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Bài 1
\(a,\left|x\right|=-\left|-\frac{5}{7}\right|=>x\in\varnothing\)
\(b,\left|x+4,3\right|-\left|-2,8\right|=0\)
\(=>\left|x+4,3\right|-2,8=0\)
\(=>\left|x+4,3\right|=0+2,8=2,8\)
\(=>x+4,3=\pm2,8\)
\(=>\hept{\begin{cases}x+4,3=2,8\\x+4,3=-2,8\end{cases}=>\hept{\begin{cases}x=-1,5\\x=-7,1\end{cases}}}\)
\(c,\left|x\right|+x=\frac{2}{3}\)
\(=>\hept{\begin{cases}x+x=\frac{2}{3}\\-x+x=\frac{2}{3}\end{cases}}=>\hept{\begin{cases}x=\frac{1}{3}\\x=-\frac{1}{3}\end{cases}}\)
\(A=\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{2012}}\)
\(\frac{1}{5}A=\frac{1}{5}\left(\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{2012}}\right)=\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{2013}}\)
\(A-\frac{1}{5}A=\frac{4}{5}A=\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{2012}}-\left(\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{2013}}\right)\)
\(\frac{4}{5}A=\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{2012}}-\frac{1}{5^2}-\frac{1}{5^3}-...-\frac{1}{5^{2013}}\)
\(\frac{4}{5}A=\frac{1}{5}-\frac{1}{5^{2013}}=\frac{5^{2012}-1}{5^{2013}}\)
\(A=\frac{5^{2012}-1}{5^{2013}}:\frac{4}{5}=\frac{5^{2012}-1}{5^{2013}}\times\frac{5}{4}=\frac{5^{2012}-1}{4.5^{2012}}=\frac{1}{4}-\frac{1}{4.5^{2012}}< \frac{1}{4}\)
a, \(A=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}}{\frac{2011}{1}+\frac{2010}{2}+\frac{2009}{3}+...+\frac{1}{2011}}\)
\(A=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}}{\left(\frac{2011}{1}+1\right)+\left(\frac{2010}{2}+1\right)+\left(\frac{2009}{3}+1\right)+...+\left(\frac{1}{2011}+1\right)+1}\)
\(A=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}}{\frac{2012}{1}+\frac{2012}{2}+\frac{2012}{3}+...+\frac{2012}{2011}+\frac{2012}{2012}}\)
\(A=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}}{2012\cdot\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}+\frac{1}{2012}\right)}=\frac{1}{2012}\)
b, \(\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+....+\frac{1}{2016}+\frac{1}{2017}}{\frac{2016}{1}+\frac{2015}{2}+\frac{2014}{3}+...+\frac{2}{2015}+\frac{1}{2016}}\)
\(\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}+\frac{1}{2017}}{\left(\frac{2016}{1}+1\right)+\left(\frac{2015}{2}+1\right)+\left(\frac{2014}{3}+1\right)+...+\left(\frac{2}{2015}+1\right)+\left(\frac{1}{2016}+1\right)+1}\)
\(\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}}{\frac{2017}{1}+\frac{2017}{2}+\frac{2017}{3}+...+\frac{2017}{2015}+\frac{2017}{2016}+\frac{2017}{2017}}\)
\(\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}}{2017\cdot\left(\frac{1}{2}+\frac{1}{3}+....+\frac{1}{2015}+\frac{1}{2016}+\frac{1}{2017}\right)}=\frac{1}{2017}\)
\(B=\frac{2001}{1}+\frac{2010}{2}+\frac{2009}{3}+...+\frac{2}{2010}+\frac{1}{2001}\)
\(B=\left(2011-1-...-1\right)+\left(\frac{2010}{2}+1\right)+\left(\frac{2009}{3}+1\right)+...+\left(\frac{1}{2011}+1\right)\)
\(B=\frac{2012}{2}+\frac{2012}{3}+...+\frac{2012}{2011}+\frac{2012}{2012}\)
\(B=2012\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2011}+\frac{1}{2012}\right)\)
\(\Rightarrow\)\(\frac{B}{A}=\frac{2012\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2011}+\frac{1}{2012}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2011}+\frac{1}{2012}}=2012\)
Vậy \(\frac{B}{A}=2012\)
Chúc bạn học tốt ~
\(\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)\cdot\cdot\cdot\cdot\cdot\left(\frac{1}{2011}-1\right)\left(\frac{1}{2012}-1\right)\)
\(=\frac{-1}{2}\cdot\frac{-2}{3}\cdot\cdot\cdot\cdot\cdot\frac{-2010}{2011}\cdot\frac{-2011}{2012}\)
\(=-\left(\frac{1\cdot2\cdot3\cdot\cdot\cdot\cdot\cdot2010\cdot2011}{2\cdot3\cdot\cdot\cdot\cdot\cdot2011\cdot2012}\right)=-\frac{1}{2012}\)
= -1/2 . ( -2/3) .........( -2010/2011) ( -2011/2012)
= - 1.2.....2010.2011 / 2.3..... 2011.2012
= - 1/2012