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bit lm bài này k giup tui
\(\frac{1}{n\left(n+1\right)}=\frac{1}{n}-\frac{1}{n+1}\)
=> \(\frac{1}{x+2000}-\frac{1}{x+2001}+\frac{1}{x+2001}-\frac{1}{x+2002}+....+\frac{1}{x+2006}-\frac{1}{x+2007}=\frac{7}{8}\)
<=> \(\frac{1}{x+2000}-\frac{1}{x+2007}=\frac{7}{8}\)
<=> \(\frac{7}{\left(x+2000\right)\left(x+2007\right)}=\frac{7}{8}\Leftrightarrow\left(x+2000\right)\left(x+2007\right)=8\)
=> x = -1999 hoặc x = - 2008
\(\frac{1}{3}+\frac{1}{2.3}\left(1+2\right)+\frac{1}{3.3}\left(1+2+3\right)+...+\frac{1}{3.2015}\left(1+2+3+...+2015\right)=\frac{1}{3}\left[\frac{2}{2}+\frac{1}{2}\left(\frac{2.3}{2}\right)+\frac{1}{3}\left(\frac{3.4}{2}\right)+...+\frac{1}{2015}\left(\frac{2016.2015}{2}\right)\right]=\frac{1}{3}.\frac{1}{2}\left(2+3+4+....+2016\right)=\frac{1}{6}\left(\frac{2016.2017}{2}-1\right)\)
Ta đã biết: \(1+2+3+...+n=\frac{n.\left(n+1\right)}{2}\)
Ta có: \(A=1+\frac{1}{2}.\left(\frac{2.3}{2}\right)+\frac{1}{3}.\left(\frac{3.4}{2}\right)+...+\frac{1}{20}.\left(\frac{20.21}{2}\right)\)
\(A=1+\frac{3}{2}+\frac{4}{2}+....+\frac{21}{2}\)
\(A=\frac{1}{2}.\left(2+3+....+21\right)\)
Tổng trong ngoặc có:21-2+2=20 (số hạng)
\(=>A=\frac{1}{2}.\left(\frac{\left(21+2\right).20}{2}\right)=\frac{1}{2}.230=115\)
Vậy..........
ta có:
\(log^{\left(2a^2\right)}_2+\left(log_2^a\right)a^{log_a^{\left(log^a_1+1\right)}}+\frac{1}{2}log^2_2a^4=log_2^2+log_2^{a^2}+log_2^a\left(log^a_2+1\right)+\frac{1}{2}log^2_2a^4\)
\(=1+2log^a_2+log^a_2\left(1+log^a_2\right)+2log^2a_2\)
\(=3log^2_2a+3log^a_2+1\)
\(A=\left(1-\frac{1}{1+2}\right)\left(1-\frac{1}{1+2+3}\right)...\left(1-\frac{1}{1+2+3+...+2006}\right)\)
\(A=\left(1-\frac{1}{\frac{\left(1+2\right).2}{2}}\right)\left(1-\frac{1}{\frac{\left(1+3\right).3}{2}}\right)...\left(1-\frac{1}{\frac{\left(1+2006\right).2006}{2}}\right)\)
\(A=\frac{2}{3}.\frac{5}{6}.\frac{9}{10}...\frac{2007.2006-2}{2006.2007}=\frac{4}{6}.\frac{10}{12}.\frac{18}{20}....\frac{2007.2006-2}{2006.2007}\) (1)
xét thấy:2007.2006-2=2006.(2008-1)+2006-2008=2006.(2008-1+1)-2008=2008.(2006-1)=2008.2005 (2)
(1),(2)\(=>A=\frac{4.1}{2.3}.\frac{5.2}{3.4}.\frac{6.3}{4.5}....\frac{2008.2005}{2006.2007}\)
\(A=\frac{\left(4.5.6...2008\right)\left(1.2.3...2005\right)}{\left(2.3.4....2006\right)\left(3.4.5...2007\right)}=\frac{2008}{2006.3}=\frac{1004}{3009}\)
Vậy A=1004/3009
Theo đề
=> \(\left|2x-1\right|-\frac{1}{2}=\frac{4}{5}\) hoặc \(\left|2x-1\right|-\frac{1}{2}=-\frac{4}{5}\)
=> |2x - 1| = 13/10 hoặc |2x - 1| = -3/10 (vô lí, loại)
=> 2x - 1 = 13/10 hoặc 2x - 1 = -13/10
=> 2x = 23/10 hoặc 2x = -3/10
=> x = 23/20 hoặc x = -3/20
Vậy...
\(\left|\left|2x-1\right|-\frac{1}{2}\right|=\frac{4}{5}\)
TH1 : \(\left|2x-1\right|-\frac{1}{2}=\frac{4}{5}\Rightarrow\left|2x-1\right|=\frac{13}{10}\)
TH2 : \(\left|2x-1\right|-\frac{1}{2}=-\frac{4}{5}\Rightarrow\left|2x-1\right|=\frac{-3}{10}\) (loại )
Ta có :
\(\left|2x-1\right|=\frac{13}{10}\)
=> TH1 : \(2x-1=\frac{13}{10}\Rightarrow2x=\frac{23}{10}\Rightarrow x=\frac{23}{20}\)
TH2 : \(2x-1=\frac{-13}{10}\Rightarrow2x=\frac{-3}{10}\Rightarrow x=\frac{-3}{20}\)
Vậy x = \(\frac{23}{20}\)
hoặc x = \(\frac{-3}{20}\)
