Đáp án:

\( - \dfrac{{2\left( {\sqrt a  + 1} \right)}}{{a + \sqrt a  + 1}}\)

Giải thích các bước giải:

\(\begin{array}{l}
DK:a \ge 0;a \ne 1\\
A = \left[ {\dfrac{{\sqrt a  + 2}}{{a\sqrt a  - 1}} - \dfrac{{\sqrt a  + 1}}{{a + \sqrt a  + 1}}} \right] - \dfrac{1}{{\sqrt a  - 1}}\\
 = \left[ {\dfrac{{\sqrt a  + 2 - \left( {\sqrt a  + 1} \right)\left( {\sqrt a  - 1} \right)}}{{\left( {\sqrt a  - 1} \right)\left( {a + \sqrt a  + 1} \right)}}} \right] - \dfrac{1}{{\sqrt a  - 1}}\\
 = \dfrac{{\sqrt a  + 2 - a + 1}}{{\left( {\sqrt a  - 1} \right)\left( {a + \sqrt a  + 1} \right)}} - \dfrac{1}{{\sqrt a  - 1}}\\
 = \dfrac{{ - a + \sqrt a  + 3 - a - \sqrt a  - 1}}{{\left( {\sqrt a  - 1} \right)\left( {a + \sqrt a  + 1} \right)}}\\
 = \dfrac{{ - 2a + 2}}{{\left( {\sqrt a  - 1} \right)\left( {a + \sqrt a  + 1} \right)}}\\
 =  - \dfrac{{2\left( {a - 1} \right)}}{{\left( {\sqrt a  - 1} \right)\left( {a + \sqrt a  + 1} \right)}}\\
 =  - \dfrac{{2\left( {\sqrt a  - 1} \right)\left( {\sqrt a  + 1} \right)}}{{\left( {\sqrt a  - 1} \right)\left( {a + \sqrt a  + 1} \right)}}\\
 =  - \dfrac{{2\left( {\sqrt a  + 1} \right)}}{{a + \sqrt a  + 1}}
\end{array}\)

ĐKXĐ : $a \geq 0 ; a \neq 1$

$A=\left (\dfrac{\sqrt{a}+2}{a\sqrt{a}-1}-\dfrac{\sqrt{a}+1}{a+\sqrt{a}+1} \right )-\dfrac{1}{\sqrt{a}-1}$

$A=\left [\dfrac{\sqrt{a}+2-(\sqrt{a}+1)(\sqrt{a}-1)}{(\sqrt{a}-1)(a+\sqrt{a}+1)} \right ] - \dfrac{a+\sqrt{a}+1}{(\sqrt{a}-1)(a+\sqrt{a}+1)}$

$A=\dfrac{\sqrt{a}+2-a+1-a-\sqrt{a}-1}{(\sqrt{a}-1)(a+\sqrt{a}+1)}$

$A=\dfrac{-2a+2}{(\sqrt{a}-1)(a+\sqrt{a}+1)}$

$A=\dfrac{-2(a-1)}{(\sqrt{a}-1)(a+\sqrt{a}+1)}$

$A=\dfrac{-2(\sqrt{a}-1)(\sqrt{a}+1)}{(\sqrt{a}-1)(a+\sqrt{a}+1)}$

$A=\dfrac{-2(\sqrt{a}+1}{a+\sqrt{a}+1}$

Vậy $A=\dfrac{-2(\sqrt{a}+1}{a+\sqrt{a}+1}$