A = 2²⁰⁰⁹ - 2²⁰⁰⁸ - 2²⁰⁰⁷ - ... - 2² - 2 - 1

A = 2²⁰⁰⁹ - (2²⁰⁰⁸ + 2²⁰⁰⁷ + ... + 2² + 2 + 1)

Đặt B = 2²⁰⁰⁸ + 2²⁰⁰⁷ + ... + 2² + 2 + 1

2B = 2²⁰⁰⁹ + 2²⁰⁰⁸ + ... + 2³ + 2² + 2

2B - B = (2²⁰⁰⁹ + 2²⁰⁰⁸ + ... + 2³ + 2² + 2) - (2²⁰⁰⁸ + 2²⁰⁰⁷ + ... + 2² + 2 + 1)

B = 2²⁰⁰⁹ - 1

=> A = 2²⁰⁰⁹ - (2²⁰⁰⁹ - 1) = 2²⁰⁰⁹ - 2²⁰⁰⁹ + 1 = 0 + 1 = 1

con cho

P/s : Lớp 6 nhé bạn 

Dấu \(.\)là dấu nhân 

Đặt \(A=\frac{2008}{1}+\frac{2007}{2}+\frac{2006}{3}+...+\frac{2}{2007}+\frac{1}{2008}\)  

      \(B=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2008}+\frac{1}{2009}\)

Ta có : 

\(A=\frac{2008}{1}+\frac{2007}{2}+\frac{2006}{3}+...+\frac{2}{2007}+\frac{1}{2008}\)

\(\Rightarrow A=1+\left(\frac{2007}{2}+1\right)+\left(\frac{2006}{3}+1\right)+...+\left(\frac{2}{2007}+1\right)+\left(\frac{1}{2008}+1\right)\)

\(\Rightarrow A=\frac{2009}{2009}+\frac{2009}{2}+\frac{2009}{3}+...+\frac{2009}{2007}+\frac{2009}{2008}\)

\(\Rightarrow A=\frac{2009}{2}+\frac{2009}{3}+...+\frac{2009}{2007}+\frac{2009}{2008}+\frac{2009}{2009}\)

\(\Rightarrow A=2009.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2007}+\frac{1}{2008}+\frac{1}{2009}\right)\)

\(\Rightarrow A=2009.B\)

Nên : \(\frac{A}{B}=\frac{2009.B}{B}=2009\)

Vậy kết quả biểu thức đã cho là \(2009\)

~ Ủng hộ nhé 

\(\frac{\frac{2008}{1}+\frac{2007}{2}+\frac{2006}{3}+...+\frac{2}{2007}+\frac{1}{2008}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2008}+\frac{1}{2009}}\)

\(=\frac{\left(\frac{2007}{2}+1\right)+\left(\frac{2006}{3}+1\right)+...+\left(\frac{2}{2007}+1\right)+\left(\frac{1}{2008}+1\right)+1}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2008}+\frac{1}{2009}}\)

\(=\frac{\frac{2009}{2}+\frac{2009}{3}+...+\frac{2009}{2007}+\frac{2009}{2008}+\frac{2009}{2009}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2008}+\frac{1}{2009}}\)

\(=\frac{2009.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2007}+\frac{1}{2008}+\frac{1}{2009}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2008}+\frac{1}{2009}}\)

\(=2009\)

Kết quả bằng 2009

Xét tử

2008+2007/2+2006/3+2005/4+ ... +2/2007+1/2008

=(1+1+1+...+1)+2007/2+2006/3+2005/4+ ... +2/2007+1/2008

= 1+ (2007/2)+1+(2006/3)+1+(2005/4)+1+ ... + (2/2007)+1+(1/2008)+1

=2009/2009+2009/2+2009/3+2009/4+ ... + 2009/2007 + 2009/2008

=2009.(1/2+1/3+1/4+ ... + 1/2007+1/2008+1/2009)

Tìm max của biểu thức: 1 3 4 2 + − x x .