Đáp án:
$a)max_A=\dfrac{1}{2} \Leftrightarrow x=\dfrac{1}{2}\\ b)max_B=\dfrac{3}{4} \Leftrightarrow x=12\\ c)max_C=-\dfrac{7}{2}- \Leftrightarrow x=-\dfrac{3}{4}\\ d)max_D=5 \Leftrightarrow \left\{\begin{array}{l} x=\dfrac{3}{4}\\ y=\dfrac{-1}{2}\end{array} \right.$
Giải thích các bước giải:
$a)A=-(2x-1)^2+\dfrac{1}{2} \\ \text{Do }-(2x-1)^2 \le 0 \ \forall \ x \Rightarrow -(2x-1)^2+\dfrac{1}{2} \le \dfrac{1}{2} \ \forall \ x$
Dấu "=" xảy ra $\Leftrightarrow 2x-1=0 \Leftrightarrow x=\dfrac{1}{2}$
$b)B=\dfrac{3}{4}-\left(\dfrac{1}{2}x-6\right)^2 \\ \text{Do } -\left(\dfrac{1}{2}x-6\right)^2 \le 0 \ \forall \ x \Rightarrow \dfrac{3}{4}-\left(\dfrac{1}{2}x-6\right)^2 \le \dfrac{3}{4} \ \forall \ x$
Dấu "=" xảy ra $\Leftrightarrow \dfrac{1}{2}x-6=0 \Leftrightarrow x=12$
$c)C=-\dfrac{7}{2}-\left(x+\dfrac{3}{4}\right)^4\\ \text{Do } \left(x+\dfrac{3}{4}\right)^4\le 0 \ \forall \ x \Rightarrow -\dfrac{7}{2}-\left(x+\dfrac{3}{4}\right)^4 \le -\dfrac{7}{2} \ \forall \ x $
Dấu "=" xảy ra $\Leftrightarrow x+\dfrac{3}{4}=0 \Leftrightarrow x=-\dfrac{3}{4}$
$d)D=-(4x-3)^2-(2y+1)^4+5\\ \text{Do } -(4x-3)^2 \le 0 \ \forall \ x; -(2y+1)^4 \le 0 \ \forall \ y\\ \Rightarrow -(4x-3)^2-(2y+1)^4+5 \le 5 \ \forall \ x,y$
Dấu "=" xảy ra $\left\{\begin{array}{l} 4x-3=0\\ 2y+1=0\end{array} \right. \Leftrightarrow \left\{\begin{array}{l} x=\dfrac{3}{4}\\ y=\dfrac{-1}{2}\end{array} \right.$
