\(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+....+\dfrac{3}{40.43}\)
\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{40}-\dfrac{1}{43}\)
\(=1-\dfrac{1}{43}=\dfrac{42}{43}\)
Sorry bạn vì trả lời hơi trễ.
\(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{40.43}\)
\(\Leftrightarrow\dfrac{1}{3}\left(3-\dfrac{3}{4}+\dfrac{3}{4}-\dfrac{4}{7}+...+\dfrac{3}{40}-\dfrac{3}{43}\right)\)
\(=\dfrac{1}{3}\left(3-\dfrac{3}{43}\right)=\dfrac{1}{3}.\dfrac{126}{43}=\dfrac{42}{43}\)