a) \(A=\left(\dfrac{\sqrt{x}}{\sqrt{x}+1}+\dfrac{\sqrt{x}-1}{\sqrt{x}}\right):\dfrac{2x-1}{3x}\)
\(A=\left[\dfrac{x}{\sqrt{x}\left(\sqrt{x}+1\right)}+\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\right]\cdot\dfrac{3x}{2x-1}\)
\(A=\dfrac{x+x-1}{\sqrt{x}\left(\sqrt{x}+1\right)}\cdot\dfrac{3x}{2x-1}\)
\(A=\dfrac{2x-1}{\sqrt{x}\left(\sqrt{x}+1\right)}\cdot\dfrac{3x}{2x-1}\)
\(A=\dfrac{3x}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(A=\dfrac{3\sqrt{x}}{\sqrt{x}+1}\)
b) \(A=\dfrac{3\sqrt{x}}{\sqrt{x}+1}=\dfrac{3\sqrt{x}+3-3}{\sqrt{x}+1}=\dfrac{3\left(\sqrt{x}+1\right)-3}{\sqrt{x}+1}\)
\(=3-\dfrac{3}{\sqrt{x}+1}\)
Để A nguyên thì \(\dfrac{3}{\sqrt{x}+1}\) nguyên:
⇒ 3 ⋮ \(\sqrt{x}+1\)
⇒ \(\sqrt{x}+1\in\) Ư(3) = {1; -1; 3; -3}
Mà: \(\sqrt{x}+1\ge1\forall x\)
\(\Rightarrow\sqrt{x}\in\left\{0;2\right\}\)
\(\Rightarrow x\in\left\{0;4\right\}\)
Kết hợp với đkxđ: \(x=4\)