a) Thay x=9 vào A ta có:
\(A=\dfrac{\sqrt{9}+2}{\sqrt{9}}=\dfrac{3+2}{3}=\dfrac{5}{3}\)
b) \(P=\dfrac{A}{B}\)
\(P=\dfrac{\sqrt{x}+2}{\sqrt{x}}:\dfrac{\sqrt{x}}{\sqrt{x}-2}\)
\(P=\dfrac{\sqrt{x}+2}{\sqrt{x}}\cdot\dfrac{\sqrt{x}-2}{\sqrt{x}}\)
\(P=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\sqrt{x}\cdot\sqrt{x}}\)
\(P=\dfrac{\left(\sqrt{x}\right)^2-2^2}{x}\)
\(P=\dfrac{x-2}{x}\)
c) \(\left|P\right|>P\) khi
\(P< 0\)
\(\Rightarrow\dfrac{x-2}{x}< 0\) \(\left(x\ne0\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x< 0\\x>2\end{matrix}\right.\\\left\{{}\begin{matrix}x>0\\x< 2\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>0\\x< 2\end{matrix}\right.\)
\(\Leftrightarrow0< x< 2\)
