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a)
\(2^x\left(1+2+2^2+2^3\right)=480\)
\(2^x.15=480\Rightarrow2^x=\frac{480}{15}=32=2^5\Rightarrow x=5\)
\(a,\frac{x+2}{2010}+\frac{x+2}{2011}+\frac{x+2}{2012}=\frac{x+2}{2013}+\frac{x+2}{2014}\)
\(\Leftrightarrow\frac{x+2}{2010}+\frac{x+2}{2011}+\frac{x+2}{2012}-\frac{x+2}{2013}-\frac{x+2}{2014}=0\)
\(\Leftrightarrow\left(x+2\right)\left(\frac{1}{2010}+\frac{1}{2011}+\frac{1}{2012}-\frac{1}{2013}-\frac{1}{2014}\right)=0\)
\(\text{Mà }\frac{1}{2010}+\frac{1}{2011}+\frac{1}{2012}-\frac{1}{2013}-\frac{1}{2014}\ne0\text{ nên:}\)
\(\Leftrightarrow x+2=0\)
\(\Leftrightarrow x=-2\)
\(b,\frac{x+4}{2000}+\frac{x+3}{2001}=\frac{x+2}{2002}+\frac{x+1}{2003}\)
\(\Leftrightarrow\frac{x+4}{2000}+1+\frac{x+3}{2001}+1=\frac{x+2}{2002}+1+\frac{x+1}{2003}+1\)
\(\Leftrightarrow \frac{x+2004}{2000}+\frac{x+2004}{2001}-\frac{x+2004}{2002}-\frac{x+2004}{2003}=0\)
\(\Leftrightarrow\left(x+2004\right)\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)=0\)
\(M\text{à}:\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\ne0 n\text{ê}n:\)
\(x+2004=0\)
\(\Leftrightarrow x=-2004\)
có 2014/1+2013/2+2012/3+...+2/2013+1/2014=[1+(2013/2)]+[1+(2012/3)]+...+[1+(2/2013)]+[1+(1/2014)]+1
=2015/2+2015/3+...+2015/2014+2015/2015=2015.[1/2+1/3+..+1/2015)
vậy (1/2+1/3+...+1/2015).x=(1/2+1/3+...+1/2015).2015
x=2015
\(\frac{x+5}{2012}+\frac{x+4}{2013}=\frac{x+3}{2014}+\frac{x+2}{2015}\)
\(\Leftrightarrow\frac{x+5}{2012}+1+\frac{x+4}{2013}+1=\frac{x+3}{2014}+1+\frac{x+2}{2015}+1\)
\(\frac{x+5+2012}{2012}+\frac{x+4+2013}{2013}=\frac{x+3+2014}{2014}+\frac{x+2+2015}{2015}\)
\(\frac{x+2017}{2012}+\frac{x+2017}{2013}=\frac{x+2017}{2014}+\frac{x+2017}{2015}\)
\(\frac{x+2017}{2012}+\frac{x+2017}{2013}-\frac{x+2017}{2014}-\frac{x+2017}{2015}=0\)
\(\left(x+2017\right)\left(\frac{1}{2012}+\frac{1}{2013}-\frac{1}{2014}-\frac{1}{2015}\right)=0\)
Mà \(\frac{1}{2012}+\frac{1}{2013}-\frac{1}{2014}-\frac{1}{2015}>0\)
\(\Rightarrow x+2017=0\)
\(\Rightarrow x=-2017\)
\(\frac{x+5}{2012}+1+\frac{x+4}{2013}+1=\frac{x+3}{2014}+1+\frac{x+2}{2015}+1\)
\(\frac{x+2017}{2012}+\frac{x+2017}{2013}-\frac{x+2017}{2014}-\frac{x+2017}{2015}=0\)
\(\left(x+2017\right)\cdot\left(\frac{1}{2012}+\frac{1}{2013}-\frac{1}{2014}-\frac{1}{2015}\right)\)
Vì \(\left(\frac{1}{2012}+\frac{1}{2013}-\frac{1}{2014}-\frac{1}{2015}\right)\ne0\)
suy ra \(x+2017=0\)
suy ra \(x=-2017\)
Vậy \(x=-2017\)
\(\frac{x+4}{2011}+\frac{x+3}{2012}=\frac{x+2}{2013}+\frac{x+1}{2014}\)
\(\frac{x+4}{2011}+1+\frac{x+3}{2012}+1=\frac{x+2}{2013}+1+\frac{x+1}{2014}+1\)
\(\frac{x+4}{2011}+\frac{2011}{2011}+\frac{x+3}{2012}+\frac{2012}{2012}=\frac{x+2}{2013}+\frac{2013}{2013}+\frac{x+1}{2014}+\frac{2014}{2014}\)
\(\frac{x+2015}{2011}+\frac{x+2015}{2012}=\frac{x+2015}{2013}+\frac{x+2015}{2014}\)
\(\frac{x+2015}{2011}+\frac{x+2015}{2012}-\frac{x+2015}{2013}-\frac{x+2015}{2014}=0\)
\(\left(x+2015\right)\left(\frac{1}{2011}+\frac{1}{2012}-\frac{1}{2013}-\frac{1}{2014}\right)=0\)
vì \(\frac{1}{2011}+\frac{1}{2012}-\frac{1}{2013}-\frac{1}{2014}\ne0\)nên:
x+2015=0
x=-2015