Đặt N = 2²⁰⁰⁹ + 2²⁰⁰⁸ + 2²⁰⁰⁷ +......+ 2¹ + 2⁰

2N = 2²⁰¹⁰ + 2²⁰⁰⁹ + 2²⁰⁰⁸ +.....+ 2² + 2¹

2N - N = 2²⁰¹⁰ - 2⁰

=> N = 2²⁰¹⁰ - 1

=> M = 2²⁰¹⁰ - (2²⁰¹⁰ - 1)

=> M = 2²⁰¹⁰ - 2²⁰¹⁰ + 1

=> M = 1 

Đặt N=2²⁰⁰⁹+2²⁰⁰⁸+...+1

=>2N=2²⁰¹⁰+2²⁰⁰⁹+...+2

=>2N-N=(2²⁰¹⁰+2²⁰⁰⁹+...+2)-(2²⁰⁰⁹+2²⁰⁰⁸+...+1)

=>N=2²⁰¹⁰-1

Mà M=2²⁰¹⁰-N=2²⁰¹⁰-(2²⁰¹⁰-1)=1

Ác Mộng trả lời đúng rùi. **** thui

\(C=\frac{\frac{1}{2008}-\frac{1}{2009}-\frac{1}{2010}}{\frac{5}{2008}-\frac{5}{2009}-\frac{5}{2010}}+\frac{\frac{2}{2007}-\frac{2}{2008}-\frac{2}{2009}}{\frac{3}{2007}-\frac{3}{2008}-\frac{3}{2009}}\)

\(=\frac{\frac{1}{2008}-\frac{1}{2009}-\frac{1}{2010}}{5.\left(\frac{1}{2008}-\frac{1}{2009}-\frac{1}{2010}\right)}+\frac{2.\left(\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2009}\right)}{3.\left(\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2009}\right)}\)

\(=\frac{1}{5}+\frac{2}{3}\)

\(=\frac{13}{15}\)

Đặt \(A=2^{2009}+2^{2008}+...+2^1+2^0\)

Ta có : \(2A=2^{2010}+2^{2009}+...+2^2+2^1\)

\(\Rightarrow2A-A=2^{2010}-2^0\Rightarrow A=2^{2010}-1\)

Do đó : \(M=2^{2010}-A=2^{2010}-\left[2^{2010}-1\right]=1\)

\(M=2^{2010}-\left(2^{2009}+2^{2008}+...+2^1+2^0\right)\)

\(2^{2010}-M=2^{2009}+2^{2008}+...+2+1\)

\(2\left(2^{2010}-M\right)=2\left(2^{2009}+2^{2008}+...+2+1\right)\)

\(2\left(2^{2010}-M\right)=2^{2010}+2^{2009}+...+2^2+2\)

\(2\left(2^{2010}-M\right)-M=\left(2^{2010}+2^{2009}+...+4+2\right)-\left(2^{2009}+2^{2008}+...+2+1\right)\)

\(2^{2010}-M=2^{2010}+2^{2009}+...+4+2-2^{2009}-2^{2008}-...-2-1\)

\(2^{2010}-M=2^{2010}-1\)

=> M = 1

Cho cách giải lun

S=2²⁰¹⁰-2²⁰⁰⁹-2²⁰⁰⁸-...-2-1

=> 2S=2. 2²⁰¹⁰ -2. 2²⁰⁰⁹-2. 2²⁰⁰⁸-....-2.2-2.1

2S=2^2011-2²⁰¹⁰-2²⁰⁰⁹-....-2²-2

- S=2^2010-2²⁰⁰⁹-2²⁰⁰⁸-...-2-1

=>S=2^2011-1

cai gi ma tum lum the