|x-2011|+|x-2| = |x-2|+|2011-x|\(\ge\)|x-2+2011-x|=2009

vậy  GTNN của biểu thức: |x-2011|+|x-2| là 2009 \(\Leftrightarrow\)x=2

\(x^{2011}+x^{2011}+1+...+1\) (2009 số 1) \(\ge2011\sqrt[2011]{x^{4022}}=2011x^2\)

Tương tự:

\(2y^{2011}+2009\ge2011y^2\); \(2z^{2011}+2009\ge2011z^2\)

Cộng vế:

\(2\left(x^{2011}+y^{2011}+z^{2011}\right)+6027\ge2011\left(x^2+y^2+z^2\right)\)

\(\Rightarrow2011\left(x^2+y^2+z^2\right)\le6033\)

\(\Rightarrow x^2+y^2+z^2\le3\)

Ta co:

\(x+3y\ge6\Rightarrow y\ge2-\frac{x}{3}\)

\(\Rightarrow P\ge\frac{2}{3}x+\frac{6}{x}+2013\ge2\sqrt{\frac{2}{3}x.\frac{6}{x}}+2013=2017\)

Dau '=' xay ra khi \(x=3;y=1\)

Áp dụng BĐT \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) ta có:

\(B=\left|x-2011\right|+\left|x-2\right|\)

\(=\left|x-2011\right|+\left|2-x\right|\)

\(\ge\left|x-2011+2-x\right|=2009\)

Xảy ra khi \(2\le x\le2011\)

\(\left|x-2011\right|+\left|x-2\right|=\left|x-2011\right|+\left|2-x\right|\)

Áp dụng bđt:

\(\left|A\right|+\left|B\right|\ge\left|A+B\right|\)

\(\Rightarrow\left|x-2011\right|+\left|2-x\right|\ge\left|x-2011+2-x\right|\)

\(\Rightarrow\left|x-2011\right|+\left|2-x\right|\ge2009\)

Dấu "=" xảy ra khi:

\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x-2011\ge0\Rightarrow x\ge2011\\2-x\ge0\Rightarrow x\le2\end{matrix}\right.\\\left\{{}\begin{matrix}x-2011< 0\Rightarrow x< 2011\\2-x< 0\Rightarrow x>2\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow2< x< 2011\)

c, C=|x-1|+|x-2|+...+|x-100|=(|x-1|+|100-x|)+(|x-2|+|99-x|)+...+(|x-50|+|56-x|) \(\ge\) |x-1+100-x|+|x-2+99-x|+...+|x-50+56-x|=99+97+...+1 = 2500

Dấu "=" xảy ra khi \(\hept{\begin{cases}\left(x-1\right)\left(100-x\right)\ge0\\\left(x-2\right)\left(99-x\right)\ge0.....\\\left(x-50\right)\left(56-x\right)\ge0\end{cases}\Leftrightarrow\hept{\begin{cases}1\le x\le100\\2\le x\le99....\\50\le x\le56\end{cases}\Leftrightarrow}50\le x\le56}\)

Vậy MinC = 2500 khi 50 =< x =< 56

a. A=|x-2011|+|x-2012|=|x-2011|+|2012-x| \(\ge\) |x-2011+2012-x| = 1

Dấu "=" xảy ra khi \(\left(x-2011\right)\left(2012-x\right)\ge0\Leftrightarrow2011\le x\le2012\)

Vậy MinA = 1 khi 2011 =< x =< 2012

b, B=|x-2010|+|x-2011|+|x-2012|=(|x-2010|+|2012-x|) + |x-2011| 

Ta có: \(\left|x-2010\right|+\left|2012-x\right|\ge\left|x-2010+2012-x\right|=0\)

Mà \(\left|x-2011\right|\ge0\forall x\)

\(\Rightarrow B=\left(\left|x-2010\right|+\left|2012-x\right|\right)+\left|x-2011\right|\ge2+0=2\)

Dấu "=" xảy ra khi \(\hept{\begin{cases}\left(x-2010\right)\left(2012-x\right)\ge0\\\left|x-2011\right|=0\end{cases}\Rightarrow\hept{\begin{cases}2010\le x\le2012\\x=2011\end{cases}\Rightarrow}x=2011}\)

Vậy MinB = 2 khi x = 2011

Câu c để nghĩ 

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