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S
10 tháng 3 2019
\(\left(\frac{1}{1}+\frac{1}{3}+\frac{1}{5}+....+\frac{1}{2n-1}\right)-\left(\frac{1}{2}+\frac{1}{4}+....+\frac{1}{2n}\right)=\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{2n-1}+\frac{1}{2n}\right)-2\left(\frac{1}{2}+\frac{1}{4}+.....+\frac{1}{2n}\right)=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2n-1}+\frac{1}{2n}-\frac{1}{1}-\frac{1}{2}-....-\frac{1}{n}=\frac{1}{n+1}+\frac{1}{n+2}+....+\frac{1}{2n}\left(\text{đpcm}\right)\)
19 tháng 3 2017
a,Vế trái:
\(1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2013}-\dfrac{1}{2014}\)
\(=1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2014}-2\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+...+\dfrac{1}{2014}\right)\)
\(=1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2014}-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{1007}\right)\)
\(=\dfrac{1}{1008}+\dfrac{1}{2009}+...+\dfrac{1}{2014}\)
b,chưa có câu trả lời, sorry nha
NT
2
18 tháng 5 2017
Ta có:\(\dfrac{1}{4^2}< \dfrac{1}{2.4}\)
\(\dfrac{1}{6^2}< \dfrac{1}{4.6}\)
\(\dfrac{1}{8^2}< \dfrac{1}{6.8}\)
...
\(\dfrac{1}{\left(2n\right)^2}< \dfrac{1}{\left(2n-2\right).2n}\)
=>\(\dfrac{1}{4^2}+\dfrac{1}{6^2}+...+\dfrac{1}{\left(2n\right)^2}< \dfrac{1}{2.4}+\dfrac{1}{4.6}+...+\dfrac{1}{\left(2n-2\right)2n}=\dfrac{1}{2}\cdot\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{2n-2}-\dfrac{1}{2n}\right)=\dfrac{1}{2}\cdot\left(\dfrac{1}{2}-\dfrac{1}{2n}\right)< \dfrac{1}{2}\cdot\dfrac{1}{2}=\dfrac{1}{4}\left(đpcm\right)\)
18 tháng 5 2017
Đặt A = \(\dfrac{1}{4^2}+\dfrac{1}{6^2}+\dfrac{1}{8^2}+...+\dfrac{1}{\left(2n\right)^2}\)
\(A=\dfrac{1}{2^2}\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{n^2}\right)\)
Đặt \(B=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{n^2}\)
Ta có :
\(\dfrac{1}{2^2}< \dfrac{1}{1.2}\) ( vì 1 > 0 ; 0 < 1.2 < 22 )
\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\) ( vì 1 > 0 ; 0 < 2.3 < 32 )
\(\dfrac{1}{4^2}< \dfrac{1}{3.4}\) ( vì 1 > 0 ; 0 < 3.4 < 42 )
...
\(\dfrac{1}{n^2}< \dfrac{1}{\left(n-1\right)n}\) ( vì 1 > 0 ; 0 < ( n - 1 ) n < n2 )
\(\Rightarrow\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{n^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{\left(n-1\right)n}\)
\(\Rightarrow B< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{n-1}-\dfrac{1}{n}\)
\(\Rightarrow B< 1-\dfrac{1}{n}< 1\Rightarrow A< 1.\dfrac{1}{4}\Rightarrow A< \dfrac{1}{4}\)
DD
2
N
27 tháng 7 2017
a) \(\frac{3-2x}{5}=\frac{2}{7}\)
\(\Rightarrow7.\left(3-2x\right)=2.5\)
\(\Rightarrow21-14x=10\)
\(\Rightarrow14x=11\)
\(\Rightarrow x=\frac{11}{14}\)
27 tháng 7 2017
b) ( 5x - 6 ) : 7 = \(4\frac{1}{2}+0,25\%\)
( 5x - 6 ) : 7 = \(\frac{19}{4}\)
5x - 6 = \(\frac{19}{4}\). 7
5x - 6 = \(\frac{133}{4}\)
5x = \(\frac{133}{4}\)+ 6
5x = \(\frac{157}{4}\)
x = \(\frac{157}{4}\): 5
x = \(\frac{157}{20}\)
AH
Akai Haruma
Giáo viên
25 tháng 12 2018
Lời giải:
Ta có:
\(N=\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{(2n)^2}< \frac{1}{4^2-1}+\frac{1}{6^2-1}+\frac{1}{8^2-1}+...+\frac{1}{(2n)^2-1}(*)\)
Mà:
\(\frac{1}{4^2-1}+\frac{1}{6^2-1}+\frac{1}{8^2-1}+...+\frac{1}{(2n)^2-1}=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{(2n-1)(2n+1)}\)
\(=\frac{1}{2}\left(\frac{5-3}{3.5}+\frac{7-5}{5.7}+\frac{9-7}{7.9}+...+\frac{(2n+1)-(2n-1)}{(2n-1)(2n+1)}\right)\)
\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{2n-1}-\frac{1}{2n+1}\right)=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{2n+1}\right)\)
\(< \frac{1}{6}< \frac{1}{4}(**)\)
Từ \((*);(**)\Rightarrow N< \frac{1}{4}\) (đpcm)
8 tháng 3 2017
Ta có:
\(\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{2n^2}\)
\(=\frac{1}{2}\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{\left(2n-2\right).2n}\right)\)
\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{12}+...+\frac{1}{2n-2}-\frac{1}{2n}\right)\)
\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{2n}\right)\)
\(=\frac{1}{4}-\frac{1}{2n.2}\)
Vì \(\frac{1}{4}-\frac{1}{2n.2}< \frac{1}{4}\)
\(\Rightarrow\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{2n^2}< \frac{1}{4}\)
Vậy \(\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{2n^2}< \frac{1}{4}\) (Đpcm)
18 tháng 9 2017
a/ Ta có :
\(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...........+\dfrac{1}{n^2}\)
Ta thấy :
\(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)
\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)
.......................
