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Sửa đề \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2015}{2016}\)
\(\Leftrightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+..+\frac{2}{x\left(x+1\right)}=\frac{2015}{2016}\)
\(\Leftrightarrow\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+..+\frac{2}{x\left(x+1\right)}=\frac{2015}{2016}\)
\(\Leftrightarrow2\cdot\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2015}{2016}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2015}{2016}\div2\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2015}{4032}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2015}{4032}\Leftrightarrow\frac{1}{x+1}=\frac{1}{4032}\)
\(\Leftrightarrow x+1=4032\Rightarrow x=4031\)
#)Giải :
\(\left(\frac{2012}{2015}+\frac{2011}{2016}+\frac{2010}{2016}+\frac{2009}{2018}\right)\left(\frac{1}{6}+\frac{1}{3}+\frac{1}{2}\right)\)
\(=\left(\frac{2012}{2015}+\frac{2011}{2016}+\frac{2010}{2016}+\frac{2009}{2018}\right)\left(\frac{1}{2}+\frac{1}{2}\right)\)
\(=\left(\frac{2012}{2015}+\frac{2011}{2016}+\frac{2010}{2016}+\frac{2009}{2018}\right)\times0\)
\(=0\)
\(\left(\frac{2012}{2015}+\frac{2011}{2016}+\frac{2010}{2017}+\frac{2009}{2018}\right).\left(\frac{1}{6}+\frac{1}{3}+\frac{1}{2}\right)\)
\(=\left(\frac{2012}{2015}+\frac{2011}{2016}+\frac{2010}{2017}+\frac{2009}{2018}\right).\left(\frac{1}{6}+\frac{2}{6}+\frac{3}{6}\right)\)
=\(\left(\frac{2012}{2015}+\frac{2011}{2016}+\frac{2010}{2017}+\frac{2009}{2018}\right).0\)
\(=0\)
Câu b:
\(\frac{21}{8}:\frac{5}{6}+\frac{1}{2}:\frac{5}{6}\)
= \(\frac{63}{20}+\frac{3}{5}\)
= \(\frac{15}{4}\)
\(\left(\frac{21}{8}+\frac{1}{2}\right):\frac{5}{6}\)
\(\frac{25}{8}:\frac{5}{6}\)
\(\frac{25}{8}.\frac{6}{5}\)
\(\frac{30}{8}\)
Xét tử: \(2015+\frac{2014}{2}+\frac{2013}{3}+...+\frac{1}{2015}\)
\(=\left(1+1+...+1\right)+\frac{2014}{2}+\frac{2013}{3}+...+\frac{1}{2015}\)( trong ngoặc có 2015 số 1 )
\(=\left(1+\frac{2014}{2}\right)+\left(1+\frac{2013}{3}\right)+...+\left(1+\frac{1}{2015}\right)+1\)
\(=\frac{2016}{2}+\frac{2016}{3}+\frac{2016}{4}+...+\frac{2016}{2015}+\frac{2016}{2016}\)
\(=2016\cdot\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}\right)\)
Ghép tử và mẫu \(\frac{2016\cdot\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2016}}=2016\)
Vậy \(A=2016\)
\(1+\frac{1}{3}+\frac{1}{6}+..+\frac{2}{\left(x+1\right)\left(x+2\right)}=1\frac{2013}{2015}\)
\(\frac{2}{2}+\frac{2}{6}+\frac{2}{12}+..+\frac{2}{\left(x+1\right)\left(x+2\right)}=\frac{4028}{2015}\)
\(2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x+1}-\frac{1}{x+2}\right)=\frac{4028}{2015}\)
\(1-\frac{1}{x+2}=\frac{4028}{2015}:2\)
\(1-\frac{1}{x+2}=\frac{2014}{2015}\)
\(\frac{1}{x+2}=1-\frac{2014}{2015}\)
\(\frac{1}{x+2}=\frac{1}{2015}\)
\(\Rightarrow x+2=2015\)
\(\Rightarrow x=2013\)
\(\left(1-\frac{1}{2}\right)\times\left(1-\frac{1}{3}\right)\times\left(1-\frac{1}{4}\right)\times...\times\left(1-\frac{1}{2015}\right)\times\left(1-\frac{1}{2016}\right)\)
\(=\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\times...\times\frac{2014}{2015}\times\frac{2015}{2016}\)
\(=\frac{1}{2016}\)
Giải : Ta có (1-1/2)*(1-1/3)*(1-1/4)*....*(1-1/2015)*(1-1/2016)
= 1* -(1/2+1/3+1/4+....+1/2015+1/2016)
= 1* - (1/2+1/2016 +1/3+1/2015 +...+1/1007)
= 1* -(1/2033134)
= -1/2033134
\(\frac{1}{6}.\frac{1}{3}+\frac{17}{6}.\frac{1}{3}+\frac{2015}{2016}-1\)
\(=\frac{1}{3}\left(\frac{1}{6}+\frac{17}{6}\right)+\frac{2015}{2016}-1\)
\(=\frac{1}{3}.3+\frac{2015}{2016}-1\)
\(=1-1+\frac{2015}{2016}=\frac{2015}{2016}\)
\(\frac{1}{6}\times\frac{1}{3}+\frac{17}{6}\times\frac{1}{3}+\frac{2015}{2016}-1\)
\(=\left(\frac{1}{6}+\frac{17}{6}\right)\times\frac{1}{3}+\frac{2015}{2016}-1\)
\(=3\times\frac{1}{3}+\frac{2015}{2016}-1\)
\(=1+\frac{2015}{2016}-1\)
\(=0+\frac{2015}{2016}=\frac{2015}{2016}\)