Bài 2:
\(a,A=\dfrac{3}{9+1}=\dfrac{3}{10}\\ b,B=\dfrac{x-2-\sqrt{x}-2+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\sqrt{x}\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}-2}{\sqrt{x}}\\ c,B=\dfrac{\sqrt{x}-2}{\sqrt{x}}=\sqrt{x}-2\\ \Leftrightarrow x-2\sqrt{x}=\sqrt{x}-2\\ \Leftrightarrow x-3\sqrt{x}+2=0\\ \Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=1\\\sqrt{x}=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\left(tm\right)\\x=4\left(tm\right)\end{matrix}\right.\)
\(d,B=\dfrac{\sqrt{x}-2}{\sqrt{x}}=1-\dfrac{2}{\sqrt{x}}\in Z\\ \Leftrightarrow\sqrt{x}\inƯ\left(2\right)=\left\{1;2\right\}\left(\sqrt{x}>0\right)\\ \Leftrightarrow x\in\left\{1;4\right\}\\ e,P=2AB+\dfrac{4}{x+1}=\dfrac{2\sqrt{x}}{x+1}\cdot\dfrac{\sqrt{x}-2}{\sqrt{x}}+\dfrac{4}{x+1}=\dfrac{2\sqrt{x}-4}{x+1}+\dfrac{4}{x+1}=\dfrac{2\sqrt{x}}{x+1}\\ P=\dfrac{2}{\sqrt{x}+\dfrac{1}{\sqrt{x}}}\le\dfrac{2}{2\sqrt{\sqrt{x}\cdot\dfrac{1}{\sqrt{x}}}}=1\\ P_{max}=1\Leftrightarrow\sqrt{x}=\dfrac{1}{\sqrt{x}}\Leftrightarrow x=1\)
