Đáp án:

$a)M=\dfrac{3\sqrt{x}}{\sqrt{x}-3} \\ b)M\left(\sqrt{\sqrt{3}-\sqrt{4-2\sqrt{3}}}\right)=-\dfrac{3}{2}\\ c) \left[\begin{array}{l} x=0 \\ x=4\\x=16\\ x=36\\x=144\end{array} \right.\\ d) 0 < x <36; x\ \ne 9$

Giải thích các bước giải:

$a)M=\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}+1}{\sqrt{x}-3}+\dfrac{11\sqrt{x}-3}{x-9} (x>0;x \ne 9)\\ =\dfrac{2\sqrt{x}(\sqrt{x}-3)}{(\sqrt{x}+3)(\sqrt{x}-3)}+\dfrac{(\sqrt{x}+1)(\sqrt{x}+3)}{(\sqrt{x}-3)(\sqrt{x}+3)}+\dfrac{11\sqrt{x}-3}{(\sqrt{x}-3)(\sqrt{x}+3)} \\ =\dfrac{2\sqrt{x}(\sqrt{x}-3)+(\sqrt{x}+1)(\sqrt{x}+3)+11\sqrt{x}-3}{(\sqrt{x}-3)(\sqrt{x}+3)} \\ =\dfrac{3x+9\sqrt{x}}{(\sqrt{x}-3)(\sqrt{x}+3)} \\ =\dfrac{3\sqrt{x}(\sqrt{x}+3)}{(\sqrt{x}-3)(\sqrt{x}+3)} \\ =\dfrac{3\sqrt{x}}{\sqrt{x}-3} \\ b)x=\sqrt{\sqrt{3}-\sqrt{4-2\sqrt{3}}}\\ =\sqrt{\sqrt{3}-\sqrt{3-2\sqrt{3}+1}}\\ =\sqrt{\sqrt{3}-\sqrt{(\sqrt{3}-1)^2}}\\ =\sqrt{\sqrt{3}-(\sqrt{3}-1)}\\ =\sqrt{\sqrt{3}-\sqrt{3}+1}\\ =\sqrt{1}\\ =1\\ M\left(\sqrt{\sqrt{3}-\sqrt{4-2\sqrt{3}}}\right)=\dfrac{3\sqrt{1}}{\sqrt{1}-3} =-\dfrac{3}{2}\\ c)M=\dfrac{3\sqrt{x}}{\sqrt{x}-3} \in \mathbb{N}\\ \Leftrightarrow \dfrac{3\sqrt{x}}{\sqrt{x}-3} \in \mathbb{N}\\ \Leftrightarrow \dfrac{3\sqrt{x}-9+9}{\sqrt{x}-3} \in \mathbb{N}\\ \Leftrightarrow \dfrac{3(\sqrt{x}-3)+9}{\sqrt{x}-3} \in \mathbb{N}\\ \Leftrightarrow 3+\dfrac{9}{\sqrt{x}-3} \in \mathbb{N}\\ \Rightarrow \dfrac{9}{\sqrt{x}-3} \in \mathbb{N}\\ x \in \mathbb{N} \Rightarrow (\sqrt{x}-3)\in Ư(9)\\ \Leftrightarrow (\sqrt{x}-3)\in \{\pm1;\pm3;\pm9\}\\ \Leftrightarrow \left[\begin{array}{l} \sqrt{x}-3=-9\\ \sqrt{x}-3=-3 \\ \sqrt{x}-3=-1\\\sqrt{x}-3=1\\ \sqrt{x}-3=3\\ \sqrt{x}-3=9\end{array} \right.\\ \Leftrightarrow \left[\begin{array}{l} \sqrt{x}=-6\\ \sqrt{x}=0 \\ \sqrt{x}=2\\\sqrt{x}=4\\ \sqrt{x}=6\\ \sqrt{x}=12\end{array} \right.\\ \Leftrightarrow \left[\begin{array}{l} x=0 \\ x=4\\x=16\\ x=36\\x=144\end{array} \right.\\ d)\dfrac{1}{M}<\dfrac{1}{6}\\ \Leftrightarrow \dfrac{\sqrt{x}-3} {3\sqrt{x}}<\dfrac{1}{6}\\ \Leftrightarrow \dfrac{\sqrt{x}-3} {3\sqrt{x}}-\dfrac{1}{6}<0\\ \Leftrightarrow \dfrac{2(\sqrt{x}-3)-\sqrt{x}} {6\sqrt{x}}<0\\ \Leftrightarrow \dfrac{\sqrt{x}-6} {6\sqrt{x}}<0\\ \Leftrightarrow \sqrt{x}-6<0\\ \Leftrightarrow \sqrt{x}<6\\ \Leftrightarrow 0 \le x <36\\ \text{Kết hợp điều kiện}\Rightarrow 0 < x <36; x\ \ne 9$