Giải:

\(A=\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{40.43}+\dfrac{3}{2015.2018}\)

\(\Leftrightarrow A=\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{40}-\dfrac{1}{43}+\dfrac{1}{2015}-\dfrac{1}{2018}\)

\(\Leftrightarrow A=\dfrac{1}{1}-\dfrac{1}{43}+\dfrac{1}{2015}-\dfrac{1}{2018}\)

\(\Leftrightarrow A=\dfrac{42}{43}+\dfrac{1}{2015}-\dfrac{1}{2018}\)

\(\Leftrightarrow A=0,977240464-\dfrac{1}{2018}\)

\(\Leftrightarrow A=0,9767449238\approx0,98\)

Vậy ...

Sửa đề : \(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{40.43}\)

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

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

A = 1/1 -1/4 +1/4 - 1/7 +1/7 ........+1/40 - 1/43 

A = 1/1 - 1/43 

A = 42/43

A=1 - 1/4 + 1/4 - 1/7 + .... + 1/40 - 1/43

  = 1 - 1/43 

  = 42/43

=42/43 nha bn

Ta có :

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

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

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

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

dễ mà

Gọi tổng đó là S. Theo đề \(S=\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{40.43}=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{40}-\frac{1}{43}\)

\(S=1-\frac{1}{43}=\frac{42}{43}\)

dễ mà sai

\(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+....+\dfrac{3}{43.46}\)

\(=\dfrac{3}{1}-\dfrac{3}{4}+\dfrac{3}{4}-\dfrac{3}{7}+\dfrac{3}{7}-\dfrac{3}{10}+.....+\dfrac{3}{43}-\dfrac{3}{46}=3-\dfrac{3}{46}=\dfrac{135}{46}\)

Học tốt nha e

\(\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{40.43}\\ =1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{40}-\dfrac{1}{43}\\ =1-\dfrac{1}{43}\\ =\dfrac{42}{43}\)

e) 3/1.4 + 3/4.7 + 3/7.10+ ... + 3/40.43
= 1-1/4 + 1/4 -1/7 + 1/7-1/10+...+1/40-1/43
= 1-1/43
= 42/43