DA
1
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
22 tháng 12 2016
sao phần b k có qui luật j vậy đúng ra nó phải là 3/2014+2/2015+2/2016 chứ ( 3 phân số cuối)
S
30 tháng 7 2019
\(\frac{2016}{1}+\frac{2015}{2}+\frac{2014}{3}+.....+\frac{1}{2014}+\frac{1}{2015}+\frac{1}{2016}=\left(\frac{2015+2}{2}\right)+\left(\frac{2014+3}{3}\right)+.....\left(\frac{1+2016}{2016}\right)+\frac{2017}{2017}=\frac{2017}{2}+\frac{2017}{3}+....+\frac{2017}{2017}=2017\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.....+\frac{1}{2017}\right)\Rightarrow\frac{B}{A}=2017\)
25 tháng 1 2020
Ta có \(B=\frac{2016}{1}+\frac{2015}{2}+...+\frac{2}{2015}+\frac{1}{2016}\)
\(\Rightarrow B=1+\left(\frac{2015}{2}+1\right)+\left(\frac{2014}{3}+1\right)+...+\left(\frac{2}{2015}+1\right)+\left(\frac{1}{2016}+1\right)\)
\(\Rightarrow B=\frac{2017}{2017}+\frac{2017}{2}+\frac{2017}{3}+...+\frac{2017}{2015}+\frac{2017}{2016}\)
\(\Rightarrow B=2017.\left(\frac{1}{2017}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}+\frac{1}{2016}\right)\)
\(\Rightarrow B=2017.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}+\frac{1}{2016}+\frac{1}{2017}\right)\)
\(\Rightarrow\frac{B}{A}=\frac{2017.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}+\frac{1}{2017}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}+\frac{1}{2017}}=2017\)
Vậy \(\frac{B}{A}\)= 2017
~ Chúc bạn học tốt
T
1 tháng 1 2017
1/ ta có:
A = \(\frac{10^{2015}+1}{10^{2016}+1}\Rightarrow10A=\frac{10^{2016}+10}{10^{2016}+1}=1+\frac{9}{10^{2016}+1}\)
B = \(\frac{10^{2016}+1}{10^{2017}+1}\Rightarrow10B=\frac{10^{2017}+10}{10^{2017}+1}=1+\frac{9}{10^{2017}+1}\)
vì \(\frac{9}{10^{2016}+1}>\frac{9}{10^{2017}+1}\) => 10A > 10B
=> A > B
vậy A > B
2/ ta có: M = 5 + 52 + 53 + ... + 52016
=> 5M = 52+53+54+...+52017
=> 5M - M = (52+53+54+...+52017) - (5+52+53+...+52016)
=> 4M = 52017- 5
=> M = \(\frac{5^{2017}-5}{4}\)
vậy M = \(\frac{5^{2017}-5}{4}\)
DP
14 tháng 8 2017
a, \(A=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}}{\frac{2011}{1}+\frac{2010}{2}+\frac{2009}{3}+...+\frac{1}{2011}}\)
\(A=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}}{\left(\frac{2011}{1}+1\right)+\left(\frac{2010}{2}+1\right)+\left(\frac{2009}{3}+1\right)+...+\left(\frac{1}{2011}+1\right)+1}\)
\(A=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}}{\frac{2012}{1}+\frac{2012}{2}+\frac{2012}{3}+...+\frac{2012}{2011}+\frac{2012}{2012}}\)
\(A=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}}{2012\cdot\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}+\frac{1}{2012}\right)}=\frac{1}{2012}\)
b, \(\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+....+\frac{1}{2016}+\frac{1}{2017}}{\frac{2016}{1}+\frac{2015}{2}+\frac{2014}{3}+...+\frac{2}{2015}+\frac{1}{2016}}\)
\(\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}+\frac{1}{2017}}{\left(\frac{2016}{1}+1\right)+\left(\frac{2015}{2}+1\right)+\left(\frac{2014}{3}+1\right)+...+\left(\frac{2}{2015}+1\right)+\left(\frac{1}{2016}+1\right)+1}\)
