`a,` Khi `x=9` Ta có :
\(A=\dfrac{\sqrt{x}+3}{x-4}\\ =\dfrac{\sqrt{9}+3}{9-4}\\ =\dfrac{3+3}{5}\\ =\dfrac{6}{5}\)
\(b,B=\dfrac{-4}{x-4}+\dfrac{1}{\sqrt{x}-2}\\ =-\dfrac{4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{1}{\sqrt{x}-2}\\ =-\dfrac{4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\\ =\dfrac{-4+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\\ =\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\\ =\dfrac{1}{\sqrt{x}+2}\)
`c,` `P=A/B`
\(\Rightarrow P=\dfrac{\dfrac{\sqrt{x}+3}{x-4}}{\dfrac{1}{\sqrt{x}+2}}\\ =\dfrac{\sqrt{x}+3}{x-4}\cdot\dfrac{\sqrt{x}+2}{1}\\ =\dfrac{\sqrt{x}+3}{\sqrt{x}-2}\)
Mà `P<1` thì :
\(\dfrac{\sqrt{x}+3}{\sqrt{x}-2}< 1\\ \Leftrightarrow\dfrac{\sqrt{x}+3}{\sqrt{x}-2}< \dfrac{\sqrt{x}-2}{\sqrt{x}-2}\\ \Leftrightarrow\sqrt{x}+3< \sqrt{x}-2\\ \Leftrightarrow0< -5\)
Vậy ko có `x` thỏa mãn `P<1`
