Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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
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 \(S=\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{40.43}=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{40}-\frac{1}{43}\)
\(S=1-\frac{1}{43}=\frac{42}{43}\)
\(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+....+\dfrac{3}{43.46}\)
\(=\dfrac{3}{1}-\dfrac{3}{4}+\dfrac{3}{4}-\dfrac{3}{7}+\dfrac{3}{7}-\dfrac{3}{10}+.....+\dfrac{3}{43}-\dfrac{3}{46}=3-\dfrac{3}{46}=\dfrac{135}{46}\)
Học tốt nha e
\(\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{40.43}\\ =1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{40}-\dfrac{1}{43}\\ =1-\dfrac{1}{43}\\ =\dfrac{42}{43}\)
A = \(\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+\dfrac{3}{7\cdot10}+...+\dfrac{3}{40\cdot43}+\dfrac{3}{2015\cdot2016}\)
A = \(\left(\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+\dfrac{3}{7\cdot10}+...+\dfrac{3}{40\cdot43}\right)+\left(\dfrac{1}{2015\cdot2016}\cdot3\right)\)
A = \(\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{40}-\dfrac{1}{43}\right)+\left(\left(\dfrac{1}{2015}-\dfrac{1}{2016}\right)\cdot3\right)\)
A = \(\left(1-\dfrac{1}{43}\right)+\dfrac{1}{1354080}=\dfrac{42}{43}+\dfrac{1}{1354080}=\dfrac{56871403}{58225440}\)