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`(x-1)/2013+(x-2)/2012+(x-3)/2011=(x-4)/2010+(x-5)/2009 +(x-6)/2008`
`<=> ((x-1)/2013-1)+((x-2)/2012-1)+((x-3)/2011-1)=( (x-4)/2010-1)+((x-5)/2009-1)+((x-6)/2008-1)`
`<=> (x-2014)/2013 +(x-2014)/2012+(x-2014)/2011=(x-2014)/2010+(x-2014)/2009+(x-2014)/2008`
`<=> x-2014=0` (Vì `1/2013+1/2012+1/2011-1/2010-1/2009-1/2008 \ne 0`)
`<=>x=2014`
Vậy `S={2014}`.
\(\dfrac{x-1}{2013}+\dfrac{x-2}{2012}+\dfrac{x-3}{2011}=\dfrac{x-4}{2010}+\dfrac{x-5}{2009}+\dfrac{x-6}{2008}\)
\(\Leftrightarrow\left(\dfrac{x-1}{2013}-1\right)+\left(\dfrac{x-2}{2012}-1\right)+\left(\dfrac{x-3}{2011}-1\right)=\left(\dfrac{x-4}{2010}-1\right)+\left(\dfrac{x-5}{2009}-1\right)+\left(\dfrac{x-6}{2008}-1\right)\)
\(\Leftrightarrow\dfrac{x-2014}{2013}+\dfrac{x-2014}{2012}+\dfrac{x-2014}{2011}=\dfrac{x-2014}{2010}+\dfrac{x-2014}{2009}+\dfrac{x-2014}{2008}\)
\(\Leftrightarrow\dfrac{x-2014}{2013}+\dfrac{x-2014}{2012}+\dfrac{x-2014}{2011}-\dfrac{x-2014}{2010}-\dfrac{x-2014}{2009}-\dfrac{x-2014}{2008}=0\)
\(\Leftrightarrow\left(x-2014\right)\left(\dfrac{1}{2013}+\dfrac{1}{2012}+\dfrac{1}{2011}-\dfrac{1}{2010}-\dfrac{1}{2009}-\dfrac{1}{2008}\right)=0\)
\(\Leftrightarrow\left(x-2014\right).A=0\)
\(\text{Vì A }\ne0\)
\(\Rightarrow x-2014=0\)
\(\Leftrightarrow x=2014\)
\(\text{Vậy phương trình có tập nghiệm là }S=\left\{2014\right\}\)
Bài của bạn nè bạn gái!
\(\dfrac{x-1}{2013}+\dfrac{x-2}{2012}+\dfrac{x-3}{2011}=\dfrac{x-4}{2010}+\dfrac{x-5}{2009}+\dfrac{x-6}{2008}\)
\(\Leftrightarrow\dfrac{x-1}{2013}-1+\dfrac{x-2}{2012}-1+\dfrac{x-3}{2011}-1=\dfrac{x-4}{2010}-1+\dfrac{x-5}{2009}-1+\dfrac{x-6}{2008}-1\)
\(\Leftrightarrow\dfrac{x-2014}{2013}+\dfrac{x-2014}{2012}+\dfrac{x-2014}{2011}-\dfrac{x-2014}{2010}-\dfrac{x-2014}{2009}-\dfrac{x-2014}{2008}=0\)
\(\Leftrightarrow\left(x-2014\right)\left(\dfrac{1}{2013}+\dfrac{1}{1012}+\dfrac{1}{2011}-\dfrac{1}{2010}-\dfrac{1}{2009}-\dfrac{1}{2008}\right)=0\)
mà \(\dfrac{1}{2013}+\dfrac{1}{2012}+\dfrac{1}{2011}-\dfrac{1}{2010}-\dfrac{1}{2009}-\dfrac{10}{2008}\ne0\)
\(\Rightarrow x-2014=0\Rightarrow x=2014\)
vậy x=2014
\(\dfrac{x-1}{2013}+\dfrac{x-2}{2012}+\dfrac{x-3}{2011}=\dfrac{x-4}{2010}+\dfrac{x-5}{2009}+\dfrac{x-6}{2008}\)
\(\Leftrightarrow\dfrac{x-1}{2013}+1+\dfrac{x-2}{2012}+1+\dfrac{x-3}{2011}+1-\dfrac{x-4}{2010}+1-\dfrac{x-5}{2009}+1-\dfrac{x-6}{2008}+1=0\)
\(\Leftrightarrow\dfrac{x-2014}{2013}+\dfrac{x-2014}{2012}+\dfrac{x-2014}{2011}-\dfrac{x-2014}{2010}-\dfrac{x-2014}{2009}-\dfrac{x-2014}{2008}=0\)
