Đáp án:
$1)\\ a)\text{ĐKXĐ: } \left\{\begin{array}{l} x \ge 0\\ x \ne 4\end{array} \right.\\ b)x=16, A=3\\ c) 0 \le x <4\\ d) x>4\\ e) \left[\begin{array}{l} x=0\\ x=1 \\ x=9 \\ x=16 \\ x =36\end{array} \right.\\ f)min_B=7 \Leftrightarrow x=16$
Giải thích các bước giải:
$1)\\ A=\dfrac{\sqrt{x}+2}{\sqrt{x}-2}\\ a)\text{ĐKXĐ: }\left\{\begin{array}{l} x \ge 0\\ \sqrt{x}-2 \ne 0\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} x \ge 0\\ \sqrt{x} \ne 2\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} x \ge 0\\ x \ne 4\end{array} \right.\\ b)x=16, A=\dfrac{\sqrt{16}+2}{\sqrt{16}-2}=\dfrac{4+2}{4-2}=\dfrac{6}{2}=3\\ c)A<1\\ \Leftrightarrow \dfrac{\sqrt{x}+2}{\sqrt{x}-2}<1\\ \Leftrightarrow \dfrac{\sqrt{x}+2}{\sqrt{x}-2}-1<0\\ \Leftrightarrow \dfrac{\sqrt{x}+2-(\sqrt{x}-2)}{\sqrt{x}-2}<0\\ \Leftrightarrow \dfrac{4}{\sqrt{x}-2}<0\\ \Leftrightarrow \sqrt{x}-2<0\\ \Leftrightarrow \sqrt{x}<2\\ \Leftrightarrow 0 \le x <4\\ d)A=|A| \Leftrightarrow A \ge 0\\ \Leftrightarrow \dfrac{\sqrt{x}+2}{\sqrt{x}-2} \ge 0\\ \Leftrightarrow \sqrt{x}-2 > 0(\text{Do }\sqrt{x}+2 >0 \ \forall \ x \ge 0, x \ne 4)\\ \Leftrightarrow \sqrt{x}>2\\ \Leftrightarrow x>4\\ e)A\in \mathbb{Z}\\ \Leftrightarrow \dfrac{\sqrt{x}+2}{\sqrt{x}-2} \in \mathbb{Z}\\ \Leftrightarrow \dfrac{\sqrt{x}-2+4}{\sqrt{x}-2} \in \mathbb{Z}\\ \Leftrightarrow 1+\dfrac{4}{\sqrt{x}-2} \in \mathbb{Z}\\ \Rightarrow \dfrac{4}{\sqrt{x}-2} \in \mathbb{Z}\\ x \in \mathbb{Z} \Rightarrow (\sqrt{x}-2) \in Ư(4)\\ \Leftrightarrow (\sqrt{x}-2) \in \{\pm 1 ; \pm 2; \pm 4\}\\ \Leftrightarrow \left[\begin{array}{l} \sqrt{x}-2 =-4 \\ \sqrt{x}-2 =-2\\ \sqrt{x}-2 =-1 \\ \sqrt{x}-2 =1 \\ \sqrt{x}-2 =2 \\ \sqrt{x}-2 =4\end{array} \right. \Leftrightarrow \left[\begin{array}{l} \sqrt{x} =-2 \\ \sqrt{x}=0\\ \sqrt{x}=1 \\ \sqrt{x} =3 \\ \sqrt{x} =4 \\ \sqrt{x} =6\end{array} \right. \Leftrightarrow \left[\begin{array}{l} x=0\\ x=1 \\ x=9 \\ x=16 \\ x =36\end{array} \right.\\ f)B=\sqrt{x}+A\\ =\sqrt{x}+\dfrac{\sqrt{x}+2}{\sqrt{x}-2}\\ =\dfrac{\sqrt{x}(\sqrt{x}-2)+\sqrt{x}+2}{\sqrt{x}-2}\\ =\dfrac{x-\sqrt{x}+2}{\sqrt{x}-2}\\ =\dfrac{x-4\sqrt{x}+4+3\sqrt{x}-6+4}{\sqrt{x}-2}\\ =\dfrac{(\sqrt{x}-2)^2+3(\sqrt{x}-2)+4}{\sqrt{x}-2}\\ =(\sqrt{x}-2)+3+\dfrac{4}{\sqrt{x}-2}\\ =(\sqrt{x}-2)+\dfrac{4}{\sqrt{x}-2}+3(x >0, \sqrt{x}-2>0)\\ \ge 2\sqrt{(\sqrt{x}-2).\dfrac{4}{\sqrt{x}-2}}+3=2.2+3=7$
Dấu "=" xảy ra $\Leftrightarrow \sqrt{x}-2=\dfrac{4}{\sqrt{x}-2}$
$\Leftrightarrow (\sqrt{x}-2)^2=4\\ \Leftrightarrow x-4\sqrt{x}+4=4\\ \Leftrightarrow x-4\sqrt{x}=0\\ \Leftrightarrow \sqrt{x}(\sqrt{x}-4)=0\\ \Leftrightarrow \left[\begin{array}{l} x=0(L)\\ x=16\end{array} \right.$
