a) Vì \(x=9\) thỏa mãn điều kiện xác định, nên thay \(x=9\) vào biểu thức \(A\) ta được:
\(A=\dfrac{9-2\sqrt{9}+1}{\sqrt{9}}=\dfrac{9-2.3+1}{3}=\dfrac{9-6+1}{3}=\dfrac{4}{3}\)
b) Với \(x>0,\) \(x\ne4,\) ta có:
\(B=\dfrac{\sqrt{x}}{\sqrt{x}+2}-\dfrac{1}{\sqrt{x}-2}-\dfrac{5\sqrt{x}+2}{4-x}\)
\(=\dfrac{\sqrt{x}}{\sqrt{x}+2}-\dfrac{1}{\sqrt{x}-2}+\dfrac{5\sqrt{x}+2}{x-4}\)
\(=\dfrac{\sqrt{x}}{\sqrt{x}+2}-\dfrac{1}{\sqrt{x}-2}+\dfrac{5\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)-\left(\sqrt{x}+2\right)+\left(5\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{x-2\sqrt{x}-\sqrt{x}-2+5\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{x+2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{\sqrt{x}}{\sqrt{x}-2}\) (điều phải chứng minh)
c) \(P=A.B=\dfrac{x-2\sqrt{x}+1}{\sqrt{x}}.\dfrac{\sqrt{x}}{\sqrt{x}-2}=\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}-2}\)
Để \(P< 0\) thì \(\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}-2}< 0\Rightarrow\left\{{}\begin{matrix}\sqrt{x}-1\ne0\\\sqrt{x}-2< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}\ne1\\\sqrt{x}< 2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne1\\0\le x< 4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne1\\0< x< 4\end{matrix}\right.\)
Mà \(x\) nguyên nên \(x\in\left\{2;3\right\}.\)
Vậy với \(x\in\left\{2;3\right\}\) thì \(P< 0.\)
