Đáp án:

$1)\\ min_A=\dfrac{11}{3}\Leftrightarrow x=\dfrac{25}{9}\\ min_B=1 \Leftrightarrow  x=0\\ min_C=4 \Leftrightarrow  x=2\\ min_D= -\dfrac{9}{2} \Leftrightarrow  x=0.$

Giải thích các bước giải:

$1)\\ A=3x-10\sqrt{x}+12(x \ge 0)\\ =3\left(x-\dfrac{10}{3}\sqrt{x}+4\right)\\ =3\left(x-2.\dfrac{5}{3}\sqrt{x}+\dfrac{25}{9}+\dfrac{11}{9}\right)\\ =3\left(x-2.\dfrac{5}{3}\sqrt{x}+\dfrac{25}{9}\right)+\dfrac{11}{3}\\ =3\left(\sqrt{x}-\dfrac{5}{3}\right)^2+\dfrac{11}{3} \ge \dfrac{11}{3} \ \forall \ x$

Dấu "=" xảy ra $\Leftrightarrow \sqrt{x}-\dfrac{5}{3}=0\Leftrightarrow \sqrt{x}=\dfrac{5}{3}\Leftrightarrow x=\dfrac{25}{9}$

$B=2x+6\sqrt{x}+1(x \ge 0)\\ =2\left(x+3\sqrt{x}+\dfrac{1}{2}\right)\\ =2\left(x+2.\dfrac{3}{2}\sqrt{x}+\dfrac{9}{4}-\dfrac{7}{4}\right)\\ =2\left(x+2.\dfrac{3}{2}\sqrt{x}+\dfrac{9}{4}\right)-\dfrac{7}{2}\\ =2\left(\sqrt{x}+\dfrac{3}{2}\right)^2-\dfrac{7}{2} \ge 2\left(\dfrac{3}{2}\right)^2-\dfrac{7}{2}=1$

Dấu "=" xảy ra $\Leftrightarrow \sqrt{x}= 0 \Leftrightarrow  x=0$

$C=x+3\sqrt{x-2}+2(x \ge 2)\\ =x-2+3\sqrt{x-2}+4\\ =x-2+2.\dfrac{3}{2}\sqrt{x-2}+\dfrac{9}{4}+\dfrac{7}{4}\\ =\left(\sqrt{x-2}+\dfrac{3}{2}\right)^2+\dfrac{7}{4} \ge\left(\dfrac{3}{2}\right)^2+\dfrac{7}{4}=4$

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

$D=\dfrac{-9}{3\sqrt{x}+2}( x \ge 0)\\ 3\sqrt{x}+2 \ge 2 \ \forall \ x \ge 0 \\ \Rightarrow \dfrac{9}{3\sqrt{x}+2} \le \dfrac{9}{2} \ \forall \ x \ge 0 \\ \Rightarrow -\dfrac{9}{3\sqrt{x}+2} \ge -\dfrac{9}{2} \ \forall \ x \ge 0 $

Dấu "=" xảy ra $\Leftrightarrow \sqrt{x}= 0 \Leftrightarrow  x=0.$