TÌm GTNN của A, biết: \(A=\left|x-2006\right|+\left|x-1\right|\)
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\(A=\left|x-2006\right|+\left|x-1\right|=\left|x-2006\right|+\left|-x+1\right|\ge\left|x-2006-x+1\right|=2005\)
dấu = xảy ra khi \(\left(x-2006\right).\left(-x+1\right)\ge0\)
\(\Rightarrow1\le x\le2006\)
Vậy Min A=2015 khi và chỉ khi \(1\le x\le2006\)
\(A=|x-2006|+|x-1|=|x-2006|+|1-x|\)
\(\Rightarrow A\ge|x-2006+1-x|=|-2005|=2005\)
\(\Rightarrow minA=2005\Leftrightarrow\left(x-2006\right).\left(1-x\right)\ge0\)
\(TH1:\hept{\begin{cases}x-2006< 0\\1-x< 0\end{cases}}\Rightarrow\hept{\begin{cases}x< 2006\\1< x\end{cases}}\Rightarrow\hept{\begin{cases}x< 2006\\x>1\end{cases}}\Rightarrow1< x< 2006\left(t/m\right)\)
\(TH2:\hept{\begin{cases}x-2006\ge0\\1-x\ge0\end{cases}}\Rightarrow\hept{\begin{cases}x\ge2006\\1\ge x\end{cases}}\Rightarrow\hept{\begin{cases}x\ge2006\\x\le1\end{cases}}\)(vô lý)
Vậy \(minA=2005\Leftrightarrow1< x< 2006\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Áp dụng BĐT \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\):
\(A=\left|x-3\right|+\left|x-1\right|+\left|x+1\right|+\left|x+3\right|\)
\(=\left|3-x\right|+\left|x+3\right|+\left|1-x\right|+\left|x+1\right|\)
\(\ge\left|3-x+x+3\right|+\left|1-x+x+1\right|=8\)
\(minA=8\Leftrightarrow\left\{{}\begin{matrix}\left(3-x\right)\left(x+3\right)\ge0\\\left(1-x\right)\left(x+1\right)\ge0\end{matrix}\right.\Leftrightarrow-1\le x\le1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt x -2006 = y
pt <=> \(\frac{y^2-y\left(y-1\right)+\left(y-1\right)^2}{y^2+y\left(y-1\right)+\left(y-1\right)^2}=\frac{19}{49}\)
<=> \(\frac{y^2-y^2+y+y^2-2y+1}{y^2+y^2-y+y^2-2y+1}=\frac{19}{49}\)
<=> \(\frac{y^2-y+1}{3y^2-3y+1}=\frac{19}{49}\)
<=> \(49y^2-49y+49=57y^2-57y+19\)
<=> \(8y^2-8y-30=0\)
<=> \(4y^2-4y+15=0\)
Giải tiếp nha
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(x+3=t\ne0\Rightarrow x=t-3\)
\(A=\dfrac{\left(t+2\right)\left(t-4\right)}{t^2}=\dfrac{t^2-2t-8}{t^2}=-\dfrac{8}{t^2}-\dfrac{2}{t}+1=-8\left(\dfrac{1}{t}+\dfrac{1}{8}\right)^2+\dfrac{9}{8}\le\dfrac{9}{8}\)
\(A_{max}=\dfrac{9}{8}\) khi \(t=-8\) hay \(x=-11\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có tính chất :
\(\left|a\right|+\left|b\right|\ge\left|a+b\right|\)
\(\rightarrow A=\left|x+5\right|+\left|x+2\right|+\left|x-7\right|+\left|x-8\right|\ge\left|x+5+x+2+x-7+x-8\right|\)
\(\rightarrow A\ge\left|4x-8\right|\)
Vì \(\left|4x-8\right|\ge0\forall x\in R\) nên :
\(\rightarrow A\ge0\forall x\in R\)
Dấu "= " xảy ra khi :
\(\left|4x-8\right|=0\) \(\Leftrightarrow4x-8=0\)
\(\Leftrightarrow x=2\)
Vậy \(A_{min}=0\Leftrightarrow x=2\)