a) \(A=\frac{a^2+\sqrt{a}}{a-\sqrt{a}+1}-\frac{2a+\sqrt{a}}{\sqrt{a}}+1\left(ĐK:a\ge0\right)\)
\(=\frac{\sqrt{a}\left(a\sqrt{a}+1\right)}{a-\sqrt{a}+1}-\frac{\sqrt{a}\left(2\sqrt{a}+1\right)}{\sqrt{a}}+1\)
\(=\frac{\sqrt{a}\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{a-\sqrt{a}+1}-\left(2\sqrt{a}+1\right)+1\)
\(=a+\sqrt{a}-2\sqrt{a}-1+1=a-\sqrt{a}\)
b) Để A=2 \(\Leftrightarrow a-\sqrt{a}=2\)
\(\Leftrightarrow a-\sqrt{a}-2=0\)
\(\Leftrightarrow\left(\sqrt{a}+1\right)\left(\sqrt{a}-2\right)=0\)
\(\Leftrightarrow\sqrt{a}-2=0\left(Vì\sqrt{a}+1\ne0\right)\)
\(\Leftrightarrow a=4\) (TM)
Vậy a=4 thì A=2
c) \(A=a-\sqrt{a}=a-\sqrt{a}+\frac{1}{4}-\frac{1}{4}=\left(\sqrt{a}-\frac{1}{2}\right)^2-\frac{1}{4}\)
Vì: \(\left(\sqrt{a}-\frac{1}{2}\right)^2\ge0\)
=> \(\left(\sqrt{a}-\frac{1}{2}\right)^2-\frac{1}{4}\ge-\frac{1}{4}\)
Vậy GTNN của A là \(-\frac{1}{4}\) khi \(a=\frac{1}{4}\)
