Violympic toán 7

HL

\(\dfrac{x+4}{2009}\)+\(\dfrac{x+3}{2010}\)=\(\dfrac{x+2}{2011}\)+\(\dfrac{x+1}{2012}\)

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AL
14 tháng 2 2019 lúc 16:41

\(\dfrac{x+4}{2009}+\dfrac{x+3}{2010}=\dfrac{x+2}{2011}+\dfrac{x+1}{2012}\)

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

\(\Rightarrow\dfrac{x+2013}{2009}+\dfrac{x+2013}{2010}=\dfrac{x+2013}{2011}+\dfrac{x+2013}{2012}\)
\(\Rightarrow\dfrac{x+2013}{2009}+\dfrac{x+2013}{2010}-\dfrac{x+2013}{2011}-\dfrac{x+2013}{2012}=0\)

\(\Rightarrow\left(x+2013\right)\left(\dfrac{1}{2009}+\dfrac{1}{2010}-\dfrac{1}{2011}-\dfrac{1}{2012}\right)=0\)

\(\dfrac{1}{2009}+\dfrac{1}{2010}-\dfrac{1}{2011}-\dfrac{1}{2012}\ne0\)

=> x +2013 = 0

=> x = -2013

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NT
14 tháng 2 2019 lúc 17:33

\(\dfrac{x+4}{2009}+\dfrac{x+3}{2010}=\dfrac{x+2}{2011}+\dfrac{x+1}{2012}\)

\(\Leftrightarrow\dfrac{x+4}{2009}+1+\dfrac{x+3}{2010}+1=\dfrac{x+2}{2011}+1+\dfrac{x+1}{2012}+1\)

\(\Leftrightarrow\dfrac{x+2013}{2009}+\dfrac{x+2013}{2010}=\dfrac{x+2013}{2011}+\dfrac{x+2013}{2012}\)

\(\Leftrightarrow\dfrac{x+2013}{2009}+\dfrac{x+2013}{2010}-\dfrac{x+2013}{2011}-\dfrac{x+2013}{2012}=0\)

\(\Leftrightarrow\left(x+2013\right)\left(\dfrac{1}{2009}+\dfrac{1}{2010}-\dfrac{1}{2011}-\dfrac{1}{2012}\right)=0\)

\(\Leftrightarrow x+2013=0\).Do \(\dfrac{1}{2009}+\dfrac{1}{2010}-\dfrac{1}{2011}-\dfrac{1}{2012}\ne0\)

\(\Rightarrow x+2013=0\)

\(\Leftrightarrow x=-2013\)

Bình luận (1)

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