a) điều kiện xác định : \(a>0;a\ne1;a\ne2\)
ta có : \(A=\left(\dfrac{a\sqrt{a}-1}{a-\sqrt{a}}-\dfrac{a\sqrt{a}-1}{a+\sqrt{a}}\right):\dfrac{a+2}{a-2}\)
\(\Leftrightarrow A=\left(\dfrac{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}-\dfrac{\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}+1\right)}\right):\dfrac{a+2}{a-2}\)
\(\Leftrightarrow A=\left(\dfrac{a+\sqrt{a}+1}{\sqrt{a}}-\dfrac{a-\sqrt{a}+1}{\sqrt{a}}\right):\dfrac{a+2}{a-2}\) \(\Leftrightarrow A=\left(\dfrac{2\sqrt{a}}{\sqrt{a}}\right):\dfrac{a+2}{a-2}=2:\dfrac{a+2}{a-2}=2\left(\dfrac{a-2}{a+2}\right)=\dfrac{2a-4}{a+2}\)b) ta có : \(A=\dfrac{2a-4}{a+2}=\dfrac{2a+4-8}{a+2}=2-\dfrac{8}{a+2}\)
để A có giá trị nguyên \(\Leftrightarrow a+2\in\left\{\pm1;\pm2;\pm4;\pm8\right\}\)
\(\Rightarrow a\in\left\{-10;-6;-4;-3;-1;0;2;6\right\}\)
vậy ............................................................................................................................
