a: Sửa đề: \(A=\dfrac{2}{\sqrt{x}-3}+\dfrac{2\sqrt{x}}{x-4\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-1}\)
\(=\dfrac{2\sqrt{x}-2+2\sqrt{x}+x-3\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-3\right)}=\dfrac{\sqrt{x}+2}{\sqrt{x}-3}\)
c: Để A là số nguyên thì \(\sqrt{x}-3+5⋮\sqrt{x}-3\)
=>\(\sqrt{x}-3\in\left\{1;-1;5;-5\right\}\)
hay \(x\in\left\{16;4;64\right\}\)