Đáp án:

$1)A\left(\dfrac{2-\sqrt{3}}{2}\right)=\dfrac{13+15\sqrt{3}}{2}\\ 2)B=\dfrac{\sqrt{x}}{\sqrt{x}-3}\\ 3) 0 < x <9\\ 4)x \in \{4;16;36\}\\ 5)min_S=5\Leftrightarrow x=4.$

Giải thích các bước giải:

$A=\dfrac{x+7}{\sqrt{x}};B=\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{2\sqrt{x}-1}{\sqrt{x}-3}-\dfrac{2x-\sqrt{x}-3}{x-9}(x>0;x \ne 9)\\ 1)x=\dfrac{2-\sqrt{3}}{2}\\ =\dfrac{4-2\sqrt{3}}{4}\\ =\dfrac{3-2\sqrt{3}+1}{4}\\ =\left(\dfrac{\sqrt{3}-1}{2}\right)^2\\ A\left(\dfrac{2-\sqrt{3}}{2}\right)\\ =\dfrac{\dfrac{2-\sqrt{3}}{2}+7}{\sqrt{\left(\dfrac{\sqrt{3}-1}{2}\right)^2}}\\ =\dfrac{\dfrac{2-\sqrt{3}+14}{2}}{\dfrac{\sqrt{3}-1}{2}}\\ =\dfrac{16-\sqrt{3}}{\sqrt{3}-1}\\ =\dfrac{(16-\sqrt{3})(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)}\\ =\dfrac{13+15\sqrt{3}}{2}\\ 2)B=\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{2\sqrt{x}-1}{\sqrt{x}-3}-\dfrac{2x-\sqrt{x}-3}{x-9}\\ =\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{2\sqrt{x}-1}{\sqrt{x}-3}-\dfrac{2x-\sqrt{x}-3}{(\sqrt{x}+3)(\sqrt{x}-3)}\\ =\dfrac{\sqrt{x}(\sqrt{x}-3)}{(\sqrt{x}+3)(\sqrt{x}-3)}+\dfrac{(2\sqrt{x}-1)(\sqrt{x}+3)}{(\sqrt{x}-3)(\sqrt{x}+3)}-\dfrac{2x-\sqrt{x}-3}{(\sqrt{x}+3)(\sqrt{x}-3)}\\ =\dfrac{\sqrt{x}(\sqrt{x}-3)+(2\sqrt{x}-1)(\sqrt{x}+3)-(2x-\sqrt{x}-3)}{(\sqrt{x}+3)(\sqrt{x}-3)}\\ =\dfrac{x+3\sqrt{x}}{(\sqrt{x}+3)(\sqrt{x}-3)}\\ =\dfrac{\sqrt{x}(\sqrt{x}+3)}{(\sqrt{x}+3)(\sqrt{x}-3)}\\ =\dfrac{\sqrt{x}}{\sqrt{x}-3}\\ 3)B<0\\ \Leftrightarrow \dfrac{\sqrt{x}}{\sqrt{x}-3} <0\\ \Leftrightarrow \sqrt{x}-3<0(\text{Do }\sqrt{x}>0 \ \forall x >0, x \ne 9)\\ \Leftrightarrow \sqrt{x}<3\\ \Leftrightarrow 0 \le x <9\\ \text{Kết hợp điều kiện: }\Rightarrow 0 < x <9\\ 4)B \in \mathbb{Z}\\ \Leftrightarrow \dfrac{\sqrt{x}}{\sqrt{x}-3} \in \mathbb{Z}\\ \Leftrightarrow \dfrac{\sqrt{x}-3+3}{\sqrt{x}-3} \in \mathbb{Z}\\ \Leftrightarrow 1+\dfrac{3}{\sqrt{x}-3} \in \mathbb{Z}\\ \Rightarrow \dfrac{3}{\sqrt{x}-3} \in \mathbb{Z}\\ x \in \mathbb{Z} \Rightarrow (\sqrt{x}-3) \in Ư(3)\\ \Leftrightarrow (\sqrt{x}-3) \in \{\pm1;\pm3\}\\ \Leftrightarrow \left[\begin{array}{l} \sqrt{x}-3 =-3 \\ \sqrt{x}-3 =-1 \\ \sqrt{x}-3=1 \\ \sqrt{x}-3= 3\end{array} \right.\\ \Leftrightarrow \left[\begin{array}{l} x=0 (L)\\x=4 \\ x=16 \\x=36\end{array} \right.\\ \Rightarrow x \in \{4;16;36\}\\ 5)S=\dfrac{1}{B}+A\\ =\dfrac{\sqrt{x}-3}{\sqrt{x}}+\dfrac{x+7}{\sqrt{x}}\\ =\dfrac{x+\sqrt{x}+4}{\sqrt{x}}\\ =\sqrt{x}+1+\dfrac{4}{\sqrt{x}}\\ =\sqrt{x}+\dfrac{4}{\sqrt{x}}+1\\ \ge 2\sqrt{\sqrt{x}.\dfrac{4}{\sqrt{x}}}+1=5$

Dấu "=" xảy ra $\Leftrightarrow \sqrt{x}=\dfrac{4}{\sqrt{x}} \Leftrightarrow x=4.$