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NN
Nguyễn Ngọc Anh Minh
CTVHS
VIP
11 tháng 5 2022
Trong dấu ngoặc đơn có số các số hạng là
Đặt tổng các số hạng trong ngoặc đơn là A
\(\dfrac{2013-1}{2}+1=1007\) số hạng
\(A=\dfrac{3+1}{1.3}-\dfrac{5+3}{3.5}+\dfrac{7+5}{5.7}-...+\dfrac{2015+2013}{2013.2015}=\)
\(=1+\dfrac{1}{3}-\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}+\dfrac{1}{7}-...+\dfrac{1}{2013}+\dfrac{1}{2015}=1+\dfrac{1}{2015}=\dfrac{2016}{2015}\)
\(\Rightarrow M=A.\dfrac{2015}{2016}=\dfrac{2016}{2015}.\dfrac{2015}{2016}=1\) là số tự nhiên
21 tháng 3 2016
<=>2-2/3+2/3-2/5........+2n-2n+2<2015/2016
<=>2-2n+2<2015/2016
=>n+2=1/2016
=>n=2014
21 tháng 3 2016
\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+......+\frac{2}{n\left(n+2\right)}\)<\(\frac{2015}{2016}\)
VT=\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{5}-\frac{1}{n+2}\)=\(1-\frac{1}{n+2}\)
Ta có:\(1-\frac{1}{n+2}=\frac{2015}{2016}\Rightarrow\)\(\frac{1}{n+2}=1-\frac{2015}{2016}\)
\(\Rightarrow\)\(\frac{1}{n+2}=\frac{1}{2016}=n+2=2016\)
\(\Rightarrow\)\(n=2014\)
Vậy\(n=2014\)
DA
1
NN
20 tháng 4 2017
Bài này dễ ý mà, vô cùng đơn giản..........
Ta có:
\(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{x\left(x+2\right)}=\dfrac{2015}{2016}.\)
\(\dfrac{2}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{x}-\dfrac{1}{x+2}\right)=\dfrac{2015}{2016}.\)
\(1\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{x}-\dfrac{1}{x+2}\right)=\dfrac{2015}{2016}.\)
\(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{x}-\dfrac{1}{x+2}=\dfrac{2015}{2016}.\)
\(\left(\dfrac{1}{3}-\dfrac{1}{3}\right)+\left(\dfrac{1}{5}-\dfrac{1}{5}\right)+...+\left(\dfrac{1}{x}-\dfrac{1}{x}\right)+\left(1-\dfrac{1}{x+2}\right)=\dfrac{2015}{2016}.\)
\(0+0+...+0+\left(1-\dfrac{1}{x+2}\right)=\dfrac{2015}{2016}.\)
\(1-\dfrac{1}{x+2}=\dfrac{2015}{2016}.\)
\(\dfrac{1}{x+2}=1-\dfrac{2015}{2016}.\)
\(\dfrac{1}{x+2}=\dfrac{1}{2016}.\)
\(\Rightarrow x+2=2016.\)
\(\Rightarrow x=2016-2=2014.\)
Vậy \(x=2014.\)
~ Học tốt nha bn!!! ~
Bài mik đúng thì nhớ tick mik nha!!!
