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<=> 1+\(\frac{1}{2014}\)+\(\frac{1}{x}\)=\(\frac{1}{x+1}\)+1+\(\frac{1}{2013}\)
<=> \(\frac{1}{2014}\)+\(\frac{1}{x}\)=\(\frac{1}{x+1}\)+\(\frac{1}{2013}\)
<=> \(\frac{1}{x}\)-\(\frac{1}{x+1}\)=\(\frac{1}{2013}\)-\(\frac{1}{2013+1}\) => x=2013
\(\frac{2015}{2014}+\frac{1}{x}=\frac{1}{x+1}+\frac{2014}{2013}\)
\(\Leftrightarrow\frac{2015}{2014}-1+\frac{1}{x}=\frac{1}{x+1}+\frac{2014}{2013}-1\)
\(\Leftrightarrow\frac{1}{2014}+\frac{1}{x}=\frac{1}{x+1}+\frac{1}{2013}\)
\(\Leftrightarrow\frac{x+2014}{2014x}=\frac{x+2014}{2013\left(x+1\right)}\)
\(\Leftrightarrow2014x=2013x+2013\)
\(\Leftrightarrow x=2013\)
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\(A=\frac{1+2+...+2014}{2015}\)
\(A=\frac{\frac{2014\cdot2015}{2}}{2015}\)
\(A=\frac{1007\cdot2015}{2015}\)
\(A=1007\)
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= 1 + 2 + 3 + ... + 2014( 1007 số hạng) / 2015 = ( 2014 + 1 ) . 1007 / 2015 = 2015 . 1007 / 2015 = 1007
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a) Ta có \(\frac{2013}{2014}=1-\frac{1}{2014}\)và \(\frac{2014}{2015}=1-\frac{1}{2015}\)
mà \(\frac{1}{2014}>\frac{1}{2015}\)nên \(\frac{2013}{2014}< \frac{2014}{2015}\)