Question

tìm x ( x>0)

\(\frac{2015}{2014}+\frac{1}{x}=\frac{1}{x+1}+\frac{2014}{2013}\)

Answer 1

<=>  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

Answer 2

\(\frac{2015}{2014}+\frac{1}{x}=\frac{1}{x+1}+\frac{2014}{2013}\)

\(\Leftrightarrow\frac{2015}{2014}-1+\frac{1}{x}=\frac{1}{x+1}+\frac{2014}{2013}-1\)

\(\Leftrightarrow\frac{1}{2014}+\frac{1}{x}=\frac{1}{x+1}+\frac{1}{2013}\)

\(\Leftrightarrow\frac{x+2014}{2014x}=\frac{x+2014}{2013\left(x+1\right)}\)

\(\Leftrightarrow2014x=2013x+2013\)

\(\Leftrightarrow x=2013\)