51}-1\)

b) \(B=1+3+3^2+...+3^{100}\)

\(\Rightarrow3B=3+3^2+...+3^{101}\)

\(\Rightarrow2B=3B-B=3+3^2+...+3^{101}-1-3-3^2-...-3^{100}=3^{101}-1\)

\(\Rightarrow B=\dfrac{3^{101}-1}{2}\)

c) \(C=5+5^2+...+5^{30}\)

\(\Rightarrow5C=5^2+5^3+...+5^{31}\)

\(\Rightarrow4C=5C-C=5^2+5^3+...+5^{31}-5-5^2-...-5^{30}=5^{31}-5\)

\(\Rightarrow C=\dfrac{5^{31}-5}{4}\)

d) \(D=2^{100}-2^{99}+2^{98}-...+2^2-2\)

\(\Rightarrow2D=2^{101}-2^{100}+2^{99}-...+2^3-2^2\)

\(\Rightarrow3D=2D+D=2^{101}-2^{100}+2^{99}-...+2^3-2^2+2^{100}-2^{99}+...+2^2-2=2^{101}-2\)

\(\Rightarrow D=\dfrac{2^{101}-2}{3}\)

tham khảo 
https://olm.vn/hoi-dap/tim-kiem?q=A=2100-299-298-297-.........-22-2-1+.+t%C3%ADnh+A&id=52301

\(A=2^{100}-2^{99}-2^{98}-...-2\)

\(\Rightarrow-2A=-2^{101}+2^{100}+2^{99}+...+2^2\)

\(\Rightarrow A-2A=2^{100}-2^{99}-...-2-2^{101}+2^{100}+...2^2\)

\(\Rightarrow-A=2^{100}+2^{100}-2^{101}-2\)

\(\Rightarrow-A=-2\Rightarrow A=2\)

giúp minh

Ta có: \(A=2^{100}-2^{99}-2^{98}-...-2^2-2-1\)

\(\Leftrightarrow2A=2^{101}-2^{100}-2^{99}-...-2^3-2^2-2\)

\(\Leftrightarrow2A-A=2^{101}-2^{100}-2^{99}-...-2^3-2^2-2-2^{100}+2^{99}+2^{98}+...+2^2+2+1\)

\(\Leftrightarrow A=2^{101}-2\cdot2^{100}+1\)

\(\Leftrightarrow A=1\)

Thank you

a: \(A=2^{100}-2^{99}+2^{98}-2^{97}+...+2^2-2\)

=>\(2A=2^{101}-2^{100}+2^{99}-2^{98}+...+2^3-2^2\)

=>\(2A+A=2^{101}-2^{100}+2^{99}-2^{98}+...+2^3-2^2+2^{100}-2^{99}+...+2^2-2\)

=>\(3A=2^{101}-2\)

=>\(A=\dfrac{2^{101}-2}{3}\)

b: Sửa đề: \(A=\dfrac{2\cdot8^4\cdot27^2+4\cdot6^9}{2^7\cdot6^7+2^7\cdot40\cdot9^4}\)

\(A=\dfrac{2\cdot2^{12}\cdot3^6+2^2\cdot2^9\cdot3^9}{2^7\cdot2^7\cdot3^7+2^7\cdot2^3\cdot5\cdot3^8}\)

\(=\dfrac{2^{11}\cdot3^6\left(2^3+3^3\right)}{2^{10}\cdot3^7\left(2^4+5\cdot3\right)}\)

\(=\dfrac{2}{3}\cdot\dfrac{4+27}{16+15}=\dfrac{2}{3}\)

c: \(B=\dfrac{4^5\cdot9^4-2\cdot6^4}{2^{10}\cdot3^8+6^8\cdot20}\)

\(=\dfrac{2^{10}\cdot3^8-2\cdot2^4\cdot3^4}{2^{10}\cdot3^8+2^8\cdot2^2\cdot5\cdot3^8}\)

\(=\dfrac{2^5\cdot3^4\left(2^5\cdot3^4-1\right)}{2^{10}\cdot3^8\left(1+5\right)}=\dfrac{1}{2^5\cdot3^4}\cdot\dfrac{32\cdot81-1}{6}\)

\(=\dfrac{2591}{2^6\cdot3^5}\)

help