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GC
7
2 tháng 9 2019
\(\frac{x+5}{2015}+\frac{x+6}{2014}+\frac{x+7}{2013}+\frac{x+8}{2012}+\frac{x+9}{2011}+5=0\)
\(\Rightarrow1+\frac{x+5}{2015}+1+\frac{x+6}{2014}+1+\frac{x+7}{2013}+1+\frac{x+8}{2012}+1+\frac{x+9}{2011}=0\)
\(\Rightarrow\frac{x+2020}{2015}+\frac{x+2020}{2014}+\frac{x+2020}{2013}+\frac{x+2020}{2012}+\frac{x+2020}{2011}=0\)
\(\Rightarrow\left(x+2020\right)\left(\frac{1}{2015}+\frac{1}{2014}+\frac{1}{2013}+\frac{1}{2012}+\frac{1}{2011}\right)=0\)
\(\Rightarrow x+2020=0\)
\(\Rightarrow x=-2020\)
Study well
16 tháng 9 2021
\(\dfrac{x-1}{2012}+\dfrac{x-2}{2013}+\dfrac{x-3}{2014}=\dfrac{x-4}{2015}+\dfrac{x-5}{2016}+\dfrac{x-6}{2017}\)
\(\Leftrightarrow\left(\dfrac{x-1}{2012}+1\right)+\left(\dfrac{x-2}{2013}+1\right)+\left(\dfrac{x-3}{2014}+1\right)=\left(\dfrac{x-4}{2015}+1\right)+\left(\dfrac{x-5}{2016}+1\right)+\left(\dfrac{x-6}{2017}+1\right)\)
\(\Leftrightarrow\dfrac{x+2011}{2012}+\dfrac{x+2011}{2013}+\dfrac{x+2011}{2014}-\dfrac{x+2011}{2015}-\dfrac{x+2011}{2016}-\dfrac{x+2011}{2017}=0\)
\(\Leftrightarrow\left(x+2011\right)\left(\dfrac{1}{2012}+\dfrac{1}{2013}+\dfrac{1}{2014}-\dfrac{1}{2015}-\dfrac{1}{2016}-\dfrac{1}{2017}\right)=0\)
\(\Leftrightarrow x=-2011\)( do \(\dfrac{1}{2012}+\dfrac{1}{2013}+\dfrac{1}{2014}-\dfrac{1}{2015}-\dfrac{1}{2016}-\dfrac{1}{2017}\ne0\))
\(\dfrac{x+5}{2015}+\dfrac{x+6}{2014}+\dfrac{x+7}{2013}+\dfrac{x+8}{2012}=-4\\ \Leftrightarrow\left(\dfrac{x+5}{2015}+1\right)+\left(\dfrac{x+6}{2014}+1\right)+\left(\dfrac{x+7}{2013}+1\right)+\left(\dfrac{x+8}{2012}+1\right)=0\\ \Leftrightarrow\dfrac{x+2020}{2015}+\dfrac{x+2020}{2014}+\dfrac{x+2020}{2013}+\dfrac{x+2020}{2012}=0\\ \Leftrightarrow\left(x+2020\right)\left(\dfrac{1}{2015}+\dfrac{1}{2014}+\dfrac{1}{2013}+\dfrac{1}{2012}\right)=0\\ \Leftrightarrow x+2020=0\\ \Leftrightarrow x=-2020\)
Vậy x = -2020 là nghiệm của pt.
\(\dfrac{x+5}{2015}+\dfrac{x+6}{2014}+\dfrac{x+7}{2013}+\dfrac{x+8}{2012}=-4\)
\(\Leftrightarrow x+2020=0\)
hay x=-2020