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Cái này lớp 6 :
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+......+\frac{2}{x\left(x+1\right)}=\frac{2011}{2013}\)
\(\Leftrightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+....+\frac{2}{x\left(x+1\right)}=\frac{2011}{2013}\)
\(\Leftrightarrow\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+....+\frac{2}{x\left(x+1\right)}=\frac{2011}{2013}\)
\(\Leftrightarrow2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+....+\frac{1}{x\left(x+1\right)}\right)=\frac{2011}{2013}\)
\(\Leftrightarrow2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2011}{2013}\)
\(\Leftrightarrow2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2011}{2013}\)
<=> \(\frac{1}{2}-\frac{1}{x+1}=\frac{2011}{4026}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2011}{4026}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{2}{4026}=\frac{1}{2013}\)
\(\Leftrightarrow x+1=2013\)
=> x = 2012
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2011}{2013}\)
\(\Rightarrow2.\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2011}{2013}\)
\(\Rightarrow2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2011}{2013}\)
\(\Rightarrow2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2011}{2013}\)
\(\Rightarrow2.\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2011}{2013}\)
\(\Rightarrow1-\frac{2}{x+1}=\frac{2011}{2013}\)
\(\Rightarrow\frac{2}{x+1}=1-\frac{2011}{2013}\)
\(\Rightarrow\frac{2}{x+1}=\frac{2}{2013}\)
\(\Rightarrow x+1=2013\)
\(\Rightarrow x=2013-1\)
\(\Rightarrow x=2012\)
Vậy \(x=2012\)
~ Ủng hộ nhé
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2011}{2013}\)
\(\Leftrightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+....+\frac{2}{x\left(x+1\right)}=\frac{2011}{2013}\)
\(\Leftrightarrow\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{x\left(x+1\right)}=\frac{2011}{2013}\)
\(\Leftrightarrow2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2011}{2013}\)
\(\Leftrightarrow2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2011}{2013}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2011}{2013}\div2\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2011}{4026}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2011}{4026}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2013}\)
\(\Leftrightarrow x=2013-1=2012\)
Câu b:
\(\frac{21}{8}:\frac{5}{6}+\frac{1}{2}:\frac{5}{6}\)
= \(\frac{63}{20}+\frac{3}{5}\)
= \(\frac{15}{4}\)
\(\left(\frac{21}{8}+\frac{1}{2}\right):\frac{5}{6}\)
\(\frac{25}{8}:\frac{5}{6}\)
\(\frac{25}{8}.\frac{6}{5}\)
\(\frac{30}{8}\)
=> \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x.\left(x+1\right)}=\frac{2010}{2011}\)
=> \(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2010}{2011}\)
=>\(1-\frac{1}{x+1}=\frac{2010}{2011}\)
=> \(\frac{1}{x+1}=\frac{2011}{2011}-\frac{2010}{2011}=\frac{1}{2011}\)
=> x + 1 = 2011
=> x = 2010
\(1+\frac{1}{3}+\frac{1}{6}+..+\frac{2}{\left(x+1\right)\left(x+2\right)}=1\frac{2013}{2015}\)
\(\frac{2}{2}+\frac{2}{6}+\frac{2}{12}+..+\frac{2}{\left(x+1\right)\left(x+2\right)}=\frac{4028}{2015}\)
\(2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x+1}-\frac{1}{x+2}\right)=\frac{4028}{2015}\)
\(1-\frac{1}{x+2}=\frac{4028}{2015}:2\)
\(1-\frac{1}{x+2}=\frac{2014}{2015}\)
\(\frac{1}{x+2}=1-\frac{2014}{2015}\)
\(\frac{1}{x+2}=\frac{1}{2015}\)
\(\Rightarrow x+2=2015\)
\(\Rightarrow x=2013\)
\(\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot\left(1-\frac{1}{4}\right)\cdot...\cdot\left(1-\frac{1}{10}\right)=\frac{x}{2010}\)
\(\Leftrightarrow\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{9}{10}=\frac{x}{2010}\)
\(\Leftrightarrow\frac{1\cdot2\cdot3\cdot....\cdot9}{2\cdot3\cdot4\cdot....\cdot10}=\frac{x}{2010}\)
\(\Leftrightarrow\frac{1}{10}=\frac{x}{2010}\)
\(\Leftrightarrow x=\frac{2010}{10}=201\)
Ta có : \(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)......\left(1-\frac{1}{10}\right)=\frac{x}{2010}\)
=> \(\frac{1}{2}.\frac{2}{3}.\frac{3}{4}......\frac{9}{10}=\frac{x}{2010}\)
\(\Rightarrow\frac{1.2.3......9}{2.3.4.....10}=\frac{x}{2010}\)
\(\Rightarrow\frac{1}{10}=\frac{x}{2010}\)
\(\Rightarrow x=\frac{2010}{10}=201\)
\(=\frac{1}{10}\)
