Question

Answer 1

\(\dfrac{x+4}{2008}+\dfrac{x+3}{2009}=\dfrac{x+2}{2010}+\dfrac{x+1}{2011}\)

\(\Leftrightarrow x+2012=0\)

hay x=-2012

Answer 2

\(\Rightarrow\left(\dfrac{x+4}{2008}+1\right)+\left(\dfrac{x+3}{2009}+1\right)=\left(\dfrac{x+2}{2010}+1\right)+\left(\dfrac{x+1}{2011}+1\right)\\ \Rightarrow\dfrac{x+2012}{2008}+\dfrac{x+2012}{2009}=\dfrac{x+2012}{2010}+\dfrac{x+2012}{2011}\\ \Rightarrow\left(x+2012\right)\left(\dfrac{1}{2008}+\dfrac{1}{2009}-\dfrac{1}{2010}-\dfrac{1}{2011}\right)=0\\ \Rightarrow x=-2012\)

Answer 3

\(\Leftrightarrow\left(\dfrac{x+4}{2008}+1\right)+\left(\dfrac{x+3}{2009}+1\right)=\left(\dfrac{x+2}{2010}+1\right)+\left(\dfrac{x+1}{2011}+1\right)\)

\(\Leftrightarrow\dfrac{x+2012}{2008}+\dfrac{x+2012}{2009}-\dfrac{x+2012}{2010}-\dfrac{x+2012}{2011}=0\)

\(\Leftrightarrow\left(x+2012\right)\left(\dfrac{1}{2008}+\dfrac{1}{2009}-\dfrac{1}{2010}-\dfrac{1}{2011}\right)=0\)

\(\Leftrightarrow x=-2012\)( do \(\dfrac{1}{2008}+\dfrac{1}{2009}-\dfrac{1}{2010}-\dfrac{1}{2011}\ne0\))

Answer 4

\(\dfrac{x+4}{2008}+\dfrac{x+3}{2009}=\dfrac{x+2}{2010}+\dfrac{x+1}{2011}\)

\(\left(\dfrac{x+4}{2008}+1\right)+\left(\dfrac{x+3}{2009}+1\right)-\left(\dfrac{x+2}{2010}+1\right)-\left(\dfrac{x+1}{2011}+1\right)=0\)

\(\dfrac{x+2012}{2008}+\dfrac{x+2012}{2009}-\dfrac{x+2012}{2010}-\dfrac{x+2012}{2011}=0\)

\(\left(x+2012\right)\left(\dfrac{1}{2008}+\dfrac{1}{2009}-\dfrac{1}{2010}-\dfrac{1}{2011}\right)\)

⇔\(x+2012=0\)

⇔\(x=-2012\)

Answer 5

\(\dfrac{x+4}{2008}+\dfrac{x+3}{2009}=\dfrac{x+2}{2010}+\dfrac{x+1}{2011}\)

<=> \(\left(\dfrac{x+4}{2008}+1\right)+\left(\dfrac{x+3}{2009}+1\right)=\left(\dfrac{x+2}{2010}+1\right)+\left(\dfrac{x+1}{2011}+1\right)\)

<=> \(\dfrac{x+2012}{2008}+\dfrac{x+2012}{2009}-\dfrac{x+2012}{2010}-\dfrac{x+2012}{2011}=0\)

<=> \(\left(x+2012\right)\left(\dfrac{1}{2008}+\dfrac{1}{2009}-\dfrac{1}{2010}-\dfrac{1}{2011}\right)=0\)

<=> \(x+2012=0\)

<=> x = -2012