Đáp án:

$5)\\ a)min_A=-1\Leftrightarrow x=3\\ b)min_B=-\dfrac{13}{4} \Leftrightarrow x=-\dfrac{1}{2}\\ c)min_C=\dfrac{3}{4} \Leftrightarrow x=\dfrac{1}{4}\\ d)min_D= -\dfrac{73}{36}\Leftrightarrow x=-\dfrac{1}{18}\\ e)min_E= -\dfrac{121}{4}\Leftrightarrow x=\dfrac{9}{4}\\ g)min_G=\dfrac{484}{25} \Leftrightarrow x=-\dfrac{4}{25}\\ h)min_H=-\dfrac{9}{4} \Leftrightarrow x=\dfrac{1}{2}$

Giải thích các bước giải:

$5)\\ a)\\ A=x^2-6x+8\\ =x^2-6x+9-1\\ =(x-3)^2-1 \ge -1 \forall \ x \text{ vì} \ (x-3)^2\ge 0 \ \forall \ x$

Dấu "=" xảy ra $\Leftrightarrow x-3=0\Leftrightarrow x=3$

Vậy $min_A=-1\Leftrightarrow x=3$

$b)\\ B=x^2+x-3\\ =x^2+2.x.\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{13}{4}\\ =\left(x+\dfrac{1}{2}\right)^2-\dfrac{13}{4} \ge -\dfrac{13}{4} \forall \ x \text{ vì} \ \left(x+\dfrac{1}{2}\right)^2 \ge 0 \ \forall \ x$

Dấu "=" xảy ra $ \Leftrightarrow x+\dfrac{1}{2}=0\Leftrightarrow x=-\dfrac{1}{2}$

Vậy $min_B=-\dfrac{13}{4} \Leftrightarrow x=-\dfrac{1}{2}$

$c)\\ C=4x^2-2x+1\\ =(2x)^2-2.2x.\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}\\ =\left(2x-\dfrac{1}{2}\right)^2+\dfrac{3}{4} \ge \dfrac{3}{4} \forall \ x \text{ vì} \ \left(2x-\dfrac{1}{2}\right)^2 \ge 0 \ \forall \ x$

Dấu "=" xảy ra $\Leftrightarrow 2x-\dfrac{1}{2}=0\Leftrightarrow x=\dfrac{1}{4}$

Vậy $min_C=\dfrac{3}{4} \Leftrightarrow x=\dfrac{1}{4}$

$d)\\ D=9x^2+x-2\\ =(3x)^2+2.3x.\dfrac{1}{6}+\dfrac{1}{36}-\dfrac{73}{36}\\ =\left(3x+\dfrac{1}{6}\right)^2-\dfrac{73}{36} \ge -\dfrac{73}{36} \forall \ x \text{ vì} \ \left(3x+\dfrac{1}{6}\right)^2 \ge 0 \ \forall \ x$

Dấu "=" xảy ra $\Leftrightarrow 3x+\dfrac{1}{6}=0\Leftrightarrow x=-\dfrac{1}{18}$

Vậy $min_D= -\dfrac{73}{36}\Leftrightarrow x=-\dfrac{1}{18}$

$e)\\ E=(x-5)(4x+2)\\ =4x^2+2x-20x-10\\ =4x^2−18x−10\\ =(2x)^2-2.2x.\dfrac{9}{2}+\dfrac{81}{4}-\dfrac{121}{4}\\ =\left(2x-\dfrac{9}{2}\right)^2-\dfrac{121}{4} \ge -\dfrac{121}{4} \forall \ x \text{ vì} \ \left(2x-\dfrac{9}{2}\right)^2 \ge 0 \ \forall \ x$

Dấu "=" xảy ra $\Leftrightarrow 2x-\dfrac{9}{2}=0\Leftrightarrow x=\dfrac{9}{4}$

Vậy $min_E= -\dfrac{121}{4}\Leftrightarrow x=\dfrac{9}{4}$

$g)\\ G=(3x+4)^2+(4x-2)^2\\ =9x^2+24x+16+16x^2−16x+4\\ =25x^2+8x+20\\ =(5x)^2+2.5x.\dfrac{4}{5}+\dfrac{16}{25}+\dfrac{484}{25}\\ =\left(5x+\dfrac{4}{5}\right)^2+\dfrac{484}{25} \ge \dfrac{484}{25} \forall \ x \text{ vì} \ \left(5x+\dfrac{4}{5}\right)^2 \ge 0 \ \forall \ x$

Dấu "=" xảy ra $\Leftrightarrow 5x+\dfrac{4}{5}=0\Leftrightarrow x=-\dfrac{4}{25}$

Vậy $min_G=\dfrac{484}{25} \Leftrightarrow x=-\dfrac{4}{25}$

$h)\\ x-y=4\Rightarrow y=x-4\\ H=(x-2)(y+5)\\ =(x-2)(x-4+5)\\ =(x-2)(x+1)\\ =x^2+x-2x-2\\ =x^2-x-2\\ =x^2-2.x.\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{9}{4}\\ =\left(x-\dfrac{1}{2}\right)^2-\dfrac{9}{4} \ge -\dfrac{9}{4} \forall \ x \text{ vì} \ \left(x-\dfrac{1}{2}\right)^2 \ge 0 \ \forall \ x$

Dấu "=" xảy ra $\Leftrightarrow x-\dfrac{1}{2}=0\Leftrightarrow x=\dfrac{1}{2}$

Vậy $min_H=-\dfrac{9}{4} \Leftrightarrow x=\dfrac{1}{2}$