ĐKXĐ: \(x\ge0;x\ne9\)
\(P=\left(\dfrac{\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\dfrac{5\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\dfrac{6}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right):\dfrac{6}{\sqrt{x}+2}\)
\(=\left(\dfrac{\sqrt{x}-3+5\sqrt{x}+15+6}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right):\dfrac{6}{\sqrt{x}+2}\)
\(=\dfrac{6\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\dfrac{\sqrt{x}+2}{6}=\dfrac{\sqrt{x}+2}{\sqrt{x}-3}\)
b.
\(P=\dfrac{\sqrt{x}-3+5}{\sqrt{x}-3}=1+\dfrac{5}{\sqrt{x}-3}\)
P nguyên khi \(\dfrac{5}{\sqrt{x}-3}\) nguyên \(\Rightarrow\sqrt{x}-3=Ư\left(5\right)\ge-3=\left\{-1;1;5\right\}\)
\(\Rightarrow\sqrt{x}=\left\{2;4;8\right\}\)
\(\Rightarrow x=\left\{4;16;64\right\}\)
a: ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x\ne9\end{matrix}\right.\)
\(P=\left(\dfrac{1}{\sqrt{x}+3}+\dfrac{5}{\sqrt{x}-3}-\dfrac{6}{9-x}\right):\dfrac{6}{\sqrt{x}+2}\)
\(=\left(\dfrac{\sqrt{x}-3+5\left(\sqrt{x}+3\right)}{x-9}+\dfrac{6}{x-9}\right):\dfrac{6}{\sqrt{x}+2}\)
\(=\dfrac{\sqrt{x}-3+5\sqrt{x}+15+6}{x-9}\cdot\dfrac{\sqrt{x}+2}{6}\)
\(=\dfrac{6\sqrt{x}+18}{x-9}\cdot\dfrac{\sqrt{x}+2}{6}\)
\(=\dfrac{6\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{\sqrt{x}+2}{6}=\dfrac{\sqrt{x}+2}{\sqrt{x}-3}\)
b: Để P nguyên thì \(\sqrt{x}+2⋮\sqrt{x}-3\)
=>\(\sqrt{x}-3+5⋮\sqrt{x}-3\)
=>\(5⋮\sqrt{x}-3\)
=>\(\sqrt{x}-3\in\left\{1;-1;5;-5\right\}\)
=>\(\sqrt{x}\in\left\{4;2;8;-2\right\}\)
mà \(\sqrt{x}>=0\)
nên \(\sqrt{x}\in\left\{2;4;8\right\}\)
=>\(x\in\left\{4;16;64\right\}\)
