Question

[Kim So Huyn]

\(\dfrac{3}{3.5}+\dfrac{3}{5.7}+...+\dfrac{3}{47.49}\)

Answer 1

[hà mai trang]

\(=3.\left(\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{47.49}\right)\)

\(=3.\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{47}-\dfrac{1}{49}\right)\)

\(=3.\left(\dfrac{1}{3}-\dfrac{1}{49}\right)\)

\(=3.\dfrac{46}{147}\)

\(=\dfrac{46}{49}\)

Answer 2

[Võ Thị Tuyết Kha]

\(\dfrac{3}{3.5}+\dfrac{3}{5.7}+...+\dfrac{3}{47.49}\)

=\(\dfrac{3}{2}.\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{47}-\dfrac{1}{49}\right)\)

=\(\dfrac{3}{2}.\left(\dfrac{1}{3}-\dfrac{1}{49}\right)\)

=\(\dfrac{3}{2}.\dfrac{46}{147}\)

=\(\dfrac{23}{49}\)