ĐKXĐ: \(4(x^2+1)\ne 0\) (luôn đúng)

\(x^2-2011\ge -2011\)

\(\to \dfrac{x^2-2011}{4(x^2+1)}\ge \dfrac{-2011}{4}\)

\(\to \begin{cases}x^2-2011=-2011\\x^2+1=1\end{cases}\)

\(\to x^2=0\)

\(\leftrightarrow x=0\)

\(\to B_{\min}=-\dfrac{2011}{4}\)

Vậy \(B_{\min}=-\dfrac{2011}{4}\)

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 

hello

C = |x + 2011| + |x - 1|

=> C = |x + 2011| + |1 - x|

=> C > |x + 2011 + 1 - x|

=> C > |2012|

=> C > 2012          

dấu "=" xảy ra khi :

(x + 2011)(1 - x) > 0 

tự xét ra

A = \(\sqrt{\left(x-3\right)-2\sqrt{x-3}+1+2}\)

= \(\sqrt{\left(\sqrt{x-3}-1\right)^2+2}\)\(\ge\)\(\sqrt{0+2}\)=\(\sqrt{2}\)

''='' <=> x = 4

=> Min A = \(\sqrt{2}\)và x = 4

B = |x-2011| + |x-1|

TH1: x \(\le\)1

=> B = 2012 - 2x \(\ge\)2010   ''='' <=> x = 1

TH2: 1\(\le\)x\(\le\)2011

=> B = x - 1 + 2011 - x = 2010 với mọi x t/m đkiện

TH3: x \(\ge\)2011

=> B = 2x - 2012 \(\ge\)2010 ''='' <=> x = 2011

Vậy Min B = 2010 <=> 1\(\le\)x\(\le\)2011

|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

giá trị nhỏ nhất là 3

\(\left|x-2010\right|+\left|x-2012\right|=\left|x-2010\right|+\left|x-2012\right|\ge\left|x-2010-x+2012\right|=2\)

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

=> \(B\ge2\)

dấu = xảy ra khi \(\hept{\begin{cases}\left(x-2010\right).\left(-x+2012\right)\ge0\\x=2011\end{cases}}\Rightarrow\hept{\begin{cases}2010\le x\le2012\\x=2011\end{cases}\Rightarrow x=2011}\)