Ta có\(\hept{\begin{cases}\left|x-2011\right|\ge2011-x,\forall x\\\left|x-211\right|\ge x-211,\forall x\end{cases}}\)

\(\Rightarrow A\ge1800.\)Dấu "=" xảy ra khi \(\hept{\begin{cases}x-2011\le0\\x-211\ge0\end{cases}}\)

\(\Rightarrow211\le x\le2011\)

Vậy.............

Vậy GTNN của biểu thức bằng 1800 hả

thật là EASY

Chỉ với 😂😂

\(A=\dfrac{4x+3}{x^2+1}\Leftrightarrow Ax^2+A=4x+3\\ \Leftrightarrow Ax^2-4x+A-3=0\)

Coi đây là PT bậc 2 ẩn x thì PT có nghiệm

\(\Leftrightarrow\Delta=16-4A\left(A-3\right)\ge0\\ \Leftrightarrow16-4A^2+12A\ge0\\ \Leftrightarrow-A^2+3A+4\ge0\\ \Leftrightarrow-1\le A\le4\)

Vậy \(A_{max}=4;A_{min}=-1\)

\(A_{max}=4\Leftrightarrow\dfrac{4x+3}{x^2+1}=4\Leftrightarrow4x^2-4x+1=0\\ \Leftrightarrow\left(2x-1\right)^2=0\Leftrightarrow x=\dfrac{1}{2}\\ A_{min}=-1\Leftrightarrow\dfrac{4x+3}{x^2+1}=-1\Leftrightarrow x^2+1=-4x-3\Leftrightarrow x^2+4x+4=0\\ \Leftrightarrow\left(x+2\right)^2=0\Leftrightarrow x=-2\)

|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

Câu 2:

ĐKXĐ: x<>0

\(B=\dfrac{-x^2-x-1}{x^2}\)

\(=-1-\dfrac{1}{x}-\dfrac{1}{x^2}\)

\(=-\left(\dfrac{1}{x^2}+\dfrac{1}{x}+1\right)\)

\(=-\left(\dfrac{1}{x^2}+2\cdot\dfrac{1}{x}\cdot\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}\right)\)

\(=-\left(\dfrac{1}{x}+\dfrac{1}{2}\right)^2-\dfrac{3}{4}< =-\dfrac{3}{4}\forall x< >0\)

Dấu '=' xảy ra khi 1/x+1/2=0

=>1/x=-1/2

=>x=-2

\(A=\left(x-y+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)

\(A_{min}=\dfrac{3}{4}\) khi \(x-y+\dfrac{1}{2}=0\)

\(B=\dfrac{7}{-\left(x-5\right)^2-5}\ge-\dfrac{7}{5}\)

\(B_{min}=-\dfrac{7}{5}\) khi \(x=5\)

(x-y)²+lx-1l+2011>(=)0+0+2011=2011

dấu bằng xảy ra khi (x-y)²=0;lx-1l=0

lx-1l=0=>x=1

=>(1-x)²=0

=>y=1

vậy Min_M=2011 khi x=y=1

Ta có:

(x-y)²\(\ge\)0

|x-1|\(\ge\)0

2011>0

Suy ra GTNN của M=2011 tại x=1, y=1

\(A=\left(x+3\right)^2+2\ge2\\ A_{min}=2\Leftrightarrow x=-3\\ B=\left(x^2+3x+\dfrac{9}{4}\right)-\dfrac{29}{4}=\left(x+\dfrac{3}{2}\right)^2-\dfrac{29}{4}\ge-\dfrac{29}{4}\\ B_{min}=-\dfrac{29}{4}\Leftrightarrow x=-\dfrac{3}{2}\\ C=\left(9x^2-12x+4\right)+2017=\left(3x-2\right)^2+2017\ge2017\\ C_{min}=2017\Leftrightarrow x=\dfrac{2}{3}\)

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