Đáp án:

$2)\\ min_A=-2 \Leftrightarrow x=4\\ min_Q=2 \Leftrightarrow x=1\\ min_P=-\dfrac{1}{3} \Leftrightarrow x=0\\ min_B=-\dfrac{4}{3} x=\dfrac{1}{2}\\ 3)\\ max_B=\dfrac{4}{3} x=\dfrac{1}{4}$

Giải thích các bước giải:

$2)\\ A=\sqrt{x-4}-2 \ge -2 \ \forall \ x \ge 4$

Dấu "=" xảy ra $\Leftrightarrow x-4=0 \Leftrightarrow x=4$

$Q=\dfrac{x+3}{\sqrt{x}+1}\\ =\dfrac{x+2\sqrt{x}+1-2\sqrt{x}-2+4}{\sqrt{x}+1}\\ =\dfrac{(\sqrt{x}+1)^2-2(\sqrt{x}+1)+4}{\sqrt{x}+1}\\ =\sqrt{x}+1-2+\dfrac{4}{\sqrt{x}+1}\\ =\sqrt{x}+1+\dfrac{4}{\sqrt{x}+1}-2\\ \ge 2\sqrt{(\sqrt{x}+1)\dfrac{4}{\sqrt{x}+1}}-2=2.2-2=2$

Dấu "=" xảy ra

$\Leftrightarrow (\sqrt{x}+1)=\dfrac{4}{\sqrt{x}+1} \Leftrightarrow (\sqrt{x}+1)^2=4 \Leftrightarrow \left[\begin{array}{l} \sqrt{x}+1=2\\ \sqrt{x}+1=-2\end{array} \right. \Leftrightarrow x=1$

$P=\dfrac{\sqrt{x}-1}{\sqrt{x}+3}\\ =\dfrac{\sqrt{x}+3-4}{\sqrt{x}+3}\\ =1-\dfrac{4}{\sqrt{x}+3}\\ \sqrt{x}+3 \ge 3 \ \forall \ x \ge 0\\ \Rightarrow \dfrac{4}{\sqrt{x}+3} \le \dfrac{4}{3} \ \forall \ x \ge 0\\ \Rightarrow -\dfrac{4}{\sqrt{x}+3} \ge -\dfrac{4}{3} \ \forall \ x \ge 0\\ \Rightarrow 1-\dfrac{4}{\sqrt{x}+3} \ge 1-\dfrac{4}{3}=-\dfrac{1}{3} \ \forall \ x \ge 0$ Dấu "=" xảy ra $\Leftrightarrow \sqrt{x}=0 \Leftrightarrow x=0$ $B=\dfrac{-1}{x^2-x+1} =\dfrac{-1}{x^2-x+\dfrac{1}{4}+\dfrac{3}{4}} =\dfrac{-1}{\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}} \left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4} \ge \dfrac{3}{4} \ \forall \ x \Rightarrow \dfrac{1}{\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}} \le \dfrac{1}{\dfrac{3}{4}}=\dfrac{4}{3} \ \forall \ x \Rightarrow \dfrac{-1}{\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}} \ge -\dfrac{4}{3} \ \forall \ x$

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

$3)\\ B=\dfrac{1}{x-\sqrt{x}+1}\\ =\dfrac{1}{x-\sqrt{x}+\dfrac{1}{4}+\dfrac{3}{4}}\\ =\dfrac{1}{\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}}\\ \left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4} \ge \dfrac{3}{4} \ \forall \ x \ge 0\\ \Rightarrow \dfrac{1}{\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}}\le \dfrac{1}{\dfrac{3}{4}}=\dfrac{4}{3} \ \forall \ x \ge 0$

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