Đáp án:
$4)\\ a)A(9)\dfrac{6}{5}\\ b)B=\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\\ c) x>9\\ 5)\\ a)P=\dfrac{\sqrt{x}+2}{\sqrt{x}-1}\\ b) x>1, x \ne 4\\ c)P\left(\dfrac{1}{4}\right)=-5$
Giải thích các bước giải:
$4)\\ a)A=\dfrac{x+3}{2\sqrt{x}+4}\\ A(9)=\dfrac{9+3}{2\sqrt{9}+4}\\ =\dfrac{12}{2.3+4}\\ =\dfrac{6}{5}\\ b)B=\left(\dfrac{\sqrt{x}}{x-4}+\dfrac{1}{\sqrt{x}-2}\right):\dfrac{2}{\sqrt{x}-2}\\ =\left(\dfrac{\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}+2)}+\dfrac{1}{\sqrt{x}-2}\right):\dfrac{2}{\sqrt{x}-2}\\ =\left(\dfrac{\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}+2)}+\dfrac{\sqrt{x}+2}{(\sqrt{x}-2)(\sqrt{x}+2)}\right).\dfrac{\sqrt{x}-2}{2}\\ =\dfrac{\sqrt{x}+\sqrt{x}+2}{(\sqrt{x}-2)(\sqrt{x}+2)}.\dfrac{\sqrt{x}-2}{2}\\ =\dfrac{2\sqrt{x}+2}{(\sqrt{x}-2)(\sqrt{x}+2)}.\dfrac{\sqrt{x}-2}{2}\\ =\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\\ c)\dfrac{A}{B}=\dfrac{\dfrac{x+3}{2\sqrt{x}+4}}{\dfrac{\sqrt{x}+1}{\sqrt{x}+2}}\\ =\dfrac{x+3}{2\sqrt{x}+4}.\dfrac{\sqrt{x}+2}{\sqrt{x}+1}\\ =\dfrac{x+3}{2(\sqrt{x}+1)}\\ \dfrac{x+3}{2(\sqrt{x}+1)}>\dfrac{3}{2}\\ \Leftrightarrow \dfrac{x+3}{2(\sqrt{x}+1)}-\dfrac{3}{2}>0\\ \Leftrightarrow \dfrac{2(x+3)-3.2(\sqrt{x}+1)}{4(\sqrt{x}+1)}>0\\ \Leftrightarrow \dfrac{2x-6\sqrt{x}}{4(\sqrt{x}+1)}>0\\ \Leftrightarrow \dfrac{x-3\sqrt{x}}{2(\sqrt{x}+1)}>0\\ \Leftrightarrow x-3\sqrt{x}>0\\ \Leftrightarrow \sqrt{x}(\sqrt{x}-3)>0\\ \Leftrightarrow \sqrt{x}-3>0(Do \ \sqrt{x}>0 \ \forall \ x \ne 4)\\ \Leftrightarrow \sqrt{x}>3\\ \Leftrightarrow x>9\\ 5)\\ P=\dfrac{\sqrt{x}}{\sqrt{x}-2}:\left(\dfrac{x-2}{x-4}-\dfrac{1}{\sqrt{x}+2}\right)\\ =\dfrac{\sqrt{x}}{\sqrt{x}-2}:\left(\dfrac{x-2}{(\sqrt{x}-2)(\sqrt{x}+2)}-\dfrac{1}{\sqrt{x}+2}\right) \\ =\dfrac{\sqrt{x}}{\sqrt{x}-2}:\left(\dfrac{x-2}{(\sqrt{x}-2)(\sqrt{x}+2)}-\dfrac{\sqrt{x}-2}{(\sqrt{x}-2)(\sqrt{x}+2)}\right) \\ =\dfrac{\sqrt{x}}{\sqrt{x}-2}:\dfrac{x-2-(\sqrt{x}-2)}{(\sqrt{x}-2)(\sqrt{x}+2)}\\ =\dfrac{\sqrt{x}}{\sqrt{x}-2}:\dfrac{x-\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}+2)}\\ =\dfrac{\sqrt{x}}{\sqrt{x}-2}.\dfrac{(\sqrt{x}-2)(\sqrt{x}+2)}{\sqrt{x}(\sqrt{x}-1)}\\ =\dfrac{\sqrt{x}+2}{\sqrt{x}-1}\\ b)P>1\\ \Leftrightarrow \dfrac{\sqrt{x}+2}{\sqrt{x}-1}>1\\ \Leftrightarrow \dfrac{\sqrt{x}+2}{\sqrt{x}-1}-1>0\\ \Leftrightarrow \dfrac{\sqrt{x}+2-(\sqrt{x}-1)}{\sqrt{x}-1}>0\\ \Leftrightarrow \dfrac{3}{\sqrt{x}-1}>0\\ \Leftrightarrow \sqrt{x}-1>0\\ \Leftrightarrow x>1\\ \text{Kết hợp điều kiện } \Rightarrow x>1, x \ne 4\\ c)P\left(\dfrac{1}{4}\right)\\ =\dfrac{\sqrt{\dfrac{1}{4}}+2}{\sqrt{\dfrac{1}{4}}-1}\\ =\dfrac{\dfrac{1}{2}+2}{\dfrac{1}{2}-1}\\ =\dfrac{\dfrac{5}{2}}{-\dfrac{1}{2}}\\ =-5$
