Bài 2:
$\begin{array}{l}P = \dfrac{1}{\sqrt{x} - 1} + \dfrac{\sqrt{x}}{x- 1};\,\,\,\,\,\,\,Q = \dfrac{\sqrt{x}}{\sqrt{x} - 1} - 1\\a) \, x = 4 \Rightarrow \sqrt{x} = 2\\\Rightarrow Q = \dfrac{2}{2 - 1} - 1= 1\\b) \, ĐK: \, x\geq 0 ;\, x \ne 1\\P = \dfrac{\sqrt{x} + 1 + \sqrt{x}}{x - 1}=\dfrac{2\sqrt{x} +1}{x - 1}\\Q = \dfrac{\sqrt{x} - (\sqrt{x} - 1)}{\sqrt{x} -1}=\dfrac{1}{\sqrt{x} - 1}\\\Rightarrow M = P:Q = \left(\dfrac{2\sqrt{x} +1}{x - 1}\right):\left(\dfrac{1}{\sqrt{x} - 1}\right)\\=\dfrac{2\sqrt{x} +1}{(\sqrt{x} - 1)(\sqrt{x} + 1)}\cdot\left(\sqrt{x} - 1\right)\\=\dfrac{2\sqrt{x}+1}{\sqrt{x} + 1}\\c) \, M < \dfrac{3}{2}\\\Leftrightarrow \dfrac{2\sqrt{x}+1}{\sqrt{x} + 1} < \dfrac{3}{2}\\\Leftrightarrow 4\sqrt{x} + 2 < 3\sqrt{x} +3\\\Leftrightarrow \sqrt{x} < 1\\\Leftrightarrow x < 1\\\text{Kết hợp điều kiện, ta được: } \, 0 \leq x \leq 1\end{array}$
$\\$
Bài 3:
$\begin{array}{l}a) \,M = \left(\dfrac{\sqrt{x} + 2}{x + 2\sqrt{x} + 1}-\dfrac{\sqrt{x} - 2}{x - 1}\right)\cdot \dfrac{\sqrt{x} + 1}{\sqrt{x}}\\ĐK: x > 0; \, x \ne 1\\M =\left(\dfrac{\sqrt{x} + 2}{(\sqrt{x} + 1)^2}-\dfrac{\sqrt{x} - 2}{(\sqrt{x} - 1)(\sqrt{x} + 1)}\right)\cdot \dfrac{\sqrt{x} + 1}{\sqrt{x}}\\=\dfrac{(\sqrt{x} + 2)(\sqrt{x} - 1)-(\sqrt{x} - 2)(\sqrt{x} + 1)}{(\sqrt{x} - 1)(\sqrt{x} + 1)^2}\cdot \dfrac{\sqrt{x} + 1}{\sqrt{x}}\\=\dfrac{x+\sqrt{x} - 2-(x - \sqrt{x} - 2)}{(\sqrt{x} - 1)(\sqrt{x} + 1)^2}\cdot \dfrac{\sqrt{x} + 1}{\sqrt{x}}\\=\dfrac{2\sqrt{x}}{(\sqrt{x} - 1)(\sqrt{x} + 1)^2}\cdot \dfrac{\sqrt{x} + 1}{\sqrt{x}}\\=\dfrac{2}{(\sqrt{x} - 1)(\sqrt{x} + 1)}=\dfrac{2}{x - 1}\\b) \,M < -1\\\Leftrightarrow \dfrac{2}{x -1}< - 1\\\Leftrightarrow 2 < 1 -x\\\Leftrightarrow x < -1\\Mà \, x > 0; \, x \ne 1\\nên \, x \in \emptyset \\\text{Vậy không có x thỏa mãn đề bài}\end{array}$