DT
0
NT
7 tháng 5 2018
Bài 1 :
Ta có :
\(A=\frac{10^{17}+1}{10^{18}+1}=\frac{\left(10^{17}+1\right).10}{\left(10^{18}+1\right).10}=\frac{10^{18}+10}{10^{19}+10}\)
Mà : \(\frac{10^{18}+10}{10^{19}+10}>\frac{10^{18}+1}{10^{19}+1}\)
Mà \(A=\frac{10^{18}+10}{10^{19}+10}\)nên \(A>B\)
Vậy \(A>B\)
Bài 2 :
Ta có :
\(S=\frac{2013}{2014}+\frac{2014}{2015}+\frac{2015}{2016}+\frac{2016}{2013}\)
\(\Rightarrow S=\frac{2014-1}{2014}+\frac{2015-1}{2015}+\frac{2016-1}{2016}+\frac{2013+3}{2013}\)
\(\Rightarrow S=1-\frac{1}{2014}+1-\frac{1}{2015}+1-\frac{1}{2016}+1+\frac{3}{2013}\)
\(\Rightarrow S=4+\frac{3}{2013}-\left(\frac{1}{2014}+\frac{1}{2015}+\frac{1}{2016}\right)\)
Vì \(\frac{1}{2013}>\frac{1}{2014}>\frac{1}{2015}>\frac{1}{2016}\)nên \(\frac{3}{2013}-\left(\frac{1}{2014}+\frac{1}{2015}+\frac{1}{2016}\right)>0\)
Nên : \(M>4\)
Vậy \(M>4\)
Bài 3 :
Ta có :
\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+.......+\frac{1}{100^2}\)
Suy ra : \(A< \frac{1}{1.3}+\frac{1}{2.4}+\frac{1}{3.5}+....+\frac{1}{99.101}\)
\(\Rightarrow A< \frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{2.4}+......+\frac{2}{99.101}\right)\)
\(\Rightarrow A< \frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-......-\frac{1}{101}\right)\)
\(\Rightarrow A< \frac{1}{2}.\left[\left(1+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{99}\right)-\left(\frac{1}{3}+\frac{1}{4}+......+\frac{1}{101}\right)\right]\)
\(\Rightarrow A< \frac{1}{2}.\left(1+\frac{1}{2}-\frac{1}{100}-\frac{1}{101}\right)\)
\(\Rightarrow A< \frac{1}{2}.\left(1+\frac{1}{2}\right)\)
\(\Rightarrow A< \frac{3}{4}\)
Vậy \(A< \frac{3}{4}\)
Bài 4 :
\(a)A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+....+\frac{1}{2015.2017}\)
\(\Rightarrow A=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+.....+\frac{1}{2015.2017}\right)\)
\(\Rightarrow A=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.....+\frac{1}{2015}-\frac{1}{2017}\right)\)
\(\Rightarrow A=\frac{1}{2}.\left(1-\frac{1}{2017}\right)\)
\(\Rightarrow A=\frac{1}{2}.\frac{2016}{2017}\)
\(\Rightarrow A=\frac{1008}{2017}\)
Vậy \(A=\frac{1008}{2017}\)
\(b)\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+......+\frac{1}{x\left(x+2\right)}=\frac{1008}{2017}\)
\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+......+\frac{2}{x.\left(x+2\right)}=\frac{2016}{2017}\)
\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.....+\frac{1}{x}-\frac{1}{x+2}=\frac{2016}{2017}\)
\(1-\frac{1}{x+2}=\frac{2016}{2017}\)
\(\Rightarrow\frac{1}{x+2}=1-\frac{2016}{2017}\)
\(\Rightarrow\frac{1}{x+2}=\frac{1}{2017}\)
\(\Rightarrow x+2=2017\)
\(\Rightarrow x=2017-2=2015\)
Vậy \(x=2015\)
ND
0
TT
3
TN
6 tháng 6 2016
\(=2x+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{9}-\frac{1}{11}=-\frac{2016}{2017}\)
\(=2x+\frac{1}{3}-\frac{1}{11}=-\frac{2016}{2017}\)
\(2x+\frac{8}{33}=-\frac{2016}{2017}\)
\(2x=\frac{-2016}{2017}-\frac{8}{33}\)
\(2x=\frac{-2024}{2017}\)
\(x=-\frac{1012}{2017}\)
6 tháng 6 2016
\(2x+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}=\frac{-2016}{2017}\)
\(2x+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}=\frac{-2016}{2017}\)
\(2x+\frac{1}{3}-\frac{1}{11}=\frac{-2016}{2017}\)
\(2x+\frac{8}{33}=\frac{-2016}{2017}\)
\(2x=\frac{-2016}{2017}-\frac{8}{33}\)
Số dư dài quá. Đến đây bạn tự làm tiếp nhé
22 tháng 4 2023
=1-1/3+1/3-1/5+...+1/2015-1/2017
=1-1/2017
=2016/2017
\(\frac{2016}{1.3}+\frac{2016}{3.5}+\frac{2016}{5.7}+....+\frac{2016}{2015.2017}\)
\(=1008.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2015.2017}\right)\)
\(=1008.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{2015}-\frac{1}{2017}\right)\)
\(=1008.\left(1-\frac{1}{2017}\right)\)
\(=1008.\frac{2016}{2017}\)
147852.