\(\Leftrightarrow\left(x-2014\right)\left(\dfrac{1}{2013}+\dfrac{1}{2012}+\dfrac{1}{2011}-\dfrac{1}{2010}-\dfrac{1}{2009}-\dfrac{1}{2008}\ne0\right)=0\)
\(\Leftrightarrow x-2014=0\)
\(\Leftrightarrow x=2014\)
Vậy PT có nghiệm là \(x=2014\)
\(\dfrac{2-x}{2010}-1=\dfrac{1-x}{2011}-\dfrac{x}{2012}\\ \Leftrightarrow\dfrac{2-x-2010}{2010}=\dfrac{2012-2012x-2011x}{2011\cdot2012}\\ \Leftrightarrow\dfrac{-2008-x}{2010}=\dfrac{2012-4023x}{4046132}\\ \Leftrightarrow\left(-2008-x\right)4046132=\left(2012-4023x\right)2010\\ \Leftrightarrow-8124633056-4046132x=4044120-8086230x\\ \Leftrightarrow-4046132x+8086230x=4044120+8124633056\\ \Leftrightarrow4040098x=8128677176\\ \Leftrightarrow x=2012\)
\(\dfrac{2-x}{2010}-1=\dfrac{1-x}{2011}-\dfrac{x}{2012}\\ \Leftrightarrow2023066\left(2-x\right)-4066362660=2022060\left(1-x\right)-2021055x\\ \Leftrightarrow4046132-2023066x-4066362660=2022060-2022060x-2021055x\\ \Leftrightarrow-4062316528-2023066x=2022060-4043115x\\ \Leftrightarrow-2023066x+4043115x=2022060+4062316528\\ \Leftrightarrow2020049x=4064338588\\ \Leftrightarrow x=2012\)
a/ Đặt \(x^2+x+1=a\Rightarrow x^2+x+2=a+1\)
Pt trở thành \(a\left(a+1\right)-12=0\Leftrightarrow a^2+a-12=0\)
\(\Leftrightarrow a^2-3a+4a-12=0\Leftrightarrow a\left(a-3\right)+4\left(a-3\right)=0\)
\(\Leftrightarrow\left(a-3\right)\left(a+4\right)=0\Leftrightarrow\left[{}\begin{matrix}a=3\\a=-4\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x^2+x+1=3\\x^2+x+1=-4\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2+x-2=0\\x^2+x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x-1\right)\left(x+2\right)=0\\\left(x+\dfrac{1}{2}\right)^2+\dfrac{19}{4}=0\left(vn\right)\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
2/ \(\dfrac{x+1}{2014}+1+\dfrac{x+2}{2013}+1=\dfrac{x+3}{2012}+1+\dfrac{x+4}{2011}+1\)
\(\Leftrightarrow\dfrac{x+2015}{2014}+\dfrac{x+2015}{2013}=\dfrac{x+2015}{2012}+\dfrac{x+2015}{2011}\)
\(\Leftrightarrow\left(x+2015\right)\left(\dfrac{1}{2014}+\dfrac{1}{2013}-\dfrac{1}{2012}-\dfrac{1}{2011}\right)=0\)
\(\Leftrightarrow x+2015=0\) (do \(\dfrac{1}{2014}+\dfrac{1}{2013}-\dfrac{1}{2012}-\dfrac{1}{2011}\ne0\))
\(\Rightarrow x=-2015\)
\(\frac{2-x}{2011}-1=\frac{1-x}{2012}-\frac{x}{2013}\)
\(\Leftrightarrow\frac{2013-x}{2011}=\frac{2013-x}{2012}+\frac{2013-x}{2013}\)
\(\Leftrightarrow\left(2013-x\right)\left(\frac{1}{2011}-\frac{1}{2012}-\frac{1}{2013}\right)=0\)
\(\Leftrightarrow2013-x=0\) (Vì \(\frac{1}{2011}-\frac{1}{2012}-\frac{1}{2013}\ne0\forall x\))
\(\Leftrightarrow x=2013\)
\(Vậy\) \(phương\) \(trình\) \(có\) \(nghiệm\) \(x=2013\).
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