Đáp án:
$7)\\ a)min_A=-12 \Leftrightarrow x=-5\\ b)max_B=-17 \Leftrightarrow \left\{\begin{array}{l} x=\dfrac{3}{2}\\y=\dfrac{5}{4}\end{array} \right.\\ c)min_C=1 \Leftrightarrow x=\dfrac{2}{3}.$
Giải thích các bước giải:
$7)\\ a)A=|x+5|-12\\ \text{Do }|x+5| \ge 0 \ \forall \ x\\ \Rightarrow |x+5| -12 \ge -12 \ \forall \ x$
Dấu "=" xảy ra $\Leftrightarrow x+5=0 \Leftrightarrow x=-5$
Vậy $min_A=-12 \Leftrightarrow x=-5$
$b)B=-|2x-3|-|4y-5|-17\\ \text{Do }|2x-3| \ge 0 \ \forall \ x \Rightarrow -|2x-3| \le 0 \ \forall \ x \\ |4y-5| \ge 0 \ \forall \ y \Rightarrow -|4y-5| \le 0 \ \forall \ y \\ \Rightarrow-|2x-3|-|4y-5| \le 0 \ \forall \ x,y\\ \Rightarrow-|2x-3|-|4y-5|-17 \le -17 \ \forall \ x,y$
Dấu "=" xảy ra $\left\{\begin{array}{l} 2x-3=0\\4y-5=0\end{array} \right. \Leftrightarrow \left\{\begin{array}{l} x=\dfrac{3}{2}\\y=\dfrac{5}{4}\end{array} \right.$
Vậy $max_B=-17 \Leftrightarrow \left\{\begin{array}{l} x=\dfrac{3}{2}\\y=\dfrac{5}{4}\end{array} \right.$
$c)C=2|3x-2|+1\\ \text{Do } |3x-2| \ge 0 \ \forall \ x\\ \Rightarrow 2|3x-2| \ge 0 \ \forall \ x\\ \Rightarrow 2|3x-2| +1\ge 1 \ \forall \ x$
Dấu "=" xảy ra $\Leftrightarrow 3x-2=0 \Leftrightarrow x=\dfrac{2}{3}$
Vậy $min_C=1 \Leftrightarrow x=\dfrac{2}{3}.$
