\(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.......+\frac{1}{205}}{\frac{204}{1}+\frac{203}{2}+\frac{202}{3}+....+\frac{1}{204}}\)
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\frac{1}{3^2}-\frac{1}{3^4}+\frac{1}{3^6}-\frac{1}{3^8}+...+\frac{1}{3^{202}}-\frac{1}{3^{204}}\)
\(9A=\frac{1}{3}-\frac{1}{3^2}+\frac{1}{3^4}-\frac{1}{3^6}+...+\frac{1}{3^{200}}-\frac{1}{3^{202}}\)
\(9A+A=\left(\frac{1}{3}-\frac{1}{3^{^2}}+...+\frac{1}{3^{200}}-\frac{1}{3^{202}}\right)+\left(\frac{1}{3^2}-\frac{1}{3^4}+...+\frac{1}{3^{202}}-\frac{1}{3^{204}}\right)\)
\(10A=\frac{1}{3}-\frac{1}{3^{204}}\)
A = (1/3 - 1/3204) : 10
Vậy A = (1/3 - 1/3204) : 10.
A= \(\frac{1}{3^2}-\frac{1}{3^4}+\frac{1}{3^6}-\frac{1}{3^8}+...+\frac{1}{3^{202}}-\frac{1}{3^{204}}\left(1\right)\\ \)
\(\frac{1}{3^2}A=\frac{1}{3^2}\left(\frac{1}{3^2}-\frac{1}{3^4}+\frac{1}{3^6}-\frac{1}{3^8}+...+\frac{1}{3^{202}}-\frac{1}{3^{204}}\right)\)
\(\frac{1}{3^2}A=\frac{1}{3^4}-\frac{1}{3^6}+\frac{1}{3^8}-\frac{1}{3^{10}}+...+\frac{1}{3^{204}}-\frac{1}{3^{206}}\left(2\right)\)
Từ (1) và (2) vế theo vế ta có :\(A-\frac{1}{3^2}A=\frac{8}{9}A=\frac{1}{3^2}-\frac{1}{3^{206}}\)
\(\Rightarrow A=\left(\frac{1}{3^2}-\frac{1}{3^{206}}\right):\frac{8}{9}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a)\(\frac{2}{6}+\frac{2}{12}+...+\frac{2}{x\left(x+1\right)}=\frac{2}{2013}\)
\(\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{x\left(x+1\right)}=\frac{2}{2013}\)
\(2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2}{2013}\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{1}{2013}\)
đề sai
b)\(\frac{x+4}{2000}+1+\frac{x+3}{2001}+1=\frac{x+2}{2002}+1+\frac{x+1}{2003}+1\)
\(\frac{x+2004}{2000}+\frac{x+2004}{2001}=\frac{x+2004}{2002}+\frac{x+2004}{2003}\)
\(\frac{x+2004}{2000}+\frac{x+2004}{2001}-\frac{x+2004}{2002}-\frac{x+2004}{2003}=0\)
\(\left(x+2004\right)\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)=0\)
\(x+2004=0\).Do \(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\ne0\)
\(x=-2004\)
c)\(\frac{x+5}{205}-1+\frac{x+4}{204}-1+\frac{x+3}{203}-1=\frac{x+166}{366}-1+\frac{x+167}{367}-1+\frac{x+168}{368}-1\)
\(\frac{x-200}{205}+\frac{x-200}{204}+\frac{x-200}{203}=\frac{x-200}{366}+\frac{x-200}{367}+\frac{x-200}{368}\)
\(\frac{x-200}{205}+\frac{x-200}{204}+\frac{x-200}{203}-\frac{x-200}{366}-\frac{x-200}{367}-\frac{x-200}{368}=0\)
\(\left(x-200\right)\left(\frac{1}{205}+\frac{1}{204}+\frac{1}{203}-\frac{1}{366}-\frac{1}{367}-\frac{1}{368}\right)=0\)
\(x-200=0\).Do\(\frac{1}{205}+\frac{1}{204}+\frac{1}{203}-\frac{1}{366}-\frac{1}{367}-\frac{1}{368}\ne0\)
\(x=200\)
d)chịu
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}}{\frac{2010}{1}+\frac{2009}{2}+...+\frac{1}{2010}}\)
\(\Rightarrow A=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}}{\left(1+1+1+...+1\right)+\frac{2009}{2}+\frac{2008}{3}+...+\frac{1}{2010}}\)
\(\Rightarrow A=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}}{1+\left(1+\frac{2009}{2}\right)+\left(1+\frac{2008}{3}\right)+...+\left(1+\frac{1}{2010}\right)}\)
\(\Rightarrow A=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}}{\frac{2011}{2}+\frac{2011}{3}+...+\frac{2011}{2010}+\frac{2011}{2011}}\)
\(\Rightarrow A=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}}{2011.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}\right)}\)
\(\Rightarrow A=\frac{1}{2011}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2011}}{\frac{2010}{1}+\frac{2009}{2}+\frac{2008}{3}+...+\frac{1}{2010}}\)
\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2011}}{\left(\frac{2009}{2}+1\right)+\left(\frac{2008}{3}+1\right)+....+\left(\frac{1}{2010}+1\right)}\)
\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2011}}{\frac{2011}{2}+\frac{2011}{3}+....+\frac{2011}{2010}}\)
\(=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}}{2011\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}\right)}\)
\(=\frac{1}{2011}\)
\(\)Đặt \(A=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}...+\frac{1}{205}}{\frac{204}{1}+\frac{203}{2}+\frac{202}{3}+...+\frac{1}{204}}=\frac{B}{C}\)
Biến đổi C:
\(C=\left(\frac{204}{1}+1\right)+\left(\frac{203}{2}+1\right)+\left(\frac{202}{3}+1\right)+...+\left(\frac{1}{204}+1\right)-204\)
\(=205+\frac{205}{2}+\frac{205}{3}+..+\frac{205}{204}+\frac{205}{205}-205\)
\(=205.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{205}\right)\)
\(\Rightarrow A=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{205}}{205.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{205}\right)}=\frac{1}{205}\)