Question

giải dùm mình bài toán này nha. nhớ trả lời sớm nha. cảm ơn

\(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{40.43}\)

Answer 1

\(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{40.43}\)

\(=3\left(\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+...+\dfrac{1}{40.43}\right)\)

\(=3.\dfrac{1}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{40}-\dfrac{1}{43}\right)\)

\(=1-\dfrac{1}{43}\)

\(=\dfrac{42}{43}\)

Answer 2

\(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{40.43}\)

\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{40}-\dfrac{1}{43}\)

\(=1-\dfrac{1}{43}=\dfrac{42}{43}\)

Answer 3

\(\dfrac{3}{1.4}+\dfrac{4}{4.7}+...+\dfrac{3}{30.43}\)

\(=\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{40}-\dfrac{1}{43}\)

\(=\dfrac{1}{1}-\dfrac{1}{43}\)

\(=\dfrac{43}{43}-\dfrac{1}{43}\)

\(=\dfrac{42}{43}\)

Answer 4

\(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{40.43}\)

\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{40}-\dfrac{1}{43}\)

\(=1-\dfrac{1}{43}\)

\(=\dfrac{42}{43}\)