\(\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}}{\frac{2017}{1}+\frac{2017}{2}+\frac{2017}{3}+...+\frac{2017}{2015}+\frac{2017}{2016}+\frac{2017}{2017}}\)
\(\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}}{2017\cdot\left(\frac{1}{2}+\frac{1}{3}+....+\frac{1}{2015}+\frac{1}{2016}+\frac{1}{2017}\right)}=\frac{1}{2017}\)
14 tháng 12 2017
\(B=\dfrac{2016}{1}+\dfrac{2015}{2}+\dfrac{2014}{3}+...+\dfrac{3}{2014}+\dfrac{2}{2015}+\dfrac{1}{2016}\)
\(B=2016+\dfrac{2015}{2}+\dfrac{2014}{3}+....+\dfrac{3}{2014}+\dfrac{2}{2015}+\dfrac{1}{2016}\)
\(B=1+\left(\dfrac{2015}{2}+1\right)+\left(\dfrac{2014}{3}+1\right)+...+\left(\dfrac{3}{2014}+1\right)+\left(\dfrac{2}{2015}+1\right)+\left(\dfrac{1}{2016}+1\right)\)
\(B=\dfrac{2017}{2017}+\dfrac{2017}{2}+\dfrac{2017}{3}+....+\dfrac{2017}{2014}+\dfrac{2017}{2015}+\dfrac{2017}{2016}\)
\(B=2017\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2014}+\dfrac{1}{2015}+\dfrac{1}{2016}+\dfrac{1}{2017}\right)\)
\(\dfrac{B}{A}=\dfrac{2017\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2014}+\dfrac{1}{2015}+\dfrac{1}{2016}+\dfrac{1}{2017}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+....+\dfrac{1}{2014}+\dfrac{1}{2015}+\dfrac{1}{2016}+\dfrac{1}{2017}}=2017\)
14 tháng 12 2017
\(\dfrac{B}{A}=\dfrac{\dfrac{2016}{1}+\dfrac{2015}{2}+\dfrac{2014}{3}+...+\dfrac{3}{2014}+\dfrac{2}{2015}+\dfrac{1}{2016}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+...+\dfrac{1}{2016}+\dfrac{1}{2017}}\)
\(=\dfrac{1+\left(\dfrac{2015}{2}+1\right)+\left(\dfrac{2014}{3}+1\right)+...+\left(\dfrac{2}{2015}+1\right)+\left(\dfrac{1}{2016}+1\right)}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+...+\dfrac{1}{2016}+\dfrac{1}{2017}}\)
\(=\dfrac{\dfrac{2017}{2017}+\left(\dfrac{2015}{2}+\dfrac{2}{2}\right)+\left(\dfrac{2014}{3}+\dfrac{3}{3}\right)+...+\left(\dfrac{1}{2016}+\dfrac{2016}{2016}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2016}+\dfrac{1}{2017}}\)
\(=\dfrac{2017\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2016}+\dfrac{1}{2017}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2016}+\dfrac{1}{2017}}\)
\(=2017\)
Vậy \(\dfrac{B}{A}=2017\)
1 tháng 11 2017
\(N=\frac{1}{2016}+\frac{2}{2015}+\frac{3}{2014}+...+\frac{2015}{2}+\frac{2016}{1}\)
\(N=1+\left(\frac{1}{2016}+1\right)+\left(\frac{2}{2015}+1\right)+\left(\frac{3}{2014}+1\right)+...+\left(\frac{2015}{2}+1\right)\)
\(N=\frac{2017}{2017}+\frac{2017}{2016}+\frac{2017}{2015}+\frac{2017}{2014}+...+\frac{2017}{2}\)
\(N=2017.\left(\frac{1}{2017}+\frac{1}{2016}+\frac{1}{2015}+\frac{1}{2014}+...+\frac{1}{2}\right)\)
\(\Rightarrow\frac{M}{N}=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}+\frac{1}{2017}}{2017.\left(\frac{1}{2017}+\frac{1}{2016}+\frac{1}{2015}+\frac{1}{2014}+...+\frac{1}{2}\right)}=\frac{1}{2017}\)
\(3A=3^{2017}-3^{2-16}+...-3^2+3\)
\(\Rightarrow A+3A=3^{2017}+1\)
\(\Rightarrow A=\frac{3^{2017}+1}{4}\)
\(5B=5^{2017}+5^{2016}+...+5\)
\(\Rightarrow5B-B=5^{2017}-1\)
\(\Rightarrow B=\frac{5^{2017}-1}{4}\)