Đáp án:

$a) \text{ĐKXĐ}:  \left\{\begin{array}{l} x \ge 0 \\ x \ne 1 \end{array} \right.\\ b)M=\sqrt{x}-x\\  d)max_M=\dfrac{1}{4}\Leftrightarrow x=\dfrac{1}{4}$

$e)x$ là số chính phương.

Giải thích các bước giải:

$a)M=\left(\dfrac{\sqrt{x}-2}{x-1}-\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right).\dfrac{x^2-2x+1}{2}\\ \text{ĐKXĐ}: \left\{\begin{array}{l} x \ge 0 \\ x-1 \ne 0\\ x+2\sqrt{x}+1 \ne 0  \end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} x \ge 0 \\ x \ne 1\\ (\sqrt{x}+1)^2 \ne 0  \end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} x \ge 0 \\ x \ne 1 \end{array} \right.\\ b)M=\left(\dfrac{\sqrt{x}-2}{x-1}-\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right).\dfrac{x^2-2x+1}{2}\\ =\left(\dfrac{\sqrt{x}-2}{x-1}-\dfrac{\sqrt{x}+2}{(\sqrt{x}+1)^2}\right).\dfrac{(x-1)^2}{2}\\ =\dfrac{(\sqrt{x}-2)(\sqrt{x}+1)^2-(\sqrt{x}+2)(x-1)}{(x-1)(\sqrt{x}+1)^2}.\dfrac{(x-1)^2}{2}\\ =\dfrac{-2x-2\sqrt{x}}{(x-1)(\sqrt{x}+1)^2}.\dfrac{(x-1)^2}{2}\\ =\dfrac{-2\sqrt{x}(\sqrt{x}+1)}{(\sqrt{x}+1)(\sqrt{x}-1)(\sqrt{x}+1)^2}.\dfrac{(\sqrt{x}+1)^2(\sqrt{x}-1)^2}{2}\\ =-\sqrt{x}(\sqrt{x}-1)\\ =\sqrt{x}-x\\ c)0<x<1\\ \Rightarrow \sqrt{x}>x\\ \Leftrightarrow \sqrt{x}-x>0\\ \Leftrightarrow M>0\\ d)M=\sqrt{x}-x\\ =-x+\sqrt{x}-\dfrac{1}{4}+\dfrac{1}{4}\\ =-\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{1}{4} \le \dfrac{1}{4} \ \forall \ x$

Dấu "=" xảy ra $\Leftrightarrow \sqrt{x}-\dfrac{1}{2}=0 \Leftrightarrow x=\dfrac{1}{4}$

$e)M=\sqrt{x}-x \in \mathbb{Z}\\ x \in \mathbb{Z} \Rightarrow \sqrt{x} \in \mathbb{Z}$

$\Rightarrow x$ là số chính phương.