Question

(x+1)/2009+ (x+2)/2008= (x+2007)/3 + (x+2006)/4

Answer 1

\(\dfrac{x+1}{2009}+\dfrac{x+2}{2008}=\dfrac{x+2007}{3}+\dfrac{x+2006}{4}\)

<=>\(\dfrac{x+1}{2009}+1+\dfrac{x+2}{2008}+1=\dfrac{x+2007}{3}+1+\dfrac{x+2006}{4}+1\)

<=>\(\dfrac{x+2010}{2009}+\dfrac{x+2010}{2008}=\dfrac{x+2010}{3}+\dfrac{x+2010}{4}\)

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

vì 1/2009+1/2008-1/3-1/4=0

=>x+2010=0

=>x=-2010

Answer 2

Giải:

\(\dfrac{x+1}{2009}+\dfrac{x+2}{2008}=\dfrac{x+2007}{3}+\dfrac{x+2006}{4}\)

\(\Leftrightarrow\dfrac{x+1}{2009}+1+\dfrac{x+2}{2008}+1=\dfrac{x+2007}{3}+1+\dfrac{x+2006}{4}+1\)

\(\Leftrightarrow\dfrac{x+1+2009}{2009}+\dfrac{x+2+2008}{2008}=\dfrac{x+2007+3}{3}+\dfrac{x+2006+4}{4}\)

\(\Leftrightarrow\dfrac{x+2010}{2009}+\dfrac{x+2010}{2008}=\dfrac{x+2010}{3}+\dfrac{x+2010}{4}\)

\(\Leftrightarrow\dfrac{x+2010}{2009}+\dfrac{x+2010}{2008}-\dfrac{x+2010}{3}-\dfrac{x+2010}{4}=0\)

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

Vì \(\Leftrightarrow\dfrac{1}{2009}+\dfrac{1}{2008}-\dfrac{1}{3}-\dfrac{1}{4}\ne0\)

Nên \(x+2010=0\)

\(\Leftrightarrow x=-2010\)

Vậy ...