Đáp án:
$min_A= \ge 9 \Leftrightarrow x=\dfrac{1}{7}\\ max_B= -1 \Leftrightarrow x=\dfrac{2}{3}\\ min_C=-16 \Leftrightarrow x=-4\\ min_D= 31 \Leftrightarrow x=0\\ min_E= 11 \Leftrightarrow\left\{\begin{array}{l}x=0\\ y=0\end{array} \right.\\ max_F= 36 \Leftrightarrow x=6$
Giải thích các bước giải:
$A=49x^2-14x+10\\ =(7x)^2-2.7x+1+9\\ =(7x-1)^2+9 \ge 9 \ \forall \ x$
Dấu "=" xảy ra $\Leftrightarrow 7x-1=0 \Leftrightarrow x=\dfrac{1}{7}$
$B=-9x^2+12x-5\\ =-(3x)^2+2.3x.2-4-1\\ =-(3x-2)^2-1 \le -1 \ \forall \ x$
Dấu "=" xảy ra $\Leftrightarrow 3x-2 =0 \Leftrightarrow x=\dfrac{2}{3}$
$C=x^2+8x\\ =x^2+8x+16-16\\ =(x+4)^2-16 \ge -16 \ \forall \ x$
Dấu "=" xảy ra $\Leftrightarrow x+4 =0 \Leftrightarrow x=-4$
$D=20x^2+31 \ge 31 \ \forall \ x$
Dấu "=" xảy ra $\Leftrightarrow x=0$
$E=40x^2+50y^2+11 \ge 11 \ \forall x,y$
Dấu "=" xảy ra $\left\{\begin{array}{l}x=0\\y=0\end{array} \right.$
$F=-x^2+12x\\ =-x^2+2.6.x-36+36\\ =-(x-6)^2+36 \le 36 \ \forall \ x$
Dấu "=" xảy ra $\Leftrightarrow x-6=0 \Leftrightarrow x=6$
