a, Ta có: \(A=\left|x+2\right|+\left|x-5\right|=\left|x+2\right|+\left|5-x\right|\)
Áp dụng bất đẳng thức \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) có:
\(A=\left|x+2\right|+\left|5-x\right|\ge\left|x+2+5-x\right|=\left|7\right|=7\)
Dấu " = " khi \(\left\{{}\begin{matrix}x+2\ge0\\5-x\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge-2\\x\le5\end{matrix}\right.\)
Vậy \(MIN_A=7\) khi \(-2\le x\le5\)
b, Ta có: \(\left\{{}\begin{matrix}\left|2x-1\right|\ge0\\\left|2y+3\right|\ge0\end{matrix}\right.\Leftrightarrow\left|2x-1\right|+\left|2y+3\right|\ge0\)
\(\Leftrightarrow B=\left|2x-1\right|+\left|2y+3\right|-2017\ge-2017\)
Dấu " = " khi \(\left\{{}\begin{matrix}\left|2x-1\right|=0\\\left|2y+3\right|=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{-3}{2}\end{matrix}\right.\)
Vậy \(MIN_B=-2017\) khi \(x=\dfrac{1}{2}\) và \(y=\dfrac{-3}{2}\)
cảm ơn 
