a) Điều kiện xác định : \(a>0\); \(a\ne1\)
b) Ta có :
\(A=\left(\frac{a-\sqrt{a}}{\sqrt{a}-1}-\frac{\sqrt{a}+1}{a+\sqrt{a}}\right):\frac{\sqrt{a}+1}{a}=\left(\frac{\sqrt{a}.\left(\sqrt{a}-1\right)}{\sqrt{a}-1}-\frac{\sqrt{a}+1}{\sqrt{a}.\left(\sqrt{a}+1\right)}\right).\frac{a}{\sqrt{a}+1}\)
\(=\left(\sqrt{a}-\frac{1}{\sqrt{a}}\right).\frac{a}{\sqrt{a}+1}=\frac{a-1}{\sqrt{a}}.\frac{a}{\sqrt{a}+1}=\frac{a.\left(\sqrt{a}-1\right).\left(\sqrt{a}+1\right)}{\sqrt{a}.\left(\sqrt{a}+1\right)}\)
\(=\sqrt{a}.\left(\sqrt{a}-1\right)=a-\sqrt{a}\)
c)
Ta có : \(A=a-\sqrt{a}=\left(a-2.\frac{1}{2}.\sqrt{a}+\frac{1}{4}\right)-\frac{1}{4}=\left(\sqrt{a}-\frac{1}{2}\right)^2-\frac{1}{4}\)
Vì \(a>0\)và \(a\ne1\)nên \(\left(\sqrt{a}-\frac{1}{2}\right)^2\ge0\)
\(\Rightarrow\) \(A=\left(\sqrt{a}-\frac{1}{2}\right)^2-\frac{1}{4}\ge-\frac{1}{4}\)
Vậy \(Min_A=-\frac{1}{4}\) khi và chỉ khi \(\sqrt{a}-\frac{1}{2}=0\Rightarrow\sqrt{a}=\frac{1}{2}\Rightarrow a=\frac{1}{4}\)
