H=2²⁰¹⁰-2²⁰⁰⁹-2²⁰⁰⁸-..-2-1

=>2H=2²⁰¹¹-2²⁰¹⁰-2²⁰⁰⁹-...-2²-2

=>2H-H=2²⁰¹¹-2²⁰¹⁰-2²⁰⁰⁹-..-2²-2²⁰¹⁰+2²⁰⁰⁹+2²⁰⁰⁸+..+2+1

=>H=2²⁰¹¹-2²⁰¹¹+1=1

=>2010^H=2010¹=2010

-2010 , ủng hộ mk nha

Ta có:

   2H=2²⁰¹¹-2²⁰¹⁰-2²⁰⁰⁹-...-2²-2

- 

     H=2²⁰¹⁰-2²⁰⁰⁹-2²⁰⁰⁸-...-2-1

----------------------------------------------------

=>2H-H=H=2²⁰¹¹-2²⁰¹⁰-2²⁰¹⁰-1=2²⁰¹¹-(2²⁰¹⁰+2²⁰¹⁰)-1=2²⁰¹¹-2²⁰¹⁰.2-1=2²⁰¹¹-2²⁰¹¹-1=0-1=-1

=>2010^H=2010^-1=1/2010

Chúc bạn học giỏi nha!!!

K cho mik vs nhé Không Cần Biết

H = 2²⁰¹¹-1

Ta có:

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

Đặt \(A=2^{2009}+2^{2008}+...+2+1\)

\(\Rightarrow2A=2^{20010}+2^{2009}+...+2^2+2\)

\(\Rightarrow2A-A=\left(2^{20010}+2^{2009}+...+2^2+2\right)-\left(2^{2009}+2^{2008}+...+2+1\right)\)\(\Rightarrow A=\left(2^{2010}-1\right)+\left(2^{2009}-2^{2009}\right)+\left(2^{2008}-2^{2008}\right)+...+\left(2-2\right)\)\(\Rightarrow A=2001-1\)

\(\Rightarrow H=2^{2010}-\left(2^{2010}-1\right)\)

\(\Rightarrow H=2^{2010}-2^{2010}+1=1\)

Thay \(H=1\) vào biểu thức \(2010^H\)

\(\Rightarrow2010^H=2010^1=1\)

Vậy \(2010^H=1\)

Ta có:

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

Đặt \(A=2^{2009}+2^{2008}+...+2+1\)

\(\Rightarrow2A=2^{20010}+2^{2009}+...+2^2+2\)

\(\Rightarrow2A-A=\left(2^{20010}+2^{2009}+...+2^2+2\right)-\left(2^{2009}+2^{2008}+...+2+1\right)\)\(\Rightarrow A=\left(2^{2010}-1\right)+\left(2^{2009}-2^{2009}\right)+\left(2^{2008}-2^{2008}\right)+...+\left(2-2\right)\)\(\Rightarrow A=2001-1\)

\(\Rightarrow H=2^{2010}-\left(2^{2010}-1\right)\)

\(\Rightarrow H=2^{2010}-2^{2010}+1=1\)

Thay \(H=1\) vào biểu thức \(2010^H\)

\(\Rightarrow2010^H=2010^1=1\)

Vậy \(2010^H=1\)