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ta có 3A = 3/1.4 + 3/4.7 + ... + 3/(3n-2).(3n+1)
3A = 1-1/4 + 1/4 - 1/7 +....+ 1/(3n-2) - 1/(3n+1)
3A = 1- 1/(3n+1)
Mà 1/(3n+1) > 0 suy ra 3A < 1 suy ra A<1/3
tk giúp mình nha
\(\frac{1}{1.3}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{n\left(n+3\right)}=\frac{2018}{6057}\)
\(\Rightarrow\frac{1}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{n\left(n+3\right)}\right)=\frac{2018}{6057}\)
\(\Rightarrow\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{n}-\frac{1}{n+3}=\frac{2018}{6057}.3\)
\(\Rightarrow1-\frac{1}{n+3}=\frac{2018}{2019}\)
\(\Rightarrow\frac{1}{n+3}=1-\frac{2018}{2019}\)
\(\Rightarrow\frac{1}{n+3}=\frac{1}{2019}\)
\(\Rightarrow n+3=2019\)
\(\Rightarrow n=2016\)
Vậy n = 2016
Giải:
\(S=\dfrac{1}{1.4}-\dfrac{1}{4.7}-\dfrac{1}{7.10}-...-\dfrac{1}{97.100}\)
\(\Leftrightarrow S=-\left(-\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+...+\dfrac{1}{97.100}\right)\)
\(\Leftrightarrow S=-\dfrac{1}{3}\left(-\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{97}-\dfrac{1}{100}\right)\)
\(\Leftrightarrow S=-\dfrac{1}{3}\left(-\dfrac{1}{1}-\dfrac{1}{100}\right)\)
\(\Leftrightarrow S=-\dfrac{1}{3}\left(-\dfrac{101}{100}\right)\)
\(\Leftrightarrow S=\dfrac{101}{300}\)
Vậy ...
Bạn ơi cho mình hỏi tại sao phía trước \(-\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-.....+\dfrac{1}{97}-\dfrac{1}{100}\) lại là \(-\dfrac{1}{3}\)
=> 3x/4+3/4.7+3/7.10+...+3/100.103=306/103(nhân cả 2 vế của đt lên 2)
=>3x/4+(1/4-1/7)+(1/7-1/10)+...+(1/100-1/103)=306/103
=>3x/4+1/4-1/103+=306/103
=>3x/4+99/412=306/103
=>3x/4=306/103-99/412=1125/412
=>x=1125/412:3/4
=>x=1125/309
( nếu thấy đúng thì tick cho mk nha
Bài 1a) \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2018.2019}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+....+\dfrac{1}{2018}-\dfrac{1}{2019}\)
\(=1-\dfrac{1}{2019}=\dfrac{2018}{2019}\)
b) \(S=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{2017.2019}\)
\(2S=\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{2017.2019}\)
\(2S=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2017}-\dfrac{1}{2019}\)
\(2S=1-\dfrac{1}{2019}=\dfrac{2018}{2019}\)
\(S=\dfrac{1009}{2019}\)
Còn lại bạn làm tương tự hết nhé .
Ta có: \(C=\frac{1}{3.4}+\frac{1}{5.6}+\frac{1}{7.8}+...+\frac{1}{51.52}\)C bé hơn\(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{50.52}=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{50}-\frac{1}{52}\right)\)
C bé hơn \(\frac{1}{2}\left(\frac{1}{2}-\frac{1}{52}\right)\)bé hơn\(\frac{1}{2}.\frac{1}{2}=\frac{1}{4}\)(đpcm)
xin lỗi nha mk ko biết viết kí hiệu bé hơn
a: =>|2x-4|<1
=>2x-4>-1 và 2x-4<1
=>2x>3 và 2x<5
=>3/2<x<5/2
b: |2x+1/2|>=3
=>2x+1/2>=3 hoặc 2x+1/2<=-3
=>2x>=5/2 hoặc 2x<=-7/2
=>x>=5/4 hoặc x<=-7/4
c: |4x-7|>x
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x< 0\\4x-7>0\end{matrix}\right.\\\left\{{}\begin{matrix}x>0\\\left(4x-7\right)^2-x^2>0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>0\\\left(3x-7\right)\left(5x-7\right)>0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>\dfrac{7}{3}\\0< x< \dfrac{7}{5}\end{matrix}\right.\)
\(\dfrac{1}{4\cdot7}+\dfrac{1}{7\cdot10}+...+\dfrac{1}{61\cdot64}\)
\(=\left(\dfrac{3}{4\cdot7}+\dfrac{3}{7\cdot10}+...+\dfrac{3}{61\cdot64}\right):3\)
\(=\left(\dfrac{7-4}{4\cdot7}+\dfrac{10-7}{7\cdot10}+...+\dfrac{64-61}{61\cdot64}\right):3\)
\(=\left(\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{61}-\dfrac{1}{64}\right):3\)
\(=\left(\dfrac{1}{4}-\dfrac{1}{64}\right):3\)
\(=\dfrac{15}{64}:3\)
\(=\dfrac{15}{192}=\dfrac{5}{64}\)
\(\dfrac{1}{12}=\dfrac{5}{60}\)
Vì \(64>60\) nên \(\dfrac{5}{64}< \dfrac{5}{60}\) hay \(\dfrac{1}{4\cdot7}+\dfrac{1}{7\cdot10}+...+\dfrac{1}{61\cdot64}< \dfrac{1}{12}\)