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Bỏ dấu giá trị tuyệt đối:
x \(\le\) 2008 | 2008 < x < 2009 | 2009 \(\le\) x < 2010 | 2010\(\le\)x < 2011 | x \(\ge\) 2011 | |
|x- 2008| | 2008-x | x-2008 | x-2008 | x-2008 | x-2008 |
|x-2009| | 2009-x | 2009-x | x-2009 | x-2009 | x-2009 |
|x-2010| | 2010-x | 2010 - x | 2010 - x | x - 2010 | x - 2010 |
|x-2011| | 2011 - x | 2011 - x | 2011 - x | 2011 - x | x - 2001 |
=>
+) Nếu x \(\le\) 2008 => A = 2008 - x + 2009 - x + 2010 - x + 2011 - x + 2008 = 10 046 - 4x \(\ge\) 10 046 - 4.2008 = 2014
+) Nếu 2008 < x < 2009 => A = x - 2008 + 2009 - x + 2010 - x + 2011 - x + 2008 = 6030 - 2x > 6030 - 2.2009 = 2012
+) Nếu 2009 \(\le\) x < 2010 => A = x - 2008 + x - 2009 + 2010 - x + 2011 - x + 2008 = 2012
+) Nếu 2010 \(\le\) x < 2011 => A = x - 2008 + x - 2009 + x - 2010 + 2011 - x + 2008 = 2x - 2008 \(\ge\) 2.2010 - 2008 = 2012
+) Nếu x \(\ge\) 2011 => A = x - 2008 + x - 2009 + x - 2010 + x - 2011 + 2008 = 4x - 6030 \(\ge\) 4.2011 - 6030 = 2014
Từ các trường hợp trên => A nhỏ nhất bằng 2012 khi x = 2009 ; hoặc x = 2010
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!x-2007!+!x-2010!>=3 đẳng thức khi 2007<=x<=2008
!x-2007!+!x-2008!+!x-2010!>=3 đẳng thức khi !x-2008!=0
=> nghiệm duy nhất x=2008 và y=2009
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\(\dfrac{x-1}{2011}+\dfrac{x-2}{2010}-\dfrac{x-3}{2009}=\dfrac{x-4}{2008}\)
<=> \(\left(\dfrac{x-1}{2011}-1\right)+\left(\dfrac{x-2}{2010}-1\right)-\left(\dfrac{x-3}{2009}-1\right)=\left(\dfrac{x-4}{2008}-1\right)\)
<=> \(\dfrac{x-2012}{2011}+\dfrac{x-2012}{2010}-\dfrac{x-2012}{2009}-\dfrac{x-2012}{2008}=0\)
<=> \(\left(x-2012\right)\left(\dfrac{1}{2011}+\dfrac{1}{2010}-\dfrac{1}{2009}-\dfrac{1}{2008}\right)=0\)
<=> x - 2012 = 0
<=> x = 2012
+) Xét \(x\ge2010\) có:
\(2012=x-2010+x-2008\)
\(\Rightarrow2x-4018=2012\)
\(\Rightarrow2x=6030\)
\(\Rightarrow x=3015\) ( t/m )
+) Xét \(2008\le x< 2010\) ta có:
\(2012=2010-x+x-2008\)
\(\Rightarrow2=2012\) ( loại )
+) Xét \(x< 2008\) có:
\(2012=2010-x+2008-x\)
\(\Rightarrow2012=4018-2x\)
\(\Rightarrow2x=2006\)
\(\Rightarrow x=1003\left(tm\right)\)
Vậy x= 3015 hoặc x = 1003