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\(2^x+2^{x+1}+2^{x+2}+2^{x+3}=480\)
\(\Rightarrow2^x\cdot1+2^x\cdot2^1+2^x\cdot2^2+2^x\cdot2^3=480\)
\(\Rightarrow2^x\left(1+2^1+2^2+2^3\right)=480\)
\(\Rightarrow2^x\cdot15=480\)
\(\Rightarrow2^x=32\Rightarrow2^x=2^5\Rightarrow x=5\)
b) \(\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}+\frac{1}{2013}\right)x=\frac{2012}{1}+\frac{2011}{2}+...+\frac{2}{2011}+\frac{1}{2012}\)
\(\Rightarrow\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}+\frac{1}{2013}\right)x=\left(\frac{2011}{2}+1\right)+...+\left(\frac{2}{2011}+1\right)+\left(\frac{1}{2012}+1\right)+1\)
\(\Rightarrow\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}+\frac{1}{2013}\right)x=\frac{2013}{2}+...+\frac{2013}{2011}+\frac{2013}{2012}+\frac{2013}{2013}\)
\(\Rightarrow\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}+\frac{1}{2013}\right)x=2013\left(\frac{1}{2}+...+\frac{1}{2012}+\frac{1}{2013}\right)\)
\(\Rightarrow x=2013.\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}+\frac{1}{2013}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}+\frac{1}{2013}}\)
\(\Rightarrow x=2013\)
Vậy \(x=2013\)
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
MS=\(\left(1+\frac{2011}{2}\right)+\left(1+\frac{2010}{3}\right)+...+\left(1+\frac{2}{2011}\right)+\left(1+\frac{1}{2012}\right)\)
=\(2013\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}+\frac{1}{2012}\right)\)
=> \(x.\frac{1}{2013}=1\)
=>x=2013
\(D=\frac{3}{2^2}.\frac{8}{3^2}.\frac{15}{4^2}...\frac{9999}{100^2}\)
\(=\frac{1.3}{2^2}.\frac{2.4}{3^2}.\frac{3.5}{4^2}...\frac{99.101}{100^2}\)
\(=\frac{1.2...99}{2.3...100}.\frac{3.4....101}{2.3....100}=\frac{1}{100}.\frac{101}{2}=\frac{101}{200}\)
1 b) Đặt A=\(\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+...+\frac{1}{66}+\frac{1}{78}\)
=> \(\frac{A}{2}=\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{132}+\frac{1}{156}=\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{11.12}+\frac{1}{12.13}\)
\(=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}=\frac{1}{3}-\frac{1}{13}\)
=> \(A=\frac{2}{3}-\frac{2}{13}\)\(=\frac{20}{39}\)
Ta có: \(\frac{x}{6}+\frac{x}{10}+\frac{x}{15}+\frac{x}{21}+...+\frac{x}{78}=\frac{220}{39}\)
<=> \(x\left(\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+\frac{1}{15}+...+\frac{1}{78}\right)=\frac{220}{39}\Leftrightarrow x.\frac{20}{39}=\frac{220}{39}\Leftrightarrow x=11\)
a.N=1-5-9+13+17-21+...+2001-2005-2009+2013+2017
N = ( 1 - 5 - 9 + 13 ) + ( 17 - 21 - 25 + 29 ) + .... + ( 2001 - 2005 - 2009 + 2013 ) + 2017
N = 0 + 0 + ... + 0 + 2017
N = 2017
a)
\(2^x\left(1+2+2^2+2^3\right)=480\)
\(2^x.15=480\Rightarrow2^x=\frac{480}{15}=32=2^5\Rightarrow x=5\)
Chính Xác 100% là X=5
k cho mink nhé các pạn