$a)A=|x|+|8-x| \ge |x+8-x|=8$
Dấu "=" xảy ra
$\Leftrightarrow \left[\begin{array}{l} \left\{\begin{array}{l} x>0\\ 8-x>0\end{array} \right.\\ \left\{\begin{array}{l} x<0\\8-x<0\end{array} \right.\end{array} \right.\Leftrightarrow \left[\begin{array}{l} \left\{\begin{array}{l} 0<x<8\end{array} \right.\\ \left\{\begin{array}{l} x<0\\x>8\end{array} \right.(L)\end{array} \right.\Leftrightarrow 0<x<8$
Vậy $min_A=8$ xảy ra khi $0<x<8$
$b)B= |x-3|+|x-5|= |x-3|+|5-x|\ge |x-3+5-x|=2$
Dấu "=" xảy ra
$\Leftrightarrow \left[\begin{array}{l} \left\{\begin{array}{l} x-3>0\\ 5-x>0\end{array} \right.\\ \left\{\begin{array}{l} x-3<0\\5-x<0\end{array} \right.\end{array} \right.\Leftrightarrow \left[\begin{array}{l} \left\{\begin{array}{l} 3<x<5\end{array} \right.\\ \left\{\begin{array}{l} x<3\\x>5\end{array} \right.(L)\end{array} \right.\Leftrightarrow 3<x<5$
Vậy $min_B=2$ xảy ra khi $3<x<5$
$c)|x-2|+|x-3|+|x-4|+|x-5|\\=|x-2|+|5-x|+|3-x|+|x-4|\\\ge |x-2+5-x|+|3-x+x-4|\\=3+1\\=4$
Dấu "=" xảy ra
$\Leftrightarrow \left\{\begin{array}{l} \left[\begin{array}{l} \left\{\begin{array}{l} x-2>0\\ 5-x>0\end{array} \right.\\ \left\{\begin{array}{l} x-2<0\\5-x<0\end{array} \right.\end{array} \right.\\\left[\begin{array}{l}\left\{\begin{array}{l} 3-x>0\\x-4>0\end{array} \right.\\ \left\{\begin{array}{l} 3-x<0\\x-4<0\end{array} \right.\end{array} \right.\end{array} \right.\Leftrightarrow\left\{\begin{array}{l} 2<x<5\\ 3<x<4\end{array} \right.\Leftrightarrow3<x<4$
Vậy $min_C=4$ xảy ra khi $3<x<4$
$d)D=3|x-1|+ (x-y)^2 -7\\ |x-1| \ge 0 \, \forall x\\ (x-y)^2\ge 0 \, \forall x,y\\ D=3|x-1|+ (x-y)^2 -7 \ge -7\forall x,y$
Dấu "=" xảy ra
$\Leftrightarrow \left\{\begin{array}{l} x-1=0\\x-y=0\end{array} \right.\Leftrightarrow \left\{\begin{array}{l} x=1\\y=1\end{array} \right.$
Vậy $min_D=-7$ xảy ra khi $x=y=1$
