ĐK \(a\ge0\)
\(A=\frac{\sqrt{a}\left(a\sqrt{a}+1\right)}{a-\sqrt{a}+1}-\frac{\sqrt{a}\left(2\sqrt{a}+1\right)}{\sqrt{a}}+1\)
\(=\frac{\sqrt{a}\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{a-\sqrt{a}+1}-\left(2\sqrt{a}+1\right)+1\)
\(=a+\sqrt{a}-2\sqrt{a}-1+1=a-\sqrt{a}\)
b. \(A=2\Rightarrow a-\sqrt{a}-2=0\Rightarrow\left(\sqrt{a}-2\right)\left(\sqrt{a}+1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}\sqrt{a}=2\\\sqrt{a}=-1\left(l\right)\end{cases}\Rightarrow a=4}\)
Vậy a=4 thì A=2
c. \(A=a-\sqrt{a}=\left(\sqrt{a}-\frac{1}{2}\right)^2-\frac{1}{4}\ge-\frac{1}{4}\forall a\Rightarrow A\ge-\frac{1}{4}\)
Vậy \(MinA=-\frac{1}{4}\Rightarrow\sqrt{a}=\frac{1}{2}\Rightarrow a=\frac{1}{4}\)
