a) Thay `x=4` vào B ta có:
\(B=\dfrac{1}{\sqrt{4}+3}=\dfrac{1}{5}\)
b)
\(A=\dfrac{x-6}{x+3\sqrt{x}}-\dfrac{1}{\sqrt{x}}\\ =\dfrac{x-6}{\sqrt{x}\left(\sqrt{x}+3\right)}-\dfrac{\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}+3\right)}\\ =\dfrac{x-6-\sqrt{x}-3}{\sqrt{x}\left(\sqrt{x}-3\right)}\\ =\dfrac{x-\sqrt{x}-9}{\sqrt{x}\left(\sqrt{x}+3\right)}\\=>P=A+B\\ =\dfrac{x-\sqrt{x}-9}{\sqrt{x}\left(\sqrt{x}+3\right)}+\dfrac{1}{\sqrt{x}+3}\\ =\dfrac{x-\sqrt{x}-9+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+3\right)}\\ =\dfrac{x-9}{\sqrt{x}\left(\sqrt{x}+3\right)}\\ =\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\sqrt{x}\left(\sqrt{x}+3\right)}\\ =\dfrac{\sqrt{x}-3}{\sqrt{x}}\)
c) Ta có:
\(P=\dfrac{\sqrt{x}-3}{\sqrt{x}}=1-\dfrac{3}{\sqrt{x}}\)
Để P nguyên thì \(\sqrt{x}\inƯ\left(3\right)=\left\{1;-1;3;-3\right\}\)
Mà: \(\sqrt{x}>0\)
=> \(\sqrt{x}\in\left\{1;3\right\}=>x\in\left\{1;9\right\}\)
