\(1,x=16\left(tm\right)\Rightarrow A=\dfrac{\sqrt{16}+2}{\sqrt{16}-3}=\dfrac{4+2}{4-3}=6\)
\(2,B=\dfrac{\sqrt{x}+5}{\sqrt{x}+1}+\dfrac{7-\sqrt{x}}{x-1}\left(dk:x\ge0,x\ne1,x\ne9\right)\\ =\dfrac{\left(\sqrt{x}+5\right)\left(\sqrt{x}-1\right)+7-\sqrt{x}}{x-1}\\ =\dfrac{x+4\sqrt{x}-5+7-\sqrt{x}}{x-1}\\ =\dfrac{x+3\sqrt{x}+2}{x-1}\\ =\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\\ =\dfrac{\sqrt{x}+2}{\sqrt{x}-1}\)
\(3,B>\dfrac{1}{2}\Leftrightarrow\dfrac{\sqrt{x}+2}{\sqrt{x}-1}>\dfrac{1}{2}\\ \Rightarrow2\sqrt{x}+4-\sqrt{x}+1>0\Rightarrow\sqrt{x}>-5\left(LD\right)\)
Vậy với điều kiện \(x\ge0,x\ne1,x\ne9\) thì mọi giá trị x còn lại thỏa mãn đề bài.
\(4,\dfrac{A}{B}=2\Leftrightarrow\dfrac{\sqrt{x}+2}{\sqrt{x}-3}.\dfrac{\sqrt{x}-1}{\sqrt{x}+2}=\dfrac{\sqrt{x}-1}{\sqrt{x}-3}=2\)
\(\Rightarrow\sqrt{x}-1-2\sqrt{x}+6=0\Rightarrow-\sqrt{x}=-5\Rightarrow x=25\left(tm\right)\)
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