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a) \(\frac{x+4}{2009}+1+\frac{x+3}{2010}+1=\frac{x+2}{2011}+1+\frac{x+1}{2012}\)
\(\frac{x+4+2009}{2009}+\frac{x+3+2010}{2010}=\frac{x+2+2011}{2011}+\frac{x+2+2012}{2012}\)
\(\frac{x+2013}{2009}+\frac{x+2013}{2010}-\frac{x+2013}{2011}-\frac{x+2013}{2012}=0\)
\(\left(x+2013\right).\left(\frac{1}{2009}+\frac{1}{2010}-\frac{1}{2011}-\frac{1}{2012}\right)=0\) (1)
Vì \(\left(\frac{1}{2009}+\frac{1}{2010}-\frac{1}{2011}-\frac{1}{2012}\right)\ne0\)
Nên biểu thức (1) xảy ra khi \(x+2013=0\)
\(x=-2013\)
b) \(\left(x-2011\right)\left(\frac{1}{2010}+\frac{1}{2011}+\frac{1}{2012}-\frac{1}{2013}-\frac{1}{2014}\right)=0\) (2)
Vì \(\left(\frac{1}{2010}+\frac{1}{2011}+\frac{1}{2012}-\frac{1}{2013}-\frac{1}{2014}\right)\ne0\)
Nên biểu thức (2) xảy ra khi \(x-2011=0\)
\(x=2011\)
2012 . | x - 2011| + (x-2011)2 = 2013 . | 2011 - x|
|x-2011|.|x-2011| + 2012 . | x - 2011| - 2013 . | 2011- x| =0
|x - 2011|.| x - 2011| + 2012 .| x - 2011| - 2013 | x - 2011| = 0
| x- 2011| .| x -2011| - | x - 2011| = 0
| x - 2011|. { | x - 2011| - 1} = 0
\(\left[{}\begin{matrix}\left|x-2011\right|=0\\\left|x-2011\right|-1=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=2011\\x=2012\\x=2010\end{matrix}\right.\)
Kết luận x \(\in\) { 2010; 2011; 2012}
\(\frac{x+1}{2013}+\frac{x+2}{2012}+\frac{x+3}{2011}=-3\)
\(=>\frac{x+1}{2013}+\frac{x+2}{2012}+\frac{x+3}{2011}+3=0\)
\(=>\frac{x+1}{2013}+1+\frac{x+2}{2012}+1+\frac{x+3}{2011}+1=0\)
\(=>\frac{x+2014}{2013}+\frac{x+2014}{2012}+\frac{x+2014}{2011}=0\)
\(=>\left(x+2014\right)\left(\frac{1}{2013}+\frac{1}{2012}+\frac{1}{2011}\right)=0\)
=>x+2014=0 (vì \(\frac{1}{2013}+\frac{1}{2012}+\frac{1}{2011}\) khác 0)
=>x=-2014
(x + 4)/2010 + (x+3)/2011 = (x+2)/2012 + (x+1)/2013
<=> [(x + 4)/2010 + 1] + [(x+3)/2011 + 1] = [(x+2)/2012 + 1] + [(x+1)/2013 + 1]
<=> (x + 2014)/2010 + (x + 2014)/2011 = (x + 2014)/2012 + (x + 2014)/2013
<=> (x + 2014)/2010 + (x + 2014)/2011 - (x + 2014)/2012 - (x + 2014)/2013 = 0
<=> (x + 2014).(1/2010 + 1/2011 - 1/2012 - 1/2013) = 0
Ta thấy (1/2010 + 1/2011 - 1/2012 - 1/2013) ≠ 0
Vậy suy ra x = -2014
\(\frac{x-1}{2013}+\frac{x-2}{2012}-\frac{x-3}{2011}=\frac{x-4}{2010}\)
\(\frac{x-1}{2013}+\frac{x-2}{2012}-\frac{x-3}{2011}-\frac{x-4}{2010}=0\)
\(\frac{x-1}{2013}-1+\frac{x-2}{2012}-1-\frac{x-3}{2011}+1-\frac{x-4}{2010}+1=0\)
\(\left(\frac{x-1}{2013}-1\right)+\left(\frac{x-2}{2012}-1\right)-\left(\frac{x-3}{2011}-1\right)-\left(\frac{x-4}{2010}-1\right)=0\)
\(\frac{x-2014}{2013}+\frac{x-2014}{2012}-\frac{x-2014}{2011}-\frac{x-2014}{2010}=0\)
\(\left(x-2014\right).\left(\frac{1}{2013}+\frac{1}{2012}-\frac{1}{2011}-\frac{1}{2010}\right)=0\)
\(x-2014=0:\left(\frac{1}{2013}+\frac{1}{2012}-\frac{1}{2011}-\frac{1}{2010}\right)\)
\(x-2014=0\)
\(x=2014\)
Nhớ tk cho mình nha =3
A.R.M.Y FIGHTING!!!!
\(\Leftrightarrow2011.\left|x-2011\right|+\left(x-2011\right)^2=2013.\left|x-2011\right|\)
\(\Leftrightarrow\left(x-2011\right)^2=2.\left|x-2011\right|\Rightarrow\orbr{\begin{cases}x-2011=0\\2=\left|x-2011\right|\end{cases}}\Rightarrow\orbr{\begin{cases}x=2013\\x=2009\end{cases}}\text{hoặc }x=2011\)
Vậy ....
=>2012|x-2011|-|x-2011|+(x-2011)^2+2013>0
=>2011|x-2011|+(x-2011)^2+2013>0(luôn đúng)