Question

[Bùi Thị Bích Hồng]

(1/2+1/3-1/4+...+1/2009-1/2010):(1/2006+1/2007+...+1/2010)

Answer 1

[Phạm Tuấn Đạt]

Đặt \(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2009}-\frac{1}{2010}\)

\(\Rightarrow A=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{2009}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2010}\right)\)

\(\Rightarrow A=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2010}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2010}\right)\)

\(\Rightarrow A=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2010}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2005}\right)\)

\(\Rightarrow A=\frac{1}{2006}+\frac{1}{2007}+...+\frac{1}{2010}\)

\(\Rightarrow\frac{A}{\frac{1}{2006}+\frac{1}{2007}+...+\frac{1}{2010}}=1\)

Answer 2

[Arima Kousei]

Bạn Phạm Tuấn Đạt làm đúng rồi 

Dấu \(.\)là dấu nhân 

Đặt \(A=\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2009}-\frac{1}{2010}\)

       \(B=\frac{1}{2006}+\frac{1}{2007}+...+\frac{1}{2010}\)

Ta có : 

\(A=\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2009}-\frac{1}{2010}\)

\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2009}-\frac{1}{2010}\)

\(\Rightarrow A=\left(1+\frac{1}{3}+...+\frac{1}{2009}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2010}\right)\)

\(\Rightarrow A=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2009}+\frac{1}{2010}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}...+\frac{1}{2010}\right)\)

\(\Rightarrow A=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2010}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2005}\right)\)

\(\Rightarrow A=\frac{1}{2006}+\frac{1}{2007}+...+\frac{1}{2010}\)

\(\Rightarrow A=B\)

Nên : 

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

Vậy giá trị của biểu thức trên là \(1\)