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Đặt \(A=1+2+2^2+2^3+...+2^{2015}\)
\(\Rightarrow2A=2+2^2+2^3+2^4+....+2^{2015}+2^{2016}\)
Suy ra \(A=2^{2016}-1\)
Khi đó \(2^x.\left(2^{2016}-1\right)+1=2^{2016}\)
\(\Rightarrow2^x.\left(2^{2016}-1\right)=2^{2016}-1\)
\(\Rightarrow2^x=1\Rightarrow x=0\)
Vậy x=0
#Mon
\(\dfrac{x-1}{2016}+\dfrac{x-2}{2015}-\dfrac{x-3}{2014}=\dfrac{x-4}{2013}\)
\(\Leftrightarrow\dfrac{x-1}{2016}+\dfrac{x-2}{2015}=\dfrac{x-4}{2013}+\dfrac{x-3}{2014}\)
\(\Leftrightarrow\left(\dfrac{x-1}{2016}-1\right)+\left(\dfrac{x-2}{2015}-1\right)=\left(\dfrac{x-4}{2013}-1\right)+\left(\dfrac{x-3}{2014}-1\right)\)
\(\Leftrightarrow\dfrac{x-2017}{2016}+\dfrac{x-2017}{2015}=\dfrac{x-2017}{2013}+\dfrac{x-2017}{2014}\)
\(\Leftrightarrow\dfrac{x-2017}{2016}+\dfrac{x-2017}{2015}-\dfrac{x-2017}{2013}-\dfrac{x-2017}{2014}=0\)
\(\Leftrightarrow x-2017.\left(\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}-\dfrac{1}{2013}\right)=0\)
\(\text{Mà }\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}-\dfrac{1}{2103}\ne0\Rightarrow x-2017=0\)
\(\Leftrightarrow x=2017\) \(\text{Vậy }x=2017\)
\(\dfrac{x+4}{2014}+\dfrac{x+3}{2015}=\dfrac{x+2}{2016}+\dfrac{x+1}{2017}\)
\(\dfrac{x+4}{2014}+1+\dfrac{x+3}{2015}+1=\dfrac{x+2}{2016}+1+\dfrac{x+1}{2017}+1\)
\(\dfrac{x+2018}{2014}+\dfrac{x+2018}{2015}=\dfrac{x+2018}{2016}+\dfrac{x+2018}{2017}\)
\(\left(x+2018\right)\left(\dfrac{1}{2014}+\dfrac{1}{2015}-\dfrac{1}{2016}-\dfrac{1}{2017}\right)=0\\ x+2018=0\\ x=-2018\)
\(\text{Đầu bài viết khó nhìn thí mồ!! viết lại nhé!!}\)
\(\frac{x+1}{2013}+\frac{x+2}{2012}+\frac{x+3}{2011}=\frac{x-1}{2015}+\frac{x-2}{2016}+\frac{x-3}{2017}\)
\(\Rightarrow\frac{x+1}{2013}+1+\frac{x+2}{2012}+1+\frac{x+3}{2011}+1=\frac{x-1}{2015}+1+\frac{x-2}{2016}+1+\frac{x-3}{2017}+1\)
\(\Leftrightarrow\frac{x+2014}{2013}+\frac{x+2014}{2012}+\frac{x+2014}{2011}-\frac{x+2014}{2015}-\frac{x+2014}{2016}-\frac{x+2014}{2017}=0\)
\(\Leftrightarrow\left(x+2014\right)\left(\frac{1}{2013}+\frac{1}{2012}+\frac{1}{2011}-\frac{1}{2015}-\frac{1}{2016}-\frac{1}{2017}\right)=0\)
\(\text{Mà }\frac{1}{2013}+\frac{1}{2012}+\frac{1}{2011}-\frac{1}{2015}-\frac{1}{2016}-\frac{1}{2017}\ne0\)
\(\text{Nên }x+2014=0\Leftrightarrow x=-2014\)