$A=\text{ }3,7+\left| \text{ }4,3-x\text{ } \right|$
Ta có : $\left| 4,3-x\text{ } \right|\ge 0\,\,\forall x$$\Rightarrow 3,7+\left| \text{ }4,3-x\text{ } \right|\ge 3,7\forall x$
Dấu “ = ” xảy ra $\Leftrightarrow 4,3-x=0\Leftrightarrow x=4,3$
Vậy GTNN của A = 3,7 khi $x=4,3$

$B=\text{ }\left| \text{3x }+8,4 \right|-14$
Ta có : $\left| 3x+8,4\text{ } \right|\ge 0\,\,\forall x$$\Rightarrow \left| \text{ }3x\text{ +8,4 } \right|-14\ge -14\forall x$
Dấu “ = ” xảy ra $\Leftrightarrow 3x+8,4=0\Leftrightarrow x=2,8$
Vậy GTNN của B = -14 khi $x=2,8$

$C=\text{ }\left| \text{ }4x\text{ }-\text{ }3\text{ } \right|\text{ }+\text{ }\left| \text{ }5y\text{ }+\text{ }7,5\text{ } \right|\text{ }+\text{ }17,5$
Ta có : $\left\{ \begin{align}& \left| \text{ }4x\text{ }-\text{ }3\text{ } \right|\ge 0\forall x \\
& \left| \text{ }5y\text{ }+\text{ }7,5\text{ } \right|\ge 0\forall y \\ \end{align} \right.$$\Rightarrow \left| \text{ }4x\text{ }-\text{ }3\text{ } \right|\text{ }+\text{ }\left| \text{ }5y\text{ }+\text{ }7,5\text{ } \right|\text{ }+\text{ }17,5\ge 17,5$
Suy ra $\left\{ \begin{align}& 4x\text{ }-\text{ }3=0 \\ & 5y\text{ }+\text{ }7,5=0 \\ \end{align} \right.$$\Leftrightarrow \left\{ \begin{align}& x=\frac{3}{4} \\ & y=\frac{3}{2} \\ \end{align} \right.$
Vậy GTNN của C = 17,5 khi $\left( x,y \right)=\left( \frac{3}{4};\frac{3}{2} \right)$

d, Áp dụng bất đẳng thức $\left| a \right|+\left| b \right|\ge \left| a+b \right|$ ta có: $D=\text{ }\left| \text{ }x\text{ }-\text{ }2018\text{ } \right|\text{ }+\text{ }\left| \text{ }x\text{ }-\text{ }2017\text{ } \right|=\text{ }\left| \text{ }x\text{ }-\text{ }2018\text{ } \right|\text{ }+\text{ }\left| \text{ }2017-x\text{ } \right|\ge \left| x-2008+2007-x \right|=1$
Dấu “ = ” xảy ra $\Leftrightarrow x-2018=2017-x\Leftrightarrow x=\frac{4035}{2}$
Vậy GTNN của D = 1 khi $x=\frac{4035}{2}$