Đáp án:

$\displaystyle \begin{array}{{>{\displaystyle}l}} Câu\ 1:\\ a.A=2\\ b.\ P=\frac{\sqrt{x} -2}{\sqrt{x} +2}\\ c.\ x=\{9;1;16;0;36\}\\ Câu\ 2:\\ a.\ m=-1;\ ( x,y) =\left(\frac{41}{3} ;\frac{17}{3}\right)\\ b.\ m=\frac{23}{3} \ \ hoặc\ m=1 \end{array}$

Giải thích các bước giải:

$\displaystyle \begin{array}{{>{\displaystyle}l}} Câu\ 1:\\ a.\ x=16\Rightarrow A=\frac{4}{4-2} =2\\ b.\ A-B=\frac{\sqrt{x}}{\sqrt{x} -2} -\frac{2}{\sqrt{x} +2} -\frac{4\sqrt{x}}{x-4}\\ ĐK\ x\geqslant 0,\ x\neq 4\\ =\frac{\sqrt{x}\left(\sqrt{x} +2\right) -2\left(\sqrt{x} -2\right) -4\sqrt{x}}{\left(\sqrt{x} -2\right)\left(\sqrt{x} +2\right)}\\ =\frac{x+2\sqrt{x} -2\sqrt{x} +4-4\sqrt{x}}{\left(\sqrt{x} +2\right)\left(\sqrt{x} -2\right)}\\ =\frac{x-4\sqrt{x} +4}{\left(\sqrt{x} +2\right)\left(\sqrt{x} -2\right)} =\frac{\left(\sqrt{x} -2\right)^{2}}{\left(\sqrt{x} -2\right)\left(\sqrt{x} +2\right)}\\ =\frac{\sqrt{x} -2}{\sqrt{x} +2}\\ c.\ P=1-\frac{4}{\sqrt{x} +2}\\ Để\ P\in \mathbb{Z} \Leftrightarrow \frac{4}{\sqrt{x} +2} \in \mathbb{Z}\\ \Rightarrow \sqrt{x} +2=\{\pm 1;\pm 2;\pm 4\}\\ \Rightarrow x=\{9;1;16;0;36;4\} \ kết\ hợp\ ĐK\\ \Rightarrow x=\{9;1;16;0;36\}\\ Câu\ 2:\ \\ a.\ Với\ m=-1,\ HPT\ trở\ thành:\ \{_{x-y=8}^{-x+4y=9}\\ \Leftrightarrow x=\frac{41}{3} \ và\ y=\frac{17}{3}\\ Vậy\ m=-1;\ ( x,y) =\left(\frac{41}{3} ;\frac{17}{3}\right)\\ b.\ \\ mx+4y=9\ ( 1)\\ x+my=8\ ( 2)\\ Từ\ ( 2) \Rightarrow x=8-my\ ( **) ,\ thay\ vào\ ( 1) :\\ 8m-m^{2} y+4y=9\\ \Leftrightarrow y\left( 4-m^{2}\right) =9-8m\ ( *)\\ Xét\ m=2,\ ( *) :0=-7\ ( vô\ lí)\\ Xét\ m=-2,\ ( *) :0=25\ ( vô\ lí)\\ \Rightarrow m\neq \pm 2\\ Khi\ đó:\ y=\frac{8m-9}{m^{2} -4} ,\ thay\ vào\ ( **) :\\ x=8-\frac{8m^{2} -9m}{m^{2} -4} =\frac{9m-32}{m^{2} -4}\\ Ta\ có:\ \ 2x+y+\frac{38}{m^{2} -4} =3\\ \Leftrightarrow 2.\frac{9m-32}{m^{2} -4} +\frac{8m-9}{m^{2} -4} +\frac{38}{m^{2} -4} =3\\ \Leftrightarrow 26m-35=3\left( m^{2} -4\right)\\ \Leftrightarrow m=\frac{23}{3} \ ( TM) \ hoặc\ m=1\ ( TM) \end{array}$