\(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.......+\frac{1}{205}}{\frac{204}{1}+\frac{203}{2}+\frac{202}{3}+....+\frac{1}{204}}\)
Tính nhanh B=\(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{205}}{\frac{204}{1}+\frac{203}{2}+\frac{202}{3}+...+\frac{1}{204}}\)
\(A=\frac{1}{3^2}-\frac{1}{3^4}+\frac{1}{3^6}-\frac{1}{3^8}+...+\frac{1}{3^{202}}-\frac{1}{3^{204}}\)
\(A=\frac{1}{3^2}-\frac{1}{3^4}+\frac{1}{3^6}-\frac{1}{3^8}+...+\frac{1}{3^{202}}-\frac{1}{3^{204}}\)
\(9A=\frac{1}{3}-\frac{1}{3^2}+\frac{1}{3^4}-\frac{1}{3^6}+...+\frac{1}{3^{200}}-\frac{1}{3^{202}}\)
\(9A+A=\left(\frac{1}{3}-\frac{1}{3^{^2}}+...+\frac{1}{3^{200}}-\frac{1}{3^{202}}\right)+\left(\frac{1}{3^2}-\frac{1}{3^4}+...+\frac{1}{3^{202}}-\frac{1}{3^{204}}\right)\)
\(10A=\frac{1}{3}-\frac{1}{3^{204}}\)
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 :\(A-\frac{1}{3^2}A=\frac{8}{9}A=\frac{1}{3^2}-\frac{1}{3^{206}}\)
\(\Rightarrow A=\left(\frac{1}{3^2}-\frac{1}{3^{206}}\right):\frac{8}{9}\)
\(A=\frac{1}{3^2}-\frac{1}{3^4}+\frac{1}{3^6}+\frac{1}{3^8}+...+\frac{1}{3^{202}}-\frac{1}{3^{204}}\)
\(\frac{1}{3^2}A=\frac{1}{3^2}\left(\frac{1}{3^2}-\frac{1}{3^4}+\frac{1}{3^6}-\frac{1}{3^8}+...+\frac{1}{3^{202}}-\frac{1}{3^{204}}\right)\)
\(\frac{1}{3^2}A=\frac{1}{3^4}-\frac{1}{3^6}+\frac{1}{3^8}-\frac{1}{3^{10}}+...+\frac{1}{3^{204}}-\frac{1}{3^{206}}\)
Do đó , \(A-\frac{1}{3^2}A=\frac{8}{9}A=\frac{1}{3^2}-\frac{1}{3^{206}}\)
\(\Rightarrow A=\left(\frac{1}{3^2}-\frac{1}{3^{205}}\right)\div\frac{8}{9}\)
M = \(\frac{3^{205}+28}{3^{203}+2};\) N = \(\frac{3^{204}+19}{3^{202}+1}\)
So sanh M và N
Bài \(1:\)TÌM \(x:\)
\(a,\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x.\left(x+1\right)}=\frac{2}{2013}\)
\(b,\frac{x+4}{2000}+\frac{x+3}{2001}=\frac{x+2}{2002}+\frac{x+1}{2003}\)
\(c,\frac{x+5}{205}+\frac{x+4}{204}+\frac{x+3}{203}=\frac{x+166}{366}+\frac{x+167}{367}+\frac{x+168}{368}\)
\(d,\) \(x.\)\(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2011}+\frac{1}{2012}}{\frac{2011}{1}+\frac{2011}{2}+\frac{2010}{3}+\frac{2009}{4}+...+\frac{2}{2011}+\frac{1}{2012}}=1\)
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
\(\frac{10+\frac{9}{2}+\frac{8}{3}+\frac{7}{4}+ \frac{6}{5}+\frac{5}{6}+\frac{4}{7}+\frac{3}{8}+\frac{2}{9}+\frac{1}{10}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{10}+\frac{1}{11}}\)
\(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{2011}}{\frac{2010}{1}+\frac{2009}{2}+\frac{2008}{3}+....+\frac{1}{2010}}\)
\(A=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}}{\frac{2010}{1}+\frac{2009}{2}+...+\frac{1}{2010}}\)
\(\Rightarrow A=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}}{\left(1+1+1+...+1\right)+\frac{2009}{2}+\frac{2008}{3}+...+\frac{1}{2010}}\)
\(\Rightarrow A=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}}{1+\left(1+\frac{2009}{2}\right)+\left(1+\frac{2008}{3}\right)+...+\left(1+\frac{1}{2010}\right)}\)
\(\Rightarrow A=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}}{\frac{2011}{2}+\frac{2011}{3}+...+\frac{2011}{2010}+\frac{2011}{2011}}\)
\(\Rightarrow A=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}}{2011.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}\right)}\)
\(\Rightarrow A=\frac{1}{2011}\)
\(\frac{\frac{1}{2}+\frac{1}{3}-\frac{1}{4}}{\left(\frac{1}{2}+\frac{1}{3}-\frac{1}{4}\right)-\frac{1}{2}.\frac{1}{3}.\frac{1}{4}}\)+\(\frac{\frac{1}{9}-\frac{1}{7}-\frac{1}{11}}{\frac{4}{9}-\frac{4}{7}-\frac{4}{11}}.\frac{1}{4}\)
rút gọn
\(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{2011}}{\frac{2010}{1}+\frac{2009}{2}+\frac{2008}{3}+....+\frac{1}{2010}}\)
giup minh voi
\(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2011}}{\frac{2010}{1}+\frac{2009}{2}+\frac{2008}{3}+...+\frac{1}{2010}}\)
\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2011}}{\left(\frac{2009}{2}+1\right)+\left(\frac{2008}{3}+1\right)+....+\left(\frac{1}{2010}+1\right)}\)
\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2011}}{\frac{2011}{2}+\frac{2011}{3}+....+\frac{2011}{2010}}\)
\(=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}}{2011\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}\right)}\)
\(=\frac{1}{2011}\)
tìm x, x\(\in\) Q:
\(\frac{\left(x+\frac{3}{4}\right).\frac{7}{2}-\frac{1}{6}}{-\left(\frac{4}{5}+\frac{1}{3}\right).\frac{1}{2}+1}=2\frac{33}{52}\)
\(\frac{\left(5-\frac{2}{7}\right).\frac{7}{9}:\frac{3}{5}}{\left(3x-\frac{5}{6}\right):\frac{1}{7}}=5\frac{5}{21}\)