Cho A=1/1.2+1/3.4+....+1/2017.2018
B=1/1010+1/1011+......+1/2018
So sánh A và B
Những câu hỏi liên quan
Tính C=(1/1.2+1/3.4+1/5.6+...+1/2017.2018)-(1/1010+1/1011+1/1012+...+1/2017)
\(C=\left(\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\frac{1}{5\cdot6}+...+\frac{1}{2017\cdot2018}\right)-\)\(\left(\frac{1}{1010}+\frac{1}{1011}+\frac{1}{1012}+...+\frac{1}{2017}\right)\)
Đặt \(A=\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\frac{1}{5\cdot6}+...+\frac{1}{2017\cdot2018}\)
\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{2017}-\frac{1}{2018}\)
\(\Rightarrow A=\left(1+\frac{1}{3}+...+\frac{1}{2017}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2018}\right)\)
\(\Rightarrow A=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2017}+\frac{1}{2018}\right)-2\cdot\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2018}\right)\)
\(\Rightarrow A=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2017}+\frac{1}{2018}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2009}\right)\)
\(\Rightarrow A=\frac{1}{1010}+\frac{1}{1011}+\frac{1}{1012}+..+\frac{1}{2017}\)
\(\Rightarrow C=\left(\frac{1}{101}+\frac{1}{1011}+\frac{1}{1012}+...+\frac{1}{2018}\right)-\left(\frac{1}{1010}+\frac{1}{1012}+...+\frac{1}{2017}\right)\)
\(\Rightarrow C=\frac{1}{2018}\)
cho \(a=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+....+\frac{1}{2017.2018}\) ; \(b=\frac{1}{1010}+\frac{1}{1011}+\frac{1}{1012}+...+\frac{1}{2018}\) . Tính (a-b)^2019
giúp mk vs mn ơi. mình cần gấp chiều mai nộp òi
cho A=1/1.2+1/3.4+1/5.6+....+1/2021.2022 và B=1011+1010/1012+1009/1013+1008/1014+...+2/2020+1/2021 Chứng minh rằng : B/A là số nguyên
\(\left(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+.....+\frac{1}{2017.2018}\right)-\left(\frac{1}{1010}+\frac{1}{1011}+\frac{1}{1012}+.....+\frac{1}{2017}\right)\)
Đặt S = ( 1/1.2 + 1/3.4 + 1/5.6 + ... + 1/2017.2018 )
Đặt A = ( 1/1.2 + 1/3.4 + ... + 1/2017.2018)
= 1 - 1/2 + 1/3 - 1/4 + ... + 1/2017 - 1/2018
= ( 1 + 1/3 + ... + 1/2017 ) - ( 1/2 + 1/4 + ... + 1/2018 )
= ( 1 + 1/2 + ... + 1/2018 ) - 2 ( 1/2 + 1/4 + ... + 1/2018) )
= ( 1 + 1/2 + ... + 1/2018 ) - ( 1 + 1/2 + ... + 1/1009 )
= 1/1010 + 1/1011 + ... + 1/2018
=> A - ( 1/1010 + 1/1011 + ... + 1/2017 ) = 1/2018
=> S = 1/2018
Vậy S = 1/2018
thanks bạn nhiều
tính hợp lí:
\(\left(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{2017.2018}\right)-\left(\frac{1}{1010}+\frac{1}{1011}+...+\frac{1}{2018}\right)\)
\(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{2019.2020}\)
\(B=\frac{1}{1010}+\frac{1}{1011}+...+\frac{1}{2020}\)
So Sánh A Và B
thôi mik làm đc rồi
Cho A = 1/1.2 + 1/3.4 + 1/5.6 +....+ 1/2017.2018
B = 1/1009 + 1/1010 + 1/1011 +...+ 1/2018
Tính B - A ??
Ai đúng và sớm nhất thì mình tick nha ^^
A=1/1.2+1/3.4+...+1/2017.2018
A=1-1/2+1/3-1/4+1/5-....+1/2017-1/2018
Bạn để riêng 2 nhóm có dấu trừ và cộng
A=(1+1/3+1/5+...+1/2017) - (1/2+1/4+1/6+...+1/2018)
A= M - N
A= M+N-2N
M=1+1/3+1/5+...+1/2017
Đúng 0
Bình luận (0)
bằng \(-\frac{1}{2018}\)
So sánh:
A = \(\frac{1}{1.2}\) + \(\frac{1}{3.4}\) + \(\frac{1}{5.6}\)+....+\(\frac{1}{2017.2018}\) và B = \(\frac{1}{1010}\)+\(\frac{1}{1011}\)+...+\(\frac{1}{2018}\)
Thực hiện phép tính :(hợp lí )
B=(\(\dfrac{1}{1.2}\)+\(\dfrac{1}{3.4}\)+...+\(\dfrac{1}{2017.2018}\)) - ( \(\dfrac{1}{1010}\)+\(\dfrac{1}{1011}\)+...+\(\dfrac{1}{2018}\))