Đáp án:
Giải thích các bước giải:
$4.\\a) P = \dfrac{x + 2}{x\sqrt{x} - 1} + \dfrac{\sqrt{x} + 1}{x + \sqrt{x} + 1} - \dfrac{1}{\sqrt{x} - 1}\\= \dfrac{x + 2}{x\sqrt{x} - 1} + \dfrac{(\sqrt{x} + 1) (\sqrt{x} - 1)}{x\sqrt{x} - 1} - \dfrac{x + \sqrt{x} + 1}{x\sqrt{x} - 1}\\=\dfrac{x + 2 + x - 1 - x -\sqrt{x} - 1}{x\sqrt{x} - 1}\\=\dfrac{x - \sqrt{x}}{x\sqrt{x} - 1}\\ = \dfrac{\sqrt{x}(\sqrt{x} - 1)}{(\sqrt{x} - 1)(x + \sqrt{x} + 1)}\\= \dfrac{\sqrt{x}}{x + \sqrt{x} + 1}\\b)\text{thay x = $\dfrac{1}{4}$ vào ta có:}\\P = \dfrac{\sqrt{\dfrac{1}{4}}}{\dfrac{1}{4} + \sqrt{\dfrac{1}{4}} + 1}\\ = \dfrac{2}{7}\\c) \text{để P < $\dfrac{1}{3}$:}\\\Leftrightarrow\dfrac{\sqrt{x}}{x + \sqrt{x} - 1} < \dfrac{1}{3}\\\Leftrightarrow 3\sqrt{x} < x + \sqrt{x} + 1\\\Leftrightarrow x - 2\sqrt{x} + 1 > 0\\\Leftrightarrow (\sqrt{x} - 1) > 0\\\Leftrightarrow \sqrt{x} - 1 > 0\\\Leftrightarrow \sqrt{x} < 1 \\ \Leftrightarrow x > 1$
chúc bạn học tốt T^T
