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\(\dfrac{x-1}{2012}+\dfrac{x-2}{2011}+\dfrac{x-3}{2010}+...+\dfrac{x-2012}{1}=2012\)
<=>\(\dfrac{x-1}{2012}-1+\dfrac{x-2}{2011}-1+\dfrac{x-3}{2010}-1+...+\dfrac{x-2012}{1}-1=0\)
<=>\(\dfrac{x-2013}{2012}+\dfrac{x-2013}{2011}+\dfrac{x-2013}{2010}+...+\dfrac{x-2013}{1}=0\)
<=>\(\left(x-2013\right)\left(\dfrac{1}{2012}+\dfrac{1}{2011}+...+1\right)=0\)
do 1/2012+1/2011....+1 khác 0 =>x-2013=0<=>x=2013
vậy..........................
\(\dfrac{x-1}{2012}+\dfrac{x-2}{2011}+\dfrac{x-3}{2010}+...+\dfrac{x-2012}{1}=2012\)
\(\left(\dfrac{x-1}{2012}+\dfrac{x-2}{2011}+\dfrac{x-3}{2010}+...+\dfrac{x-2012}{1}\right)-2012=0\)
\(\Rightarrow\dfrac{x-2013}{2012}+\dfrac{x-2013}{2011}+\dfrac{x-2013}{2010}+...+\dfrac{x-2013}{1}=0\)
\(\Rightarrow x-2013\left(\dfrac{1}{2012}+\dfrac{1}{2011}+\dfrac{1}{2010}+...+\dfrac{1}{1}\right)=0\)
Vì \(x-2013\left(\dfrac{1}{2012}+\dfrac{1}{2011}+\dfrac{1}{2010}+...+\dfrac{1}{1}\right)=0\)nên x - 2013 hoặc \(\dfrac{1}{2012}+\dfrac{1}{2011}+\dfrac{1}{2010}+...+\dfrac{1}{1}\) = 0. Nhưng \(\dfrac{1}{2012}+\dfrac{1}{2011}+\dfrac{1}{2010}+...+\dfrac{1}{1}\ne0\) nên x - 2013 = 0. Vì vậy x = 2013.
Vậy...
\(\dfrac{x-1}{2012}+\dfrac{x-2}{2011}+\dfrac{x-3}{2010}+...+\dfrac{x-2012}{1}=2012\)
\(\Leftrightarrow\dfrac{x-1}{2012}-1+\dfrac{x-2}{2011}-1+...+\dfrac{x-2012}{1}-1=0\)
\(\Leftrightarrow\dfrac{x-2013}{2012}+\dfrac{x-2013}{2011}+...+\dfrac{x-2013}{1}=0\)
\(\Leftrightarrow\left(x-2013\right)\left(\dfrac{1}{2012}+\dfrac{1}{2011}+...+\dfrac{1}{1}\right)=0\)
Dễ thấy: \(\dfrac{1}{2012}+\dfrac{1}{2011}+...+\dfrac{1}{1}>0\)
\(\Rightarrow x-2013=0\Rightarrow x=2013\)
Sao lại trừ 1 vậy bạn ??? mình không hiểu cho lắm mong bạn giúp đỡ
Lời giải:
Ta có:
\(\frac{x-1}{2012}+\frac{x-2}{2011}+\frac{x-3}{2010}+...+\frac{x-2012}{1}=2012\)
\(\Leftrightarrow \left(\frac{x-1}{2012}-1\right)+\left(\frac{x-2}{2011}-1\right)+\left(\frac{x-3}{2010}-1\right)+...+\left(\frac{x-2012}{1}-1\right)=0\)
\(\Leftrightarrow \frac{x-2013}{2012}+\frac{x-2013}{2011}+...+\frac{x-2013}{1}=0\)
\(\Leftrightarrow (x-2013)\left(\frac{1}{2012}+\frac{1}{2011}+...+1\right)=0\)
Dễ thấy \(\frac{1}{2012}+\frac{1}{2011}+...+1\neq 0\Rightarrow x-2013=0\)
\(\Leftrightarrow x=2013\)
Vậy PT có nghiệm \(x=2013\)
