Đáp án:
$a)E=\dfrac{x}{\sqrt{x}-1}\\ b)x>1\\ c)min_E=4 \Leftrightarrow x=4\\ d) x \in \{4\}\\ e)E(2)=2\sqrt{2}+2\\ g) \left[\begin{array}{l} x=9\\ x=\dfrac{9}{4}\end{array} \right.$
Giải thích các bước giải:
$E=\dfrac{x+\sqrt{x}}{x-2\sqrt{x}+1}:\left(\dfrac{\sqrt{x}+1}{\sqrt{x}}-\dfrac{1}{1-\sqrt{x}}+\dfrac{2-x}{x-\sqrt{x}}\right)\\ ĐKXĐ: \left\{\begin{array}{l} x \ge 0\\ x-2\sqrt{x}+1 \ne 0 \\ \sqrt{x} \ne 0 \\ 1-\sqrt{x} \ne 0 \\ x-\sqrt{x} \ne 0 \\\left(\dfrac{\sqrt{x}+1}{\sqrt{x}}-\dfrac{1}{1-\sqrt{x}}+\dfrac{2-x}{x-\sqrt{x}}\right) \ne 0\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} x \ge 0\\ (\sqrt{x}-1)^2 \ne 0 \\ \sqrt{x} \ne 0 \\ \sqrt{x} \ne 1 \\ \sqrt{x}(\sqrt{x}-1) \ne 0 \\\left(\dfrac{(\sqrt{x}+1)(\sqrt{x}-1)}{\sqrt{x}(\sqrt{x}-1)}+\dfrac{\sqrt{x}}{(\sqrt{x}-1)\sqrt{x}}+\dfrac{2-x}{\sqrt{x}(\sqrt{x}-1)}\right) \ne 0\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} x > 0\\ x \ne 1 \\\dfrac{(\sqrt{x}+1)(\sqrt{x}-1)+\sqrt{x}+2-x}{\sqrt{x}(\sqrt{x}-1)}\ \ne 0\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} x > 0\\ x \ne 1 \\\dfrac{\sqrt{x}+1}{\sqrt{x}(\sqrt{x}-1)}\ \ne 0\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} x > 0\\ x \ne 1 \end{array} \right.\\ a)E=\dfrac{x+\sqrt{x}}{x-2\sqrt{x}+1}:\left(\dfrac{\sqrt{x}+1}{\sqrt{x}}-\dfrac{1}{1-\sqrt{x}}+\dfrac{2-x}{x-\sqrt{x}}\right)\\ =\dfrac{\sqrt{x}(\sqrt{x}+1)}{(\sqrt{x}-1)^2}:\dfrac{\sqrt{x}+1}{\sqrt{x}(\sqrt{x}-1)}\\ =\dfrac{\sqrt{x}(\sqrt{x}+1)}{(\sqrt{x}-1)^2}.\dfrac{\sqrt{x}(\sqrt{x}-1)}{\sqrt{x}+1}\\ =\dfrac{x}{\sqrt{x}-1}\\ b)E>1\\ \Leftrightarrow \dfrac{x}{\sqrt{x}-1}>1\\ \Leftrightarrow \dfrac{x}{\sqrt{x}-1}-1>0\\ \Leftrightarrow \dfrac{x-\sqrt{x}+1}{\sqrt{x}-1}>0\\ \Leftrightarrow \dfrac{x-\sqrt{x}+\dfrac{1}{4}+\dfrac{3}{4}}{\sqrt{x}-1}>0\\ \Leftrightarrow \dfrac{\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}}{\sqrt{x}-1}>0\\ \Leftrightarrow \sqrt{x}-1>0\\ \Leftrightarrow x>1\\ c)x>1 \Rightarrow \sqrt{x}-1>0\\ E=\dfrac{x}{\sqrt{x}-1}\\ =\dfrac{x-2\sqrt{x}+1+2\sqrt{x}-2+1}{\sqrt{x}-1}\\ =\dfrac{(\sqrt{x}-1)^2+2(\sqrt{x}-1)+1}{\sqrt{x}-1}\\ =\sqrt{x}-1+2+\dfrac{1}{\sqrt{x}-1}\\ =\sqrt{x}-1+\dfrac{1}{\sqrt{x}-1}+2\\ \ge 2\sqrt{(\sqrt{x}-1).\dfrac{1}{\sqrt{x}-1}}+2(Cauchy)\\ =4\\ Dấu "=" xảy ra \Leftrightarrow \sqrt{x}-1=\dfrac{1}{\sqrt{x}-1} \Leftrightarrow x=4\\ d)E=\dfrac{x}{\sqrt{x}-1} \in \mathbb{Z}\\ \Leftrightarrow \dfrac{x-\sqrt{x}+\sqrt{x}-1+1}{\sqrt{x}-1} \in \mathbb{Z}\\ \Leftrightarrow \dfrac{\sqrt{x}(\sqrt{x}-1)+\sqrt{x}-1+1}{\sqrt{x}-1} \in \mathbb{Z}\\ \Leftrightarrow \sqrt{x}+1+\dfrac{1}{\sqrt{x}-1} \in \mathbb{Z}\\ x in \mathbb{Z} \Rightarrow \dfrac{1}{\sqrt{x}-1} \in \mathbb{Z}\\ x in \mathbb{Z} \Rightarrow (\sqrt{x}-1) \in Ư(1)\\ \Leftrightarrow (\sqrt{x}-1) \in \{pm 1\}\\ \Rightarrow x \in \{0;4\}\\ \text{Kết hợp điều kiện} \Rightarrow x \in \{4\}\\ e)|2x+1|=5\\ \Leftrightarrow \left[\begin{array}{l} 2x+1=5 \\ 2x+1 =-5\end{array} \right.\\ \Leftrightarrow \left[\begin{array}{l} 2x=4 \\ 2x=-6\end{array} \right.\\ \Leftrightarrow \left[\begin{array}{l} x=2 \\ x=-3\end{array} \right.\\ E(2)=\dfrac{2}{\sqrt{2}-1}=\dfrac{2(\sqrt{2}+1)}{(\sqrt{2}-1)(\sqrt{2}+1)}=2\sqrt{2}+2\\ \text{$x=-3$ làm cho E không xác định}\\ g)E=\dfrac{9}{2}\\ \Leftrightarrow \dfrac{x}{\sqrt{x}-1}=\dfrac{9}{2}\\ \Leftrightarrow \dfrac{x}{\sqrt{x}-1}-\dfrac{9}{2}=0\\ \Leftrightarrow \dfrac{2x-9(\sqrt{x}-1)}{2(\sqrt{x}-1)}=0\\ \Leftrightarrow \dfrac{2x-9\sqrt{x}+9}{2(\sqrt{x}-1)}=0\\ \Leftrightarrow \dfrac{2x-6\sqrt{x}-3\sqrt{x}+9}{2(\sqrt{x}-1)}=0\\ \Leftrightarrow \dfrac{2\sqrt{x}(\sqrt{x}-3)-3(\sqrt{x}-3)}{2(\sqrt{x}-1)}=0\\ \Leftrightarrow \dfrac{(\sqrt{x}-3)(2\sqrt{x}-3)}{2(\sqrt{x}-1)}=0\\ \Leftrightarrow \left[\begin{array}{l} \sqrt{x}-3=0\\ 2\sqrt{x}-3=0\end{array} \right.\\ \Leftrightarrow \left[\begin{array}{l} x=9\\ x=\dfrac{9}{4}\end{array} \right.$
