Ta có :

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

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

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

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

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

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

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

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

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

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

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

3/1.4+3/4.7+3/7.10+......+3/40.43

=1/1-1/4+1/4-1/7+1/7-1/10+......+1/40-1/43

triệt tiêu hết cho nhau ta còn:

1/1-1/43=43/43-1/43=42/43

nhớ cho mình nhé

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

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

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

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

=1/1-1/4+1/4-1/7+1/7-1/10+...+1/40-1/41

=1-1/41=40/41

= 3(1/1.4+1/4.7+1/7.10+.......+1/40.43)

=3(1-1/4+1/4-1/7+1/7-1/10+....+1/40-1/43)

=3(1-1/43)

=3.42/43

=126/43