Question

[Vương Gia Lương]

Tính tỉ số  biết:

Answer 1

[Online]

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

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

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

Có số số hạng là :

( 2009 - 2 ) : 1 + 1 = 2008 ( số hạng )

\(\Leftrightarrow\frac{1}{A}=\frac{\left(2009+2\right).2008}{2}=2019044\)

\(\Leftrightarrow\frac{1}{A}=2019044\)

\(\Leftrightarrow A=\frac{1}{2019044}\)

Answer 2

[Quỳnh Anh]

Trả lời:

Ta có: \(B=\frac{2008}{1}+\frac{2007}{2}+\frac{2006}{3}+...+\frac{2}{2007}+\frac{1}{2008}\)

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

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

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

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

\(\Rightarrow B=2009\cdot A\)

\(\Rightarrow\frac{B}{A}=2009\)

\(\Rightarrow\frac{A}{B}=\frac{1}{2009}\)