Đáp án:

$3)\\ a) ĐKXĐ: \left\{\begin{array}{l} x > 0\\ x \ne 1 \\ x \ne 4 \end{array} \right.\\ B=\dfrac{\sqrt{x}-2}{3\sqrt{x}}\\ b)B(9-4\sqrt{5})=-\dfrac{3+2\sqrt{5}}{3}\\ c)\Leftrightarrow x>4\\ 4)\\a) \forall \ m \ne 2, m \ne 3\\ b)\left[\begin{array}{l} m>3\\ m<2\end{array} \right.\\ c) \Leftrightarrow  2<m<3$

Giải thích các bước giải:

$3)\\ B=\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}}\right):\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}-\dfrac{\sqrt{x}+2}{\sqrt{x}-1}\right)\\ a)ĐKXĐ:  \left\{\begin{array}{l} x \ge 0\\ \sqrt{x}-1 \ne 0 \\ \sqrt{x} \ne 0 \\ \sqrt{x}-2 \ne 0\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} x \ge 0\\ \sqrt{x} \ne 1 \\ \sqrt{x} \ne 0 \\ \sqrt{x} \ne 2 \end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} x \ge 0\\ x \ne 1 \\ x \ne 0 \\ x \ne 4 \end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} x > 0\\ x \ne 1 \\ x \ne 4 \end{array} \right.\\ B=\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}}\right):\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}-\dfrac{\sqrt{x}+2}{\sqrt{x}-1}\right)\\ =\left(\dfrac{\sqrt{x}}{(\sqrt{x}-1)\sqrt{x}}-\dfrac{\sqrt{x}-1}{\sqrt{x}(\sqrt{x}-1)}\right):\left(\dfrac{(\sqrt{x}+1)(\sqrt{x}-1)}{(\sqrt{x}-2)(\sqrt{x}-1)}-\dfrac{(\sqrt{x}+2)(\sqrt{x}-2)}{(\sqrt{x}-1)(\sqrt{x}-2)}\right)\\ =\dfrac{\sqrt{x}-(\sqrt{x}-1)}{\sqrt{x}(\sqrt{x}-1)}:\dfrac{(\sqrt{x}+1)(\sqrt{x}-1)-(\sqrt{x}+2)(\sqrt{x}-2)}{(\sqrt{x}-1)(\sqrt{x}-2)}\\ =\dfrac{1}{\sqrt{x}(\sqrt{x}-1)}:\dfrac{x-1-(x-4)}{(\sqrt{x}-1)(\sqrt{x}-2)}\\ =\dfrac{1}{\sqrt{x}(\sqrt{x}-1)}:\dfrac{3}{(\sqrt{x}-1)(\sqrt{x}-2)}\\ =\dfrac{1}{\sqrt{x}(\sqrt{x}-1)}.\dfrac{(\sqrt{x}-1)(\sqrt{x}-2)}{3}\\ =\dfrac{\sqrt{x}-2}{3\sqrt{x}}\\ b)x=9-4\sqrt{5}\\ =5-2.2.\sqrt{5}+4\\ =(\sqrt{5}-2)^2\\ B(9-4\sqrt{5})\\ =\dfrac{\sqrt{(\sqrt{5}-2)^2}-2}{3\sqrt{(\sqrt{5}-2)^2}}\\ =\dfrac{(\sqrt{5}-2)-2}{3(\sqrt{5}-2)}\\ =\dfrac{\sqrt{5}-4}{3(\sqrt{5}-2)}\\ =\dfrac{(\sqrt{5}-4)(\sqrt{5}+2)}{3(\sqrt{5}-2)(\sqrt{5}+2)}\\ =-\dfrac{3+2\sqrt{5}}{3}\\ c)B>0\\ \Leftrightarrow \dfrac{\sqrt{x}-2}{3\sqrt{x}}>0\\ \Leftrightarrow \sqrt{x}-2>0 (Do \ \ 3\sqrt{x}>0 \ \forall x>0, x \ne 1, x\ ne 4)\\ \Leftrightarrow \sqrt{x}>2\\ \Leftrightarrow x>4\\ 4)\\ y=\dfrac{1}{m^2-5m+6}x+3\\ ĐKXĐ: m^2-5m+6 \ne 0\\ \Leftrightarrow m^2-2m-3m+6 \ne 0\\ \Leftrightarrow m(m-2)-3(m-2) \ne 0\\ \Leftrightarrow (m-3)(m-2)\ne 0\\ \Leftrightarrow m \ne 2, m \ne 3\\ a)\dfrac{1}{m^2-5m+6} \ne 0 \ \forall \ m \ne 2, m \ne 3$

$\Rightarrow $Hàm số luôn là hàm bậc nhất với $\ \forall \ m \ne 2, m \ne 3$

$b)$Hàm số đồng biến

$\Rightarrow \dfrac{1}{m^2-5m+6}>0\\ \Leftrightarrow m^2-5m+6>0\\ \Leftrightarrow (m-3)(m-2)>0\\ \Leftrightarrow  \left[\begin{array}{l} m>3\\ m<2\end{array} \right.$

$c)$Hàm số nghịch biến

$\Rightarrow \dfrac{1}{m^2-5m+6}<0\\ \Leftrightarrow m^2-5m+6<0\\ \Leftrightarrow (m-3)(m-2)<0\\ \Leftrightarrow  2<m<3\\ d)y=\dfrac{3}{2}x(d)\\ x=2 \Rightarrow y=3 \Rightarrow A(2;3) \in (d)\\ x=4 \Rightarrow y=6 \Rightarrow B(4;6) \in (d)$

Vẽ đường thẳng đi qua 2 điểm $A,B$ được đồ thị hàm số $y=\dfrac{3}{2}x$

$y=-x(d')\\ x=1 \Rightarrow y=-1 \Rightarrow C(1;-1) \in (d')\\ x=2 \Rightarrow y=-2 \Rightarrow D(2;-2) \in (d')$

Vẽ đường thẳng đi qua 2 điểm $C,D$ được đồ thị hàm số $y=-x$