tinh nhanh b=4/1.4+4/4.7+4/710+...+4/2014.2017
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
BB
8
3 tháng 3 2017
Ta có : B = \(\frac{4}{1.4}+\frac{4}{4.7}+\frac{4}{7.10}+......+\frac{4}{2014.2017}\)
\(=\frac{4}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+......+\frac{3}{2014.2017}\right)\)
\(=\frac{4}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+......+\frac{1}{2014}-\frac{1}{2017}\right)\)
\(=\frac{4}{3}.\left(1-\frac{1}{2017}\right)\)
\(=\frac{4}{3}.\frac{2016}{2017}=\frac{2688}{2017}\)
NT
4
10 tháng 5 2017
\(B=\frac{5}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+....+\frac{1}{2014}-\frac{1}{2017}\right)\)
\(B=\frac{5}{3}.\left(1-\frac{1}{2017}\right)\)
\(B=\frac{5}{3}.\frac{2016}{2017}=\frac{10080}{6051}\)
Q
10 tháng 5 2017
\(B=\frac{5}{1.4}+\frac{5}{4.7}+...+\frac{5}{2014.2017}\)
\(3M=5\left(\frac{1}{1.4}+\frac{1}{4.7}+...+\frac{1}{2014.2017}\right)\)
\(3M=5\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{2014}-\frac{1}{2017}\right)\)
\(3M=5\left(1-\frac{1}{2017}\right)\)
\(3M=5.\frac{2016}{2017}\)
\(3M=\frac{10080}{2017}\)
\(\Rightarrow M=\frac{3360}{2017}\)
B
3
G
4 tháng 2 2020
A = \(\frac{1}{1.4}\)+ \(\frac{1}{4.7}\)+\(\frac{1}{7.10}\)+...+ \(\frac{1}{2014.2017}\)
3A = \(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+....+\frac{3}{2014.2017}\)
3A = \(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+.....+\frac{1}{2014}-\frac{1}{2017}\)
3A= 1 - \(\frac{1}{2017}\)
A = \(\frac{1}{3}-\frac{1}{2017.3}\)
A = \(\frac{672}{2017}\)
YN
4 tháng 2 2020
Ta có \(A=\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{2014.2017}\)
\(\Rightarrow A=\frac{1}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{2014}-\frac{1}{2017}\right)\)
\(\Rightarrow A=\frac{1}{3}.\left(1-\frac{1}{2017}\right)\)
\(\Rightarrow A=\frac{1}{3}.\frac{2016}{2017}=\frac{672}{2017}\)
Vậy \(A=\frac{672}{2017}\)
~ Học tốt
# Chiyuki Fujito
NN
Nguyễn Ngọc Anh Minh
CTVHS
VIP
27 tháng 8 2016
\(\frac{3A}{4}=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{25.28}.\)
\(\frac{3A}{4}=\frac{4-1}{1.4}+\frac{7-4}{4.7}+\frac{10-7}{7.10}+...+\frac{28-25}{25.28}\)
\(\frac{3A}{4}=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{25}-\frac{1}{28}=1-\frac{1}{28}=\frac{27}{28}\)
\(A=\frac{27.4}{28.3}=\frac{9}{7}\)
29 tháng 7 2021
a) Ta có: \(A=\dfrac{4}{1\cdot4}+\dfrac{4}{4\cdot7}+\dfrac{4}{7\cdot10}+...+\dfrac{4}{31\cdot34}\)
\(=\dfrac{4}{3}\left(\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+\dfrac{3}{7\cdot10}+...+\dfrac{3}{31\cdot34}\right)\)
\(=\dfrac{4}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{31}-\dfrac{1}{34}\right)\)
\(=\dfrac{4}{3}\left(1-\dfrac{1}{34}\right)\)
\(=\dfrac{4}{3}\cdot\dfrac{33}{34}=\dfrac{22}{17}\)
8 tháng 5 2015
\(A=\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(2A=\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{99.100}\)
\(2A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(2A=\frac{1}{1}-\frac{1}{100}\)
\(2A=\frac{99}{100}\Rightarrow A=\frac{99}{100}:2\Rightarrow A=\frac{99}{200}\)
Câu B và C làm tương tự.
8 tháng 5 2015
bạn Nhi làm sai rồi
\(\frac{2}{2\cdot3}\) sao có thể bằng \(\frac{1}{2}-\frac{1}{3}\) được
\(\frac{1}{2\cdot3}\) mới bằng \(\frac{1}{2}-\frac{1}{3}\)
kết quả là : \(\frac{49}{100}\)
ND
Nguyễn Đức Trí
VIP
25 tháng 7 2023
Bài 1 :
\(S=1.3+3.5+5.7+...+99.101=3+15+35+...9999\)
Ta thấy :
\(3=2^2-1\)
\(15=4^2-1\)
\(35=6^2-1\)
.....
\(9999=100^2-1\)
\(\Rightarrow S=2^2+4^2+...+100^2-\left(1\right).\left(\left(100-2\right):2+1\right)\)
\(\Rightarrow S=\dfrac{100.\left(100+1\right)\left(2.100+1\right)}{6}-51\)
\(\Rightarrow S=\dfrac{100.101.201}{6}-51=338299\)
CN
1
\(B=\dfrac{4}{1\cdot4}+\dfrac{4}{4\cdot7}+...+\dfrac{4}{2014\cdot2017}\)
\(=\dfrac{4}{3}\left(\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+...+\dfrac{3}{2014\cdot2017}\right)\)
\(=\dfrac{4}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{2014}-\dfrac{1}{2017}\right)\)
\(=\dfrac{4}{3}\left(1-\dfrac{1}{2017}\right)=\dfrac{4}{3}\cdot\dfrac{2016}{2017}=\dfrac{8064}{6051}\)