Tổng \(\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+...+\dfrac{1}{43\cdot45}\) bằng
\(\dfrac{14}{45}\).\(\dfrac{44}{45}\).\(\dfrac{7}{45}\).\(\dfrac{1}{3}\).Hướng dẫn giải:\(\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+...+\dfrac{1}{43\cdot45}\\ =\dfrac{1}{2}\cdot\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{43}-\dfrac{1}{45}\right)\\ =\dfrac{1}{2}\cdot\left(\dfrac{1}{3}-\dfrac{1}{45}\right)=\dfrac{1}{2}\cdot\dfrac{14}{45}=\dfrac{7}{45}.\)