Đáp án:
$\begin{array}{l}
a)A = 1\\
b)B = b - a\left( {a,b > 0;a \ne b} \right)
\end{array}$
Giải thích các bước giải:
$\begin{array}{l}
a)A = \left( {2 + \dfrac{{3 + \sqrt 3 }}{{\sqrt 3 + 1}}} \right)\left( {2 - \dfrac{{3 - \sqrt 3 }}{{\sqrt 3 - 1}}} \right)\\
= \left( {2 + \dfrac{{\sqrt 3 \left( {\sqrt 3 + 1} \right)}}{{\sqrt 3 + 1}}} \right)\left( {2 - \dfrac{{\sqrt 3 \left( {\sqrt 3 - 1} \right)}}{{\sqrt 3 - 1}}} \right)\\
= \left( {2 + \sqrt 3 } \right)\left( {2 - \sqrt 3 } \right)\\
= {2^2} - {\left( {\sqrt 3 } \right)^2}\\
= 1\\
b)B = \left( {\dfrac{{\sqrt b }}{{a - \sqrt {ab} }} - \dfrac{{\sqrt a }}{{\sqrt {ab} - b}}} \right).\left( {a\sqrt b - b\sqrt a } \right)\left( {DK:a,b > 0;a \ne b} \right)\\
= \left( {\dfrac{{\sqrt b }}{{\sqrt a \left( {\sqrt a - \sqrt b } \right)}} - \dfrac{{\sqrt a }}{{\sqrt b \left( {\sqrt a - \sqrt b } \right)}}} \right)\sqrt {ab} \left( {\sqrt a - \sqrt b } \right)\\
= \dfrac{{\sqrt b .\sqrt b - \sqrt a .\sqrt a }}{{\sqrt {ab} \left( {\sqrt a - \sqrt b } \right)}}.\sqrt {ab} \left( {\sqrt a - \sqrt b } \right)\\
= b - a
\end{array}$
