\(A=\left|x-3\right|+\left|x+\dfrac{3}{2}\right|\)
Dễ thấy: \(\left|x-3\right|=\left|3-x\right|\)
\(\Rightarrow A=\left|3-x\right|+\left|x+\dfrac{3}{2}\right|\)
Áp dụng bất đẳng thức:
\(\left|A\right|+\left|B\right|\ge\left|A+B\right|\)
\(\Rightarrow A\ge\left|3-x+x+\dfrac{3}{2}\right|\)
\(\Rightarrow A\ge\dfrac{9}{2}\)
Dấu "=" xảy ra khi:
\(\left[{}\begin{matrix}\left\{{}\begin{matrix}3-x< 0\Rightarrow3< x\\x+\dfrac{3}{2}< 0\Rightarrow x< \dfrac{3}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}3-x\ge0\Rightarrow x\le3\\x+\dfrac{3}{2}\ge0\Rightarrow x\ge-\dfrac{3}{2}\end{matrix}\right.\end{matrix}\right.\)
Vậy xảy ra khi: \(-\dfrac{3}{2}\le x\le3\)
Áp dụng bất đẳng thức \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) có:
\(A=\left|x-3\right|+\left|x+\dfrac{3}{2}\right|=\left|3-x\right|+\left|x+\dfrac{3}{2}\right|\)
\(\ge\left|3-x+x+\dfrac{3}{2}\right|=\left|\dfrac{9}{2}\right|=\dfrac{9}{2}\)
Dấu " = " khi \(\left\{{}\begin{matrix}3-x\ge0\\x+\dfrac{3}{2}\ge0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\le3\\x\ge\dfrac{-3}{2}\end{matrix}\right.\)
Vậy \(MIN_A=\dfrac{9}{2}\) khi \(\dfrac{-3}{2}\le x\le3\)
