$B=|x-5|+|x+\dfrac{3}{4}|$
$→B=|5-x|+|x+\dfrac{3}{4}|$
$→B≥|5-x+x+\dfrac{3}{4}|=5,75$
Dấu $"="$ xảy ra $⇔(5-x)(x+\dfrac{3}{4})>0$
$⇔$ $5-x$ và $x+\dfrac{3}{4}$ cùng dấu.
$⇔$ \(\left[ \begin{array}{l}\left\{\begin{matrix} 5-x>0 & \\ x+\dfrac{3}{4}>0 & \end{matrix}\right.\\\left\{\begin{matrix} 5-x<0 & \\ x+\dfrac{3}{4}<0 & \end{matrix}\right.\end{array} \right.\)
$⇔$ \(\left[ \begin{array}{l}\left\{\begin{matrix} x<5 & \\ x>\dfrac{-3}{4} & \end{matrix}\right.\\\left\{\begin{matrix} x>5 & \\ x<\dfrac{-3}{4} \to vo \quad ly& \end{matrix}\right.\end{array} \right.\)
Vậy GTNN của B là 5,75 $⇔$ $\dfrac{-3}{4}<x<5$
