ĐKXĐ: \(x>0;x\ne1\)
\(A=\left(\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}{x\sqrt{x}-1}-\frac{\sqrt{x}-3}{x\sqrt{x}-1}\right).\left(\frac{x\sqrt{x}\left(\sqrt{x}+1\right)-\left(\sqrt{x}+1\right)}{\sqrt{x}}\right)\)
\(A=\frac{x+2\sqrt{x}-3-\sqrt{x}+3}{\left(x\sqrt{x}-1\right)}.\frac{\left(x\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}}\)
\(A=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(x\sqrt{x}-1\right)}.\frac{\left(x\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}}=\left(\sqrt{x}+1\right)^2\)
b/ \(x=3+2\sqrt{2}=\left(\sqrt{2}+1\right)^2\Rightarrow\sqrt{x}=\sqrt{2}+1\)
\(\Rightarrow A=\left(2+\sqrt{2}\right)^2=6+4\sqrt{2}\)
b/ \(\left(\sqrt{x}+1\right)^2=2\sqrt{x}+3\Leftrightarrow x+2\sqrt{x}+1=2\sqrt{x}+3\Rightarrow x=2\)
c/ Để A nguyên \(\Rightarrow\sqrt{x}\) nguyên \(\Rightarrow x\) là số chính phương
Vậy với mọi x có dạng \(x=k^2\) (\(k\in Z;k>1\)) thì A là số nguyên