\(\dfrac{1}{n^2}< \dfrac{1}{\left(n-1\right)n}\)
\(\Leftrightarrow A< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...........+\dfrac{1}{\left(n-1\right)n}\)
\(\Leftrightarrow A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+..........+\dfrac{1}{n-1}-\dfrac{1}{n}\)
\(\Leftrightarrow A< 1-\dfrac{1}{n}< 1\)
\(\Leftrightarrow A< 1\)
b/ Ta có :
\(B=\dfrac{1}{2^2}+\dfrac{1}{4^2}+\dfrac{1}{6^2}+.................+\dfrac{1}{\left(2n\right)^2}\)
\(=\dfrac{1}{4}\left(1+\dfrac{1}{2^2}+\dfrac{1}{4^2}+..........+\dfrac{1}{n^2}\right)\)
Ta thấy :
\(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)
\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)
..................
\(\dfrac{1}{n^2}< \dfrac{1}{\left(n-1\right)n}\)
\(\Leftrightarrow B< \dfrac{1}{4}\left(1+\dfrac{1}{1.2}+\dfrac{1}{2.3}+.........+\dfrac{1}{\left(n-1\right)n}\right)\)
\(\Leftrightarrow B< \dfrac{1}{4}\left(1+1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+......+\dfrac{1}{n-1}-\dfrac{1}{n}\right)\)
\(\Leftrightarrow B< \dfrac{1}{4}\left(1+1-\dfrac{1}{n}\right)\)
\(\Leftrightarrow B< \dfrac{1}{2}-\dfrac{1}{4n}< \dfrac{1}{2}\)
\(\Leftrightarrow B< \dfrac{1}{2}\)
LN
19 tháng 9 2017
\(\)\(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{n^2}\)
\(A< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{\left(n-1\right)n}\)
\(A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{n-1}-\dfrac{1}{n}\)
\(A< 1-\dfrac{1}{n}< 1\)
\(B=\dfrac{1}{2^2}+\dfrac{1}{4^2}+\dfrac{1}{6^2}+...+\dfrac{1}{2n^2}\)
\(B=\dfrac{1}{2}\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{2n^2}\right)\)
\(B=\dfrac{1}{4}+\dfrac{1}{2}\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{2n^2}\right)\)
\(B< \dfrac{1}{4}+\dfrac{1}{2}\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+.....+\dfrac{1}{\left(n-1\right)n}\right)\)
25 tháng 5 2022
a: \(\Leftrightarrow\dfrac{2}{3}x-\dfrac{5}{6}=\dfrac{-3}{2}x+\dfrac{3}{4}\)
=>13/6x=3/4+5/6
=>13/6x=9/12+10/12=19/12
hay x=19/26
b: \(\left(5x-3\right)\left(2x+5\right)=0\)
=>5x-3=0 hoặc 2x+5=0
=>x=3/5 hoặc x=-5/2
c: \(\left(\dfrac{5}{6}:x-\dfrac{5}{4}\right)^4=\dfrac{81}{16}\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{5}{6}:x-\dfrac{5}{4}=\dfrac{3}{2}\\\dfrac{5}{6}:x-\dfrac{5}{4}=-\dfrac{3}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\dfrac{5}{6}:x=\dfrac{11}{4}\\\dfrac{5}{6}:x=-\dfrac{1}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{10}{33}\\x=-\dfrac{10}{3}\end{matrix}\right.\)
d: \(\left|\dfrac{2}{5}x-\dfrac{1}{5}\right|\cdot\dfrac{5}{4}-2=\dfrac{3}{2}\)
\(\Leftrightarrow\left|\dfrac{2}{5}x-\dfrac{1}{5}\right|\cdot\dfrac{5}{4}=\dfrac{3}{2}+2=\dfrac{7}{2}\)
\(\Leftrightarrow\left|\dfrac{2}{5}x-\dfrac{1}{5}\right|=\dfrac{7}{2}:\dfrac{5}{4}=\dfrac{7}{2}\cdot\dfrac{4}{5}=\dfrac{28}{10}=\dfrac{14}{5}\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{2}{5}x-\dfrac{1}{5}=\dfrac{14}{5}\\\dfrac{2}{5}x-\dfrac{1}{5}=-\dfrac{14}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3:\dfrac{2}{5}=\dfrac{15}{2}\\x=-\dfrac{13}{5}:\dfrac{2}{5}=\dfrac{-13}{2}\end{matrix}\right.\)
TN
29 tháng 4 2017
BÀi 1
Để A \(\in\) Z
=>\(\left(n+2\right)⋮\left(n-5\right)\)
=>\([\left(n-5\right)+7]⋮\left(n-5\right)\)
=>\(7⋮\left(n-5\right)\)
=>\(n-5\in\left\{1;7;-1;-7\right\}\)
=>\(n\in\left\{6;13;4;-2\right\}\)
Vậy \(n\in\left\{6;13;4;-2\right\}\)
bạn có nhầm đề bài ko???????????????
ko