giai pt sau:|x-2013|^5+|x-2015|^7=1
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NT
1
21 tháng 3 2016
=> 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
NH
2
7 tháng 5 2017
\(\dfrac{x-5}{2012}+\dfrac{x-4}{2013}=\dfrac{x-3}{2014}+\dfrac{x-2}{2015}\)
\(\Rightarrow\left(\dfrac{x-5}{2012}-1\right)+\left(\dfrac{x-4}{2013}-1\right)=\left(\dfrac{x-3}{2014}-1\right)+\left(\dfrac{x-2}{2015}-1\right)\)
\(\Leftrightarrow\dfrac{x-2017}{2012}+\dfrac{x-2017}{2013}=\dfrac{x-2017}{2014}+\dfrac{x-2017}{2015}\)
\(\Leftrightarrow\dfrac{x-2017}{2012}+\dfrac{x-2017}{2013}-\dfrac{x-2017}{2014}-\dfrac{x-2017}{2015}=0\)
\(\Leftrightarrow\left(x-2017\right)\left(\dfrac{1}{2012}+\dfrac{1}{2013}-\dfrac{1}{2014}-\dfrac{1}{2015}\right)=0\)
\(\Rightarrow x-2017=0\Leftrightarrow x=2017\)
Vậy x = 2017
TV
2
31 tháng 8 2016
\(\frac{x-1}{2015}+\frac{x-3}{2013}=\frac{x-5}{2011}+\frac{x-7}{2009}\)
=> \(\frac{x-1}{2015}-1+\frac{x-3}{2013}-1=\frac{x-5}{2011}-1+\frac{x-7}{2009}-1\)
=> \(\frac{x-2016}{2015}+\frac{x-2016}{2013}=\frac{x-2016}{2011}+\frac{x-2016}{2009}\)
=> \(\frac{x-2016}{2009}+\frac{x-2016}{2011}-\frac{x-2016}{2013}-\frac{x-2016}{2015}=0\)
=> \(\left(x-2016\right).\left(\frac{1}{2009}+\frac{1}{2011}-\frac{1}{2013}-\frac{1}{2015}\right)\)
Vì \(\frac{1}{2009}>\frac{1}{2013};\frac{1}{2011}>\frac{1}{2015}\)
=> \(\frac{1}{2009}+\frac{1}{2011}-\frac{1}{2013}-\frac{1}{2015}\ne0\)
=> \(x-2016=0\)
=> \(x=2016\)
KT
11 tháng 2 2018
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
CC
7 tháng 6 2018
\(Pt\Leftrightarrow x^4+x^2+\dfrac{1}{4}=x^2+2013-\sqrt{x^2+2013}+\dfrac{1}{4}\\ \Leftrightarrow\left(x^2+\dfrac{1}{2}\right)^2=\left(\sqrt{x^2+2013}-\dfrac{1}{2}\right)^2\\ \Rightarrow x^2+1=\sqrt{x^2+2013}\Leftrightarrow x^4+x^2-2012=0\\ \Leftrightarrow x_{1,2}=\pm\sqrt{\dfrac{-1+\sqrt{8049}}{2}}\)