\(a,B=\left(\dfrac{\sqrt{x}}{x-4}+\dfrac{1}{\sqrt{x}-2}\right):\dfrac{\sqrt{x}+2}{x-4}\left(x\ne4;x\ge0\right)\\ B=\dfrac{\sqrt{x}+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\sqrt{x}+2}\\ B=\dfrac{2\left(\sqrt{x}+1\right)}{\sqrt{x}+2}\)
\(c,C=A\left(B-2\right)=A\cdot\dfrac{2\left(\sqrt{x}+1\right)-2\sqrt{x}-4}{\sqrt{x}+2}\\ =\dfrac{\sqrt{x}+2}{\sqrt{x}-2}\cdot\dfrac{-2}{\sqrt{x}+2}=\dfrac{-2}{\sqrt{x}-2}\)
Để \(C\in Z\Leftrightarrow-2⋮\sqrt{x}-2\)
\(\Leftrightarrow\sqrt{x}-2\inƯ\left(-2\right)=\left\{-2;-1;1;2\right\}\\ \Leftrightarrow\sqrt{x}\in\left\{0;1;3;4\right\}\\ \Leftrightarrow x\in\left\{0;1;9;16\right\}\)
a: Ta có: \(B=\left(\dfrac{\sqrt{x}}{x-4}+\dfrac{1}{\sqrt{x}-2}\right):\dfrac{\sqrt{x}+2}{x-4}\)
\(=\dfrac{2\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{x-4}{\sqrt{x}+2}\)
\(=\dfrac{2\sqrt{x}+2}{\sqrt{x}+2}\)