\(\dfrac{2-x}{2010}-1=\dfrac{1-x}{2011}-\dfrac{x}{2012}\\ \Leftrightarrow\dfrac{2-x-2010}{2010}=\dfrac{2012-2012x-2011x}{2011\cdot2012}\\ \Leftrightarrow\dfrac{-2008-x}{2010}=\dfrac{2012-4023x}{4046132}\\ \Leftrightarrow\left(-2008-x\right)4046132=\left(2012-4023x\right)2010\\ \Leftrightarrow-8124633056-4046132x=4044120-8086230x\\ \Leftrightarrow-4046132x+8086230x=4044120+8124633056\\ \Leftrightarrow4040098x=8128677176\\ \Leftrightarrow x=2012\)
\(\dfrac{2-x}{2010}-1=\dfrac{1-x}{2011}-\dfrac{x}{2012}\\ \Leftrightarrow2023066\left(2-x\right)-4066362660=2022060\left(1-x\right)-2021055x\\ \Leftrightarrow4046132-2023066x-4066362660=2022060-2022060x-2021055x\\ \Leftrightarrow-4062316528-2023066x=2022060-4043115x\\ \Leftrightarrow-2023066x+4043115x=2022060+4062316528\\ \Leftrightarrow2020049x=4064338588\\ \Leftrightarrow x=2012\)
\(\dfrac{x+1}{2012}+\dfrac{x+2}{2011}+\dfrac{x+3}{2010}=\dfrac{x-1}{2014}+\dfrac{x-2}{2015}+\dfrac{x-3}{2016}\)
\(\Leftrightarrow\left(\dfrac{x+1}{2012}+1\right)+\left(\dfrac{x+2}{2011}+1\right)+\left(\dfrac{x+3}{2010}\right)=\left(\dfrac{x-1}{2014}+1\right)+\left(\dfrac{x-2}{2015}+1\right)+\left(\dfrac{x-3}{2016}+1\right)\)
\(\Leftrightarrow\dfrac{x+2013}{2012}+\dfrac{x+2013}{2011}+\dfrac{x+2013}{2010}-\dfrac{x+2013}{2014}-\dfrac{x+2013}{2015}-\dfrac{x+2013}{2016}=0\)
\(\Leftrightarrow\left(x+2013\right)\left(\dfrac{1}{2012}+\dfrac{1}{2011}+\dfrac{1}{2010}-\dfrac{1}{2014}-\dfrac{1}{2015}-\dfrac{1}{2016}\right)=0\)
\(\Leftrightarrow x+2013=0\)
\(\Leftrightarrow x=-2013\)
\(\frac{2-x}{2011}-1=\frac{1-x}{2012}-\frac{x}{2013}\)
\(\Leftrightarrow\frac{2013-x}{2011}=\frac{2013-x}{2012}+\frac{2013-x}{2013}\)
\(\Leftrightarrow\left(2013-x\right)\left(\frac{1}{2011}-\frac{1}{2012}-\frac{1}{2013}\right)=0\)
\(\Leftrightarrow2013-x=0\) (Vì \(\frac{1}{2011}-\frac{1}{2012}-\frac{1}{2013}\ne0\forall x\))
\(\Leftrightarrow x=2013\)
\(Vậy\) \(phương\) \(trình\) \(có\) \(nghiệm\) \(x=2013\).
Chúc bạn học tốt@@
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 \(\dfrac{1}{2014}+\dfrac{1}{2013}-\dfrac{1}{2012}-\dfrac{1}{2011}\ne0\))
\(\Rightarrow x=-2015\)
1, \(\dfrac{x-3}{2011}+\dfrac{x-2}{2012}=\dfrac{x-2012}{2}+\dfrac{x-2011}{3}\\ \\ < =>\dfrac{x-3}{2011}-1+\dfrac{x-2}{2012}-1=\dfrac{x-2012}{2}-1+\dfrac{x-2011}{3}-1\\ \\ < =>\dfrac{x-2014}{2011}+\dfrac{x-2014}{2012}-\dfrac{x-2014}{2}-\dfrac{x-2014}{3}=0\\ \\ < =>\left(x-2014\right).\left(\dfrac{1}{2011}+\dfrac{1}{2012}-\dfrac{1}{2}-\dfrac{1}{3}\right)=0\\ \\ < =>x-2014=0< =>x=2014\)
2, \(x^2+1=x\\ \\ < =>x^2-x+1=0\\ \\ < =>x^2-x+\dfrac{1}{4}+\dfrac{3}{4}=0\\ \\ < =>\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}=0\)
có vế trái luôn dương, vế phải = 0 => vô nghiệm