a) ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\sqrt{x}-1\ge0\\\sqrt{x}-1\ge0\\x+\sqrt{x}+1\ge0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x\ge0\\x\sqrt{x}-1\ge0\\\sqrt{x}\ge1\\x+\sqrt{x}+1\ge0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x\ge1\\x\sqrt{x}-1\ge0\\x+\sqrt{x}+1\ge0\end{matrix}\right.\)
\(Q=\left(\dfrac{2x+1}{x\sqrt{x}-1}-\dfrac{1}{\sqrt{x}-1}\right):\left(1-\dfrac{x+4}{x+\sqrt{x}+1}\right)\)
\(Q=\left(\dfrac{2x+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{1}{\sqrt{x}-1}\right):\left(\dfrac{x+\sqrt{x}+1}{x+\sqrt{x}+1}-\dfrac{x+4}{x+\sqrt{x}+1}\right)\)
\(Q=\left(\dfrac{2x+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right):\dfrac{x+\sqrt{x}+1-x-4}{x+\sqrt{x}+1}\)
\(Q=\dfrac{2x+1-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{x+\sqrt{x}+1}{x+\sqrt{x}+1-x-4}\)
\(Q=\dfrac{x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{x+\sqrt{x}+1}{\sqrt{x}-3}\)
\(Q=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{x+\sqrt{x}+1}{\sqrt{x}-3}\)
\(Q=\dfrac{\sqrt{x}}{\sqrt{x}-3}\)
\(Q\in Z\Rightarrow\dfrac{\sqrt{x}}{\sqrt{x}-3}\in Z\\ \Rightarrow\sqrt{x}⋮\sqrt{x}-3\)
\(\Rightarrow\sqrt{x}-3+3⋮\sqrt{x}-3\)
Vì \(\sqrt{x}-3⋮\sqrt{x}-3\Rightarrow3⋮\sqrt{x}-3\Rightarrow\sqrt{x}-3\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\Rightarrow x=\left\{0;4;16;36\right\}\)
a: Ta có: \(Q=\left(\dfrac{2x+1}{x\sqrt{x}-1}-\dfrac{1}{\sqrt{x}-1}\right):\left(1-\dfrac{x+4}{x+\sqrt{x}+1}\right)\)
\(=\dfrac{2x+1-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}:\dfrac{x+\sqrt{x}+1-x-4}{x+\sqrt{x}+1}\)
\(=\dfrac{x-\sqrt{x}}{\sqrt{x}-1}\cdot\dfrac{1}{\sqrt{x}-3}\)
\(=\dfrac{\sqrt{x}}{\sqrt{x}-3}\)
b: Để Q nguyên thì \(\sqrt{x}⋮\sqrt{x}-3\)
\(\Leftrightarrow3⋮\sqrt{x}-3\)
\(\Leftrightarrow\sqrt{x}-3\in\left\{-3;-1;1;3\right\}\)
\(\Leftrightarrow\sqrt{x}\in\left\{0;2;4;6\right\}\)
hay \(x\in\left\{0;4;16;36\right\}\)
