a, đkxđ: a>0;a≠1
\(S=\left(\dfrac{1}{a-\sqrt{a}}+\dfrac{1}{\sqrt{a}-1}\right):\dfrac{\sqrt{a}+1}{a-2\sqrt{a}+1}=\dfrac{1+\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}=\dfrac{\sqrt{a}-1}{\sqrt{a}}\)
b/ so sánh: \(\dfrac{\sqrt{a}-1}{\sqrt{a}}\) và 1
Ta có: \(\dfrac{\sqrt{a}-1}{\sqrt{a}}=\dfrac{\sqrt{a}}{\sqrt{a}}-\dfrac{1}{\sqrt{a}}=1-\dfrac{1}{\sqrt{a}}< 1\)(vì \(\dfrac{1}{\sqrt{a}}>0\))
c/ \(\dfrac{\sqrt{a}-1}{\sqrt{a}}=-2\Leftrightarrow\sqrt{a}-1=-2\sqrt{a}\Leftrightarrow3\sqrt{a}=1\Leftrightarrow\sqrt{a}=\dfrac{1}{3}\Leftrightarrow a=\dfrac{1}{9}\left(TM\right)\)
d/ \(S=\dfrac{\sqrt{a}-1}{\sqrt{a}}=1-\dfrac{1}{\sqrt{a}}\)
để bt nguyên => \(\dfrac{1}{\sqrt{a}}\in Z\Leftrightarrow\sqrt{a}\inƯ\left(1\right)\)
\(\Leftrightarrow\sqrt{a}=\left\{1\right\}\Rightarrow a=1\left(KTM\right)\)
Vậy không có gt nào của a để S nguyên