Câu a : ĐK : \(x\ge0\)
Câu b : \(A=\dfrac{1}{5}\)
\(\Leftrightarrow\dfrac{\sqrt{x}-3}{\sqrt{x}+1}=\dfrac{1}{5}\)
\(\Leftrightarrow5\sqrt{x}-15=\sqrt{x}+1\)
\(\Leftrightarrow4\sqrt{x}=16\)
\(\Leftrightarrow\sqrt{x}=4\)
\(\Leftrightarrow x=16\left(TM\right)\)
Câu c : Ta có :
\(A=\dfrac{\sqrt{x}-3}{\sqrt{x}+1}=\dfrac{\left(\sqrt{x}+1\right)-4}{\sqrt{x}+1}=1-\dfrac{4}{\sqrt{x}+1}\)
Để : \(1-\dfrac{4}{\sqrt{x}+1}\) đạt MIN thì \(\dfrac{4}{\sqrt{x}+1}\) đạt MAX .
Mà :\(\dfrac{4}{\sqrt{x}+1}\le\dfrac{4}{1}=4\)
\(\Rightarrow A=1-\dfrac{4}{\sqrt{x}+1}\ge1-4=-3\)
Dấu \("="\) xảy ra khi \(x=0\)
Câu d : Để \(A\in Z\Leftrightarrow\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\in Z\Leftrightarrow1-\dfrac{4}{\sqrt{x}+1}\in Z\)
Do đó A thuộc Z thì \(\sqrt{x}+1\inƯ\left(4\right)\)
Mà : \(\sqrt{x}+1\ge1\)
\(\Rightarrow\left\{{}\begin{matrix}\sqrt{x}+1=1\\\sqrt{x}+1=2\\\sqrt{x}+1=4\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\x=1\\x=9\end{matrix}\right.\)
