=> 3x-(1/2013+2/2012+3/2011)=3x-(4/2010+5/2009+6/2008)=>6x=-4/2010-5/2009-6/2008+1/2013+2/2012+3/2011                                                                                       =>x=...                                                                          làm tiếp đi bạn

\(\dfrac{x-5}{2012}+\dfrac{x-4}{2013}=\dfrac{x-3}{2014}+\dfrac{x-2}{2015}\)

\(\Rightarrow\left(\dfrac{x-5}{2012}-1\right)+\left(\dfrac{x-4}{2013}-1\right)=\left(\dfrac{x-3}{2014}-1\right)+\left(\dfrac{x-2}{2015}-1\right)\)

\(\Leftrightarrow\dfrac{x-2017}{2012}+\dfrac{x-2017}{2013}=\dfrac{x-2017}{2014}+\dfrac{x-2017}{2015}\)

\(\Leftrightarrow\dfrac{x-2017}{2012}+\dfrac{x-2017}{2013}-\dfrac{x-2017}{2014}-\dfrac{x-2017}{2015}=0\)

\(\Leftrightarrow\left(x-2017\right)\left(\dfrac{1}{2012}+\dfrac{1}{2013}-\dfrac{1}{2014}-\dfrac{1}{2015}\right)=0\)

\(\Rightarrow x-2017=0\Leftrightarrow x=2017\)

Vậy x = 2017

cảm ơn nhé

Tìm x nha mấy bạn :<

\(\frac{x-1}{2015}+\frac{x-3}{2013}=\frac{x-5}{2011}+\frac{x-7}{2009}\)

=> \(\frac{x-1}{2015}-1+\frac{x-3}{2013}-1=\frac{x-5}{2011}-1+\frac{x-7}{2009}-1\)

=> \(\frac{x-2016}{2015}+\frac{x-2016}{2013}=\frac{x-2016}{2011}+\frac{x-2016}{2009}\)

=> \(\frac{x-2016}{2009}+\frac{x-2016}{2011}-\frac{x-2016}{2013}-\frac{x-2016}{2015}=0\)

=> \(\left(x-2016\right).\left(\frac{1}{2009}+\frac{1}{2011}-\frac{1}{2013}-\frac{1}{2015}\right)\)

Vì \(\frac{1}{2009}>\frac{1}{2013};\frac{1}{2011}>\frac{1}{2015}\)

=> \(\frac{1}{2009}+\frac{1}{2011}-\frac{1}{2013}-\frac{1}{2015}\ne0\)

=> \(x-2016=0\)

=> \(x=2016\)

Giúp mình nhé các bạn mình đang cần gấp lắm

=0

b)       \(x\left(x+1\right)\left(x+2\right)\left(x+3\right)=24\)

\(\Leftrightarrow\)\(\left(x^2+3x\right)\left(x^2+3x+2\right)-24=0\)

Đặt   \(x^2+3x=t\)   ta có:

          \(t\left(t+2\right)-24=0\)

\(\Leftrightarrow\)\(t^2+2t-24=0\)

\(\Leftrightarrow\)\(\left(1-4\right)\left(1+6\right)=0\)

đến đây bn giải tiếp

\(Pt\Leftrightarrow x^4+x^2+\dfrac{1}{4}=x^2+2013-\sqrt{x^2+2013}+\dfrac{1}{4}\\ \Leftrightarrow\left(x^2+\dfrac{1}{2}\right)^2=\left(\sqrt{x^2+2013}-\dfrac{1}{2}\right)^2\\ \Rightarrow x^2+1=\sqrt{x^2+2013}\Leftrightarrow x^4+x^2-2012=0\\ \Leftrightarrow x_{1,2}=\pm\sqrt{\dfrac{-1+\sqrt{8049}}{2}}\)