a) \(G=\frac{\sqrt{x-2\sqrt{x-1}}+\sqrt{x+2\sqrt{x-1}}}{\sqrt{1-\frac{2}{x}+\frac{1}{x^2}}}\)
Tử : \(\sqrt{x-2\sqrt{x-1}}+\sqrt{x+2\sqrt{x-1}}\)
\(=\sqrt{x-1-2\sqrt{x-1}+1}+\sqrt{x-1+2\sqrt{x-1}+1}\)
\(=\sqrt{\left(\sqrt{x-1}-1\right)^2}+\sqrt{\left(\sqrt{x-1}+1\right)^2}\)
\(=\left|\sqrt{x-1}-1\right|+\sqrt{x-1}+1\)
Mẫu : \(\sqrt{1-\frac{2}{x}+\frac{1}{x^2}}=\sqrt{\left(\frac{1}{x}-1\right)^2}=\left|\frac{1}{x}-1\right|\)
\(\Rightarrow G=\frac{\left|\sqrt{x-1}-1\right|+\sqrt{x-1}+1}{\left|\frac{1}{x}-1\right|}\)
b) \(x>2\Leftrightarrow\left\{{}\begin{matrix}x-1>1\Leftrightarrow\sqrt{x-1}>1\\\frac{1}{x}< 1\end{matrix}\right.\)
\(\Rightarrow G=\frac{\sqrt{x-1}-1+\sqrt{x-1}+1}{1-\frac{1}{x}}\)
\(G=\frac{2\sqrt{x-1}}{\frac{x-1}{x}}=\frac{2x\sqrt{x-1}}{x-1}=\frac{2x}{\sqrt{x-1}}\)
Để G nguyên thì \(2x⋮\sqrt{x-1}\)
\(\Leftrightarrow2x-2+2⋮\sqrt{x-1}\)
\(\Leftrightarrow2\left(x-1\right)+2⋮\sqrt{x-1}\)
Ta có \(2\left(x-1\right)⋮\sqrt{x-1}\)
\(\Rightarrow2⋮\sqrt{x-1}\)
\(\Rightarrow\sqrt{x-1}\inƯ\left(2\right)=\left\{1;2\right\}\)
\(\Leftrightarrow x-1\in\left\{1;4\right\}\)
\(\Leftrightarrow x\in\left\{2;5\right\}\)( thỏa )
Vậy....
