Question

[Đặng Anh Thư_Thư Đặng-A1]

 Tính: 1/2003+1/2004+1/2005

           2/2003+2/2004+2/2005     

Answer 1

[Xyz OLM]

\(\frac{\frac{1}{2003}+\frac{1}{2004}+\frac{1}{2005}}{\frac{2}{2003}+\frac{2}{2004}+\frac{2}{2005}}\)

= \(\frac{\frac{1}{2003}+\frac{1}{2004}+\frac{1}{2005}}{\frac{1}{2003}.2+\frac{1}{2004}.2+\frac{1}{2005}.2}\)

= \(\frac{1.\left(\frac{1}{2003}+\frac{1}{2004}+\frac{1}{2005}\right)}{2.\left(\frac{1}{2003}+\frac{1}{2004}+\frac{1}{2005}\right)}\)

\(=\frac{1}{2}\)