`a)`

ĐK : `{(a>=0),(sqrta-1 \ne 0),(a+sqrta \ne 0),(sqrta+1 \ne 0):}``<=> {(a>=0),(a ne 1):}`

`b)`

`A=((a-sqrta)/(sqrta-1)-(sqrta+1)/(a+sqrta)):(sqrta+1)/a`

`=((sqrta(sqrta-1))/(sqrta-1)-(sqrta+1)/(sqrta(sqrta+1))).a/(sqrta+1)`

`=(sqrta-1/(sqrta)).a/(sqrta+1)`

`=(sqrta.sqrta-1)/(sqrta).a/(sqrta+1)`

`=(a-1)/(sqrta).a/(sqrta+1)`

`=((sqrta-1)(sqrta+1))/(sqrta) . (sqrta.sqrta)/(sqrta+1)`

`=sqrta(sqrta-1)`

`=a-sqrta`

`c)`

`A=a-sqrta`   

`=a-2. sqrta . 1/2+1/4-1/4`

`=(sqrta-1/2)^2-1/4>=-1/4`

Dấu "=" xảy ra khi : `(sqrta-1/2)^2=0`

`<=> a=1/4 \ \ (tm)`

Vậy `A_(min)=-1/4 <=> a=1/4`

a) Điều kiện xác định:

$\left\{ \begin{array}{l} \sqrt a  - 1 \ne 0\\ a \ge 0\\ \sqrt a  + 1 \ne 0 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} a \ge 0\\ a \ne 1 \end{array} \right.$

b)

$\begin{array}{l} A = \left( {\dfrac{{a - \sqrt a }}{{\sqrt a  - 1}} - \dfrac{{\sqrt a  + 1}}{{a + \sqrt a }}} \right):\dfrac{{\sqrt a  + 1}}{a}\\ A = \left[ {\dfrac{{\sqrt a \left( {\sqrt a  - 1} \right)}}{{\sqrt a  - 1}} - \dfrac{{\sqrt a  + 1}}{{\sqrt a \left( {\sqrt a  + 1} \right)}}} \right].\dfrac{a}{{\sqrt a  + 1}}\\ A = \left( {\sqrt a  - \dfrac{1}{{\sqrt a }}} \right).\dfrac{a}{{\sqrt a  + 1}} = \dfrac{{a - 1}}{{\sqrt a }}.\dfrac{a}{{\sqrt a  + 1}}\\ A = \dfrac{{\left( {\sqrt a  - 1} \right)\left( {\sqrt a  + 1} \right)}}{{\sqrt a }}.\dfrac{a}{{\sqrt a  + 1}} = \sqrt a \left( {\sqrt a  - 1} \right)\\ c)A = a - \sqrt a  = a - \sqrt a  + \dfrac{1}{4} - \dfrac{1}{4} = {\left( {\sqrt a  - \dfrac{1}{2}} \right)^2} - \dfrac{1}{4} \ge  - \dfrac{1}{4}\\  \Rightarrow \min A =  - \dfrac{1}{4}\\ ' = ' \Leftrightarrow \sqrt a  - \dfrac{1}{2} = 0 \Leftrightarrow \sqrt a  = \dfrac{1}{2} \Leftrightarrow a = \dfrac{1}{4} \end{array}$