a,ĐKXĐ $a \geq 0$
$A=\dfrac{\sqrt[]a(a\sqrt[]a+1)}{a-\sqrt[]a+1}-\dfrac{\sqrt[]a(2\sqrt[]a+1)}{\sqrt[]a}+1$
$=\dfrac{\sqrt[]a(\sqrt[]a+1)(a-\sqrt[]a+1)}-(2\sqrt[]a+1)+1$
$=a+\sqrt[]a-2\sqrt[]a-1+1$
$=a-\sqrt[]a$
b, $a \geq 1⇒\sqrt[]a \geq 1⇒a \geq \sqrt[]a$
Nên $A \geq 0⇒A=|A|$
c,
$A=2⇔a-\sqrt[]a-2=0⇔\sqrt[]a=-1$ hoặc $\sqrt[]a=2$ suy ra $a=4$
d, $A=a-\sqrt[]a+\dfrac{1}{4}-\dfrac{1}{4}=(\sqrt[]a-\dfrac{1}{2})^2-\dfrac{1}{4} \geq -\dfrac{1}{4}$
Dấu $=$ xảy ra khi $a=\dfrac{1}{4}$
