Đáp án+Giải thích các bước giải:
$a.\\N = \dfrac{\sqrt{x} +1}{\sqrt{x} -3}\\\text{x=49} \Rightarrow N = \dfrac{\sqrt{49}+1}{\sqrt{49}-3}\\ \Leftrightarrow N = \dfrac{7+1}{7-3}\\N = {2}\\b.\\M = \dfrac{2\sqrt{x}}{\sqrt{x} +3} -\dfrac{\sqrt{x}}{3-\sqrt{x}}-\dfrac{3x+3}{x-9}\\M = \dfrac{2\sqrt{a}}{\sqrt{x}+3} +\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+3}{(\sqrt{x} -3)(\sqrt{x}+3)}\\M = \dfrac{2\sqrt{x}(\sqrt{x} -3)+(\sqrt{x} +3)\sqrt{x} -(3x+3)}{(\sqrt{x}-3)(\sqrt{x}+3)}\\M = \dfrac{2x-6\sqrt{x} + x + 3\sqrt{x} -3x -3}{x-9}\\M = -\dfrac{3\sqrt{x} +3}{x-9}\\c.\\M÷N> -\dfrac{1}{2}\\\Rightarrow -\dfrac{3\sqrt{x} +3}{x-9}÷\dfrac{\sqrt{x}+1}{\sqrt{x} -3}>-\dfrac{1}{2}\\(x \ne 9, x \ge 0)\\\Leftrightarrow -\dfrac{3(\sqrt{x} +1)}{(\sqrt{x}-3)(\sqrt{x}+3)}\times \dfrac{\sqrt{x} -3}{\sqrt{x} +1} \ge -\dfrac{1}{2}\\\Leftrightarrow -\dfrac{3}{\sqrt{x} +3} > -\dfrac{1}{2}\\\Leftrightarrow \dfrac{-6+\sqrt{x} +3}{2(\sqrt{x} +3)} >0\\\Leftrightarrow -3 +\sqrt{x} >0\\\Leftrightarrow x >9$
