Đáp án:

$4)\\ a)M=\dfrac{\sqrt{x}+5}{\sqrt{x}+1}\\ b)M>1 \\ c)x \in \{1;9\}\\ d)x \in \left\{9;1;\dfrac{1}{9}\right\}$

Giải thích các bước giải:

$4)\\ M=\left(\dfrac{2x+3\sqrt{x}}{x\sqrt{x}+1}+\dfrac{1}{x-\sqrt{x}+1}-\dfrac{1}{\sqrt{x}+1}\right).\dfrac{x-\sqrt{x}+1}{\sqrt{x}}\\ ĐKXĐ: \left\{\begin{array}{l} x \ge 0\\ x\sqrt{x}+1 \ne 0\\ x-\sqrt{x}+1 \ne 0 \\ \sqrt{x}+1 \ne 0 \\ \sqrt{x} \ne 0\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} x \ge 0\\ x\sqrt{x}+1 \ne 0\\ x-\sqrt{x}+\dfrac{1}{4}+\dfrac{3}{4}  \ne 0 \\ \sqrt{x}+1 \ne 0 \\ \sqrt{x} \ne 0\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} x \ge 0\\ x\sqrt{x}+1 \ne 0\\ \left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}  \ne 0 \\ \sqrt{x}+1 \ne 0 \\ \sqrt{x} \ne 0\end{array} \right.\\ \Leftrightarrow x>0\\ a)M=\left(\dfrac{2x+3\sqrt{x}}{x\sqrt{x}+1}+\dfrac{1}{x-\sqrt{x}+1}-\dfrac{1}{\sqrt{x}+1}\right).\dfrac{x-\sqrt{x}+1}{\sqrt{x}}\\ =\left(\dfrac{2x+3\sqrt{x}}{\sqrt{x}^3+1^3}+\dfrac{1}{x-\sqrt{x}+1}-\dfrac{1}{\sqrt{x}+1}\right).\dfrac{x-\sqrt{x}+1}{\sqrt{x}}\\ =\left(\dfrac{2x+3\sqrt{x}}{(\sqrt{x}+1)(x-\sqrt{x}+1)}+\dfrac{1}{x-\sqrt{x}+1}-\dfrac{1}{\sqrt{x}+1}\right).\dfrac{x-\sqrt{x}+1}{\sqrt{x}}\\ =\left(\dfrac{2x+3\sqrt{x}}{(\sqrt{x}+1)(x-\sqrt{x}+1)}+\dfrac{\sqrt{x}+1}{(\sqrt{x}+1)(x-\sqrt{x}+1)}-\dfrac{x-\sqrt{x}+1}{(\sqrt{x}+1)(x-\sqrt{x}+1)}\right).\dfrac{x-\sqrt{x}+1}{\sqrt{x}}\\ =\dfrac{2x+3\sqrt{x}+\sqrt{x}+1-(x-\sqrt{x}+1)}{(\sqrt{x}+1)(x-\sqrt{x}+1)}.\dfrac{x-\sqrt{x}+1}{\sqrt{x}}\\ =\dfrac{x+5\sqrt{x}}{(\sqrt{x}+1)(x-\sqrt{x}+1)}.\dfrac{x-\sqrt{x}+1}{\sqrt{x}}\\ =\dfrac{\sqrt{x}(\sqrt{x}+5)}{(\sqrt{x}+1)(x-\sqrt{x}+1)}.\dfrac{x-\sqrt{x}+1}{\sqrt{x}}\\ =\dfrac{\sqrt{x}+5}{\sqrt{x}+1}\\ b)M-1=\dfrac{\sqrt{x}+5}{\sqrt{x}+1}-1\\ =\dfrac{\sqrt{x}+5-(\sqrt{x}+1)}{\sqrt{x}+1}\\ =\dfrac{4}{\sqrt{x}+1} >0 \ \forall \ x>0\\ \Rightarrow M>1 \ \forall \ x>0\\ c)M \in \mathbb{Z}\\ \Leftrightarrow \dfrac{\sqrt{x}+5}{\sqrt{x}+1} \in \mathbb{Z}\\ \Leftrightarrow \dfrac{\sqrt{x}+1+4}{\sqrt{x}+1} \in \mathbb{Z}\\ \Leftrightarrow 1+\dfrac{4}{\sqrt{x}+1} \in \mathbb{Z}\\ \Rightarrow \dfrac{4}{\sqrt{x}+1} \in \mathbb{Z}\\ \Rightarrow (\sqrt{x}+1) \in Ư(4)\\ \Leftrightarrow (\sqrt{x}+1) \in \{\pm 1; \pm 2; \pm 4\}\\ \Rightarrow x \in \{0;1;9\}$

Kết hợp điều kiện $\Rightarrow x \in \{1;9\}$

$d)M=1+\dfrac{4}{\sqrt{x}+1} \in \mathbb{Z}\\ \Rightarrow \dfrac{4}{\sqrt{x}+1} \in \mathbb{Z}\\ \sqrt{x}+1 >1\ \forall \ x>0 \Rightarrow \dfrac{4}{\sqrt{x}+1}<4 \ \forall \ x>0\\ \sqrt{x}+1 >0 \ \forall \ x>0 \Rightarrow \dfrac{4}{\sqrt{x}+1} >0 \ \forall \ x>0\\\Rightarrow 0<\dfrac{4}{\sqrt{x}+1}<4\\ \dfrac{4}{\sqrt{x}+1} \in \mathbb{Z}\\ \Rightarrow \dfrac{4}{\sqrt{x}+1} \in \{1;2;3\}\\ \Rightarrow x \in \left\{9;1;\dfrac{1}{9}\right\}$

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