$\displaystyle \begin{array}{{>{\displaystyle}l}} 2.\\ a.\ ĐKXĐ:\ x\geqslant 0;\ x\neq 1\\ P=\frac{\left(\sqrt{x} +2\right)\left(\sqrt{x} +1\right) -\left(\sqrt{x} -2\right)\left(\sqrt{x} -1\right)}{\left(\sqrt{x} -1\right)^{2}\left(\sqrt{x} +1\right)} .\frac{x-1}{x+1-x+1}\\ P=\frac{x+3\sqrt{x} +2-x+3\sqrt{x} -2}{\left(\sqrt{x} -1\right)^{2}\left(\sqrt{x} +1\right)} .\frac{x-1}{2}\\ P=\frac{6\sqrt{x}}{2\left(\sqrt{x} -1\right)} =\frac{3\sqrt{x}}{\sqrt{x} -1}\\ |2\sqrt{x} -1|=3\\ TH1:\ 2\sqrt{x} -1=3\Rightarrow \sqrt{x} =2\Rightarrow P=\frac{3.2}{2-1} =6\\ TH\ 2:2\sqrt{x} -1=-3\Rightarrow \sqrt{x} =-1( vô\ lí)\\ b.\ P=3+\frac{3}{\sqrt{x} -1} \in \mathbb{Z} \Leftrightarrow \frac{3}{\sqrt{x} -1} \in \mathbb{Z} \Leftrightarrow \sqrt{x} -1=\{3\} \Leftrightarrow x=16\ ( TM)\\ 3.\\ Gọi\ độ\ dài\ 2\ cạnh\ \ góc\ vuông\ lần\ lượt\ là\ a\ và\ b\\ ( a >0;\ b >0)\\ Theo\ bài\ ra\ ta\ có\ \\ +) \ ( a+3)( b+3) =ab+36\\ \Leftrightarrow ab+3( a+b) +9=ab+36\\ \Leftrightarrow 3( a+b) =27\ ( 1)\\ +) \ ( a-2)( b-4) =ab-26\\ \Leftrightarrow ab-2b-4a+8=ab-26\\ \Leftrightarrow -4a-2b=-34\ ( 2)\\ Từ\ ( 1)( 2) \Rightarrow a=8;\ b=1\\ Vậy\ độ\ dài\ 2\ cạnh\ góc\ vuông\ lần\ lượt\ là\ 8\ và\ 1\\ 4.\\ \Delta =( m+4)^{2} -4\left( -2m^{2} +5m+3\right) =m^{2} -8m+16+8m^{2} -20m-12\\ =9m^{2} -28m+4\\ Để\ PT\ có\ 2\ nghiệm\ \Leftrightarrow 9m^{2} -28m+4\geqslant 0\Leftrightarrow m\geqslant \frac{14+4\sqrt{10}}{9} ;\ m\leqslant \frac{14-4\sqrt{10}}{9} \ ( *)\\ Theo\ Viet:\ x_{1} +x_{2} =m+4;\ x_{1} x_{2} =-2m^{2} +5m+3\\ Ta\ có\ x_{1} < 3< x_{2}\\ \Rightarrow ( x_{1} -3)( x_{2} -3) < 0\\ \Leftrightarrow x_{1} x_{2} -3( x_{1} +x_{2}) +9< 0\\ \Leftrightarrow -2m^{2} +5m+3-3m-12+9< 0\\ \Leftrightarrow -2m^{2} +2m< 0\\ \Leftrightarrow m >1;\ m< 0\ kết\ hợp\ ( *)\\ \Rightarrow m\geqslant \frac{14+4\sqrt{10}}{9} ;\ \ m< 0\\ \\ \end{array}$