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x - 3/2011 + x - 2/2012 = x - 2012/2 + x - 2011/3
( x - 3 -2011)/2011 + (x - 2-2012)/2012 = (x - 2012-2)/2 + (x - 2011-3)/3
(x-2014)/2011+(x-2014)/2012=(x-2014)/2+(x-2014)/3
(x-2014)(1/2011+1/2012-1/2-1/3)=0
x-2014=0 vì (1/2011+1/2012-1/2-1/3 khác 0
x= 2014
k cho mk nha
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(x-3/2011)-1+(x-2/2012)-1 = (x-2012/2)-1+(x-2011/3)-1
x-2014/2011+x-2014/2012 = x-2014/2+x-2014/ 3
(x-2014)(1/2011+1/2012-1/2-1/3)=0
x-2014 =0 [vì (1/2011+ 1/2012-1/2-1/3#0)]
x=2014
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Ta có : | x - 2011 |2011 + | x - 2012 |2012 \(\ge\)0
Mà | x - 2011 |2011 + | x - 2012 |2012 = 1
xét 2 TH :
TH1 : | x - 2011 |2011 = 0 ; | x - 2012 |2012 = 1
\(\Rightarrow\)x = 2011
TH2 : | x - 2011 |2011 = 1 ; | x - 2012 |2012 = 0
\(\Rightarrow\)x = 2012
vậy x = 2011 hoặc x = 2012
+) Xét x < 2011 thì \(x-2012< -1\)
\(\Rightarrow\left|x-2012\right|^{2012}>1\)
Mà \(\left|x-2011\right|^{2011}>0\forall x< 2011\)
\(\Rightarrow VT>1\left(vl\right)\)
+) Xét x = 2011 thì thỏa mãn
+) Xét 2011 < x < 2012 thì \(\hept{\begin{cases}0< x-2011< 1\\-1< x-2012< 0\end{cases}}\Rightarrow\hept{\begin{cases}\left|x-2011\right|^{2011}< x-2011\\\left|x-2012\right|^{2012}< 2012-x\end{cases}}\)
\(\Rightarrow VT< 1\left(vl\right)\)
+) Xét x = 2012 thì thỏa mãn
+) Xét x > 2012 thì \(x-2011>1\)
\(\Rightarrow\left|x-2011\right|^{2011}>1\)
và \(\left|x-2012\right|^{2012}>0\forall x>2012\)
\(\Rightarrow VT>1\)(vl)
Vậy tập nghiệm S = {2011;2012}
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Ta có:\(\frac{x-3}{2011}+\frac{x-2}{2012}=\frac{x-2012}{2}+\frac{x-2011}{3}\)
\(\Rightarrow\left(\frac{x-3}{2011}-1\right)+\left(\frac{x-2}{2012}-1\right)=\left(\frac{x-2012}{2}-1\right)+\left(\frac{x-2011}{3}-1\right)\)
\(\Rightarrow\frac{x-2014}{2011}+\frac{x-2014}{2012}=\frac{x-2014}{2}+\frac{x-2014}{3}\)
\(\Rightarrow\frac{x-2014}{2011}+\frac{x-2014}{2012}-\frac{x-2014}{2}-\frac{x-2014}{3}=0\)
\(\Rightarrow\left(x-2014\right).\left(\frac{1}{2011}+\frac{1}{2012}-\frac{1}{2}-\frac{1}{3}\right)\)
\(\Rightarrow x-2014=0\)( vì \(\frac{1}{2011}+\frac{1}{2012}-\frac{1}{2}-\frac{1}{3}\ne0\))
\(\Rightarrow x=2014\)
Vậy x= 2014.
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\(\frac{x-3}{2011}+\frac{x-2}{2012}=\frac{x-2012}{2}+\frac{x-2011}{3}\)
\(\Rightarrow\left(\frac{x-3}{2011}-1\right)+\left(\frac{x-2}{2012}-1\right)=\left(\frac{x-2012}{2}-1\right)+\left(\frac{x-2011}{3}-1\right)\)
\(\Rightarrow\frac{x-2014}{2011}+\frac{x-2014}{2012}=\frac{x-2014}{2}+\frac{x-2014}{3}\)
\(\Rightarrow x=0\)
Xét \(x>2012;x< 2011\)
\(\Rightarrow|x-2011|^{2011}+|x-2012|^{2012}>1\)
Xét \(2011< x< 2012\)
\(\Rightarrow\hept{\begin{cases}|x-2011|^{2011}< |x-2011|=x-2011\\|x-2012|^{2012}< |x-2012|=2012-x\end{cases}}\)
\(\Rightarrow|x-2011|^{2011}+|x-2012|^{2012}< x-2011+2012-x=1\)
Xét \(x=2011;x=2012\) dễ thấy nó là nghiệm của phương trình