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b./ \(\Leftrightarrow\frac{x+1}{2009}+1+\frac{x+2}{2008}+1+\frac{x+3}{2007}+1=\frac{x+10}{2000}+1+\frac{x+11}{1999}+1+\frac{x+12}{1998}+1.\)
\(\Leftrightarrow\frac{x+2010}{2009}+\frac{x+2010}{2008}+\frac{x+2010}{2007}-\frac{x+2010}{2000}-\frac{x+2010}{1999}-\frac{x+2010}{1998}=0\)
\(\Leftrightarrow\left(x+2010\right)\left(\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}-\frac{1}{2000}-\frac{1}{1999}-\frac{1}{1998}\right)=0\)(b)
Mà \(\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}-\frac{1}{2000}-\frac{1}{1999}-\frac{1}{1998}< 0\)
(b) \(\Leftrightarrow x+2010=0\Leftrightarrow x=-2010\)
a./
\(\Leftrightarrow\frac{x+1}{2}+\frac{x+1}{3}+\frac{x+1}{4}-\frac{x+1}{5}-\frac{x+1}{6}=0.\)
\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}-\frac{1}{5}-\frac{1}{6}\right)=0\)(a)
Mà \(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}-\frac{1}{5}-\frac{1}{6}>0\)
(a) \(\Leftrightarrow x+1=0\Leftrightarrow x=-1\)
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left[x+1\right]}=\frac{2007}{2009}\)
\(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left[x+1\right]}=\frac{2007}{2009}\)
\(2\left[\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{x\left[x+1\right]}\right]=\frac{2007}{2009}\)
\(2\left[\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right]=\frac{2007}{2009}\)
\(2\left[\frac{1}{2}-\frac{1}{x+1}\right]=\frac{2007}{2009}\)
\(1-\frac{2}{x+1}=\frac{2007}{2009}\)
\(\frac{2}{x+1}=1-\frac{2007}{2009}\)
\(\frac{2}{x+1}=\frac{2}{2009}\)
\(\Rightarrow x+1=2009\Leftrightarrow x=2008\)
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{\left[x\left(x+1\right)\right]:2}=\frac{1999}{2001}\)
\(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{1999}{2001}\)
\(2.\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{1999}{2001}\)
\(2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{1999}{2001}\)
\(2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{1999}{2001}\)
\(2.\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{1999}{2001}\)
\(2.\frac{1}{2}-2.\frac{1}{x+1}=\frac{1999}{2001}\)
\(1-\frac{2}{x+1}=\frac{1999}{2001}\)
\(\frac{2}{x+1}=1-\frac{1999}{2001}\)
\(\frac{2}{x+1}=\frac{2}{2001}\)
=> x + 1 = 2001
=> x = 2001 - 1
=> x = 2000
Vậy x = 2000
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{\left[x\left(x+1\right)\right]:2}=\frac{1999}{2001}\)
\(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{1999}{2001}\)
\(2.\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{1999}{2001}\)
\(2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{1999}{2001}\)\(2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{1999}{2001}\)
\(2.\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{1999}{2001}\)
\(2.\frac{1}{2}-2.\frac{1}{x+1}=\frac{1999}{2001}\)
\(1-\frac{2}{x+1}=\frac{1999}{2001}\)
\(\frac{2}{x+1}=1-\frac{1999}{2001}\)
\(\frac{2}{x+1}=\frac{2}{2001}\)
=> x + 1 = 2001
=> x = 2001 - 1
=> x = 2000
Vậy x = 2000
a) Đặt \(A=\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+.....+\frac{1}{\left(x-2\right)x}+\frac{1}{x\left(x+2\right)}\)
=> \(3A=\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+.....+\frac{3}{\left(x-2\right)x}+\frac{3}{x\left(x+2\right)}\)
=> \(3A=\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+....+\frac{1}{\left(x-2\right)}-\frac{1}{x}+\frac{1}{x}-\frac{1}{x+2}\)
=> 3A = \(\frac{1}{5}-\frac{1}{x+2}\)
=> A = \(\frac{1}{15}-\frac{1}{3x+6}\)
Mà : A = \(\frac{101}{1540}\)
=> \(\frac{1}{15}-\frac{1}{3x+6}=\frac{101}{1540}\)
=> \(\frac{1}{3x+6}=\frac{1}{15}-\frac{101}{1540}=\frac{1}{924}\)
=> 3x + 6 = 924
=> 3(x + 2) = 924
=> x + 2 = 308
=> x = 306
a) Ta có: \({{1} \over x(x+2)}= {{1} \over 3}({{1} \over x}-{{1} \over x+2})\) \(\Rightarrow\) \({{1} \over 3}({{1} \over 5}-{{1} \over 8}+{{1} \over 8}-...+{{1} \over x}-{{1} \over x+2})={{101} \over 1540} \)\(\Leftrightarrow\) \({{1} \over 3}({{1} \over 5}-{{1} \over x+2})={{101} \over 1540}\)\(\Leftrightarrow\)x+2 = 308 \(\Leftrightarrow\) x=306 Lúc sau lm hơi tắt mọi người thông cảm
=\(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+......+\frac{2}{x\left(x+1\right)}\)
=\(2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+.....+\frac{1}{x\left(x+1\right)}\right)\)
=\(2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)\)
=\(2\left(\frac{1}{2}-\frac{1}{x+1}\right)\)
\(\frac{1}{2}-\frac{1}{x+1}\)=\(\frac{2011}{4026}\)
bạn tính tiếp đi, mình bận rồi nhé, mình gợi ý hết cho bạn rồi, tự làm tiếp nhé bạn, bye
\(\Rightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{2013}{2015}\\ \Rightarrow2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2013}{2015}\\ \Rightarrow2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2013}{2015}\\ 2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2013}{2015}\\ \Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2013}{2015}:2\\ \Rightarrow-\frac{1}{x+1}=\frac{2013}{4030}-\frac{1}{2}\\ \Rightarrow-\frac{1}{x+1}=-\frac{1}{2015}\\ \Rightarrow x=2015\)
Vậy x=2015
\(\Rightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{2013}{2015}\\ \Rightarrow2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2013}{2015}\\ \Rightarrow2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2013}{2015}\\ 2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2013}{2015}\\ \Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2013}{2015}:2\\ \Rightarrow-\frac{1}{x+1}=\frac{2013}{4030}\\ \Rightarrow-\frac{1}{x+1}=-\frac{1}{2015}\\ \Rightarrow x+1=2015\\ \Rightarrow x=2014\)
Vậy x=2014
xin lỗi nhé! vừa nãy mình vội quá nên làm nhầm