1: \(A=\left(\frac{x\cdot\sqrt{x}-1}{x-\sqrt{x}}-\frac{x\sqrt{x}+1}{x+\sqrt{x}}\right):\frac{2\left(x-2\sqrt{x}+1\right)}{x-1}\)
\(=\left(\frac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}-\frac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\right):\frac{2\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\left(\frac{x+\sqrt{x}+1}{\sqrt{x}}-\frac{x-\sqrt{x}+1}{\sqrt{x}}\right):\frac{2\left(\sqrt{x}-1\right)}{\sqrt{x}+1}\)
\(=2\cdot\frac{\sqrt{x}+1}{2\left(\sqrt{x}-1\right)}=\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
2: Để A là số nguyên thì \(\sqrt{x}+1\) ⋮\(\sqrt{x}-1\)
=>\(\sqrt{x}-1+2\) ⋮\(\sqrt{x}-1\)
=>2⋮\(\sqrt{x}-1\)
=>\(\sqrt{x}-1\in\left\lbrace1;-1;2;-2\right\rbrace\)
=>\(\sqrt{x}\in\left\lbrace2;0;3\right\rbrace\)
=>x∈{0;4;9}
Kết hợp ĐKXĐ, ta được: x∈{4;9}
