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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
\(\frac{x+4}{2011}+\frac{x+3}{2012}=\frac{x+2}{2013}+\frac{x+1}{2014}\)
\(\frac{x+4}{2011}+1+\frac{x+3}{2012}+1=\frac{x+2}{2013}+1+\frac{x+1}{2014}+1\)
\(\frac{x+4}{2011}+\frac{2011}{2011}+\frac{x+3}{2012}+\frac{2012}{2012}=\frac{x+2}{2013}+\frac{2013}{2013}+\frac{x+1}{2014}+\frac{2014}{2014}\)
\(\frac{x+2015}{2011}+\frac{x+2015}{2012}=\frac{x+2015}{2013}+\frac{x+2015}{2014}\)
\(\frac{x+2015}{2011}+\frac{x+2015}{2012}-\frac{x+2015}{2013}-\frac{x+2015}{2014}=0\)
\(\left(x+2015\right)\left(\frac{1}{2011}+\frac{1}{2012}-\frac{1}{2013}-\frac{1}{2014}\right)=0\)
vì \(\frac{1}{2011}+\frac{1}{2012}-\frac{1}{2013}-\frac{1}{2014}\ne0\)nên:
x+2015=0
x=-2015
a)
\(2^x\left(1+2+2^2+2^3\right)=480\)
\(2^x.15=480\Rightarrow2^x=\frac{480}{15}=32=2^5\Rightarrow x=5\)
=\(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+......+\frac{2}{x\left(x+1\right)}\)
=\(2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+.....+\frac{1}{x\left(x+1\right)}\right)\)
=\(2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)\)
=\(2\left(\frac{1}{2}-\frac{1}{x+1}\right)\)
\(\frac{1}{2}-\frac{1}{x+1}\)=\(\frac{2011}{4026}\)
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\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{x\left(x+1\right)}=\frac{2011}{2012}\)
\(\Rightarrow\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{x\left(x+1\right)}=\frac{2011}{2012}\)
\(\Rightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...-\frac{1}{x+1}=\frac{2011}{2012}\)
\(\Rightarrow1-\frac{1}{x+1}=\frac{2011}{2012}\)
\(\Rightarrow\frac{1}{x+1}=1-\frac{2011}{2012}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{2012}\)
\(\Rightarrow x+1=2012\)
\(\Rightarrow x=2011\)
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{x\left(x+1\right)}=\frac{2011}{2012}\)
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}=\frac{2011}{2012}\)
\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2011}{2012}\)
\(1-\frac{1}{x+1}=\frac{2011}{2012}\)
\(\frac{1}{x+1}=1-\frac{2011}{2012}=\frac{1}{2012}\)
\(\Leftrightarrow x+1=2012\)
\(\Leftrightarrow x=2011\)
Vậy ...
P/s: Hoq chắc :<
\(\left(\frac{x+2015}{2014}-1\right)+\left(\frac{x+2015}{2013}-1\right)+\left(\frac{x+2015}{2012}-1\right)=3\left(\frac{x+2015}{2011}-1\right)\)
\(\frac{x+2015}{2014}+\frac{x+2015}{2013}+\frac{x+2015}{2012}=\frac{3\left(x+2015\right)}{2011}\)
\(\frac{x+2015}{2014}+\frac{x+2015}{2013}+\frac{x+2015}{2012}-\frac{x+2015}{2011}+\frac{x+2015}{2011}+\frac{x+2015}{2011}=0\)
\(\left(x+2015\right)\left(\frac{1}{2014}+\frac{1}{2013}+\frac{1}{2012}-\frac{1}{2011}+\frac{1}{2011}+\frac{1}{2011}\right)=0\)
\(x+2015=0\text{ Vì }\frac{1}{2014}+\frac{1}{2013}+\frac{1}{2012}-\frac{1}{2011}+\frac{1}{2011}+\frac{1}{2011}\ne0\)
\(x=-2015\)