a) Ta có: \(A=\left(\frac{2+\sqrt{x}}{2-\sqrt{x}}-\frac{4x}{x-4}-\frac{2-\sqrt{x}}{2+\sqrt{x}}\right):\frac{x\sqrt{x}-x}{2x-x\sqrt{x}}\)
\(=\left(\frac{\left(2+\sqrt{x}\right)^2}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}+\frac{4x}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}-\frac{\left(2-\sqrt{x}\right)^2}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\right):\frac{x\left(\sqrt{x}-1\right)}{x\left(2-\sqrt{x}\right)}\)
\(=\frac{4+4\sqrt{x}+x+4x-\left(4-4\sqrt{x}+x\right)}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}:\frac{\sqrt{x}-1}{2-\sqrt{x}}\)
\(=\frac{4+4\sqrt{x}+5x-4+4\sqrt{x}-x}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\cdot\frac{2-\sqrt{x}}{\sqrt{x}-1}\)
\(=\frac{4x+8\sqrt{x}}{2+\sqrt{x}}\cdot\frac{1}{\sqrt{x}-1}\)
\(=\frac{4\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{4\sqrt{x}}{\sqrt{x}-1}\)
b) ĐKXĐ: \(\left\{{}\begin{matrix}x>0\\x\notin\left\{1;4\right\}\end{matrix}\right.\)
Để A nhận giá trị nguyên thì \(4\sqrt{x}⋮\sqrt{x}-1\)
\(\Leftrightarrow4\sqrt{x}-4+4⋮\sqrt{x}-1\)
mà \(4\sqrt{x}-4⋮\sqrt{x}-1\forall x\)
nên \(4⋮\sqrt{x}-1\)
\(\Leftrightarrow\sqrt{x}-1\inƯ\left(4\right)\)
\(\Leftrightarrow\sqrt{x}-1\in\left\{1;-1;2;-2;4;-4\right\}\)
\(\Leftrightarrow\sqrt{x}-1\in\left\{1;2;4\right\}\)(Vì \(\sqrt{x}-1>-1\forall x\) thỏa mãn ĐKXĐ)
\(\Leftrightarrow\sqrt{x}\in\left\{2;3;5\right\}\)
\(\Leftrightarrow x\in\left\{4;9;25\right\}\)
Kết hợp ĐKXĐ, ta được: \(x\in\left\{9;25\right\}\)
Vậy: Để A nhận giá trị nguyên thì \(x\in\left\{9;25\right\}\)
