Giải thích các bước giải:

ĐKXĐ: \(\left\{ \begin{array}{l}
x,y > 0\\
x \ne y
\end{array} \right.\)

Ta có:

\(\begin{array}{l}
a,\\
{B_1} = \left( {\sqrt x  + \frac{{y - \sqrt {xy} }}{{\sqrt x  + \sqrt y }}} \right):\left( {\frac{x}{{\sqrt {xy}  + y}} + \frac{y}{{\sqrt {xy}  - x}} - \frac{{x + y}}{{\sqrt {xy} }}} \right)\\
 = \left( {\frac{{\sqrt x \left( {\sqrt x  + \sqrt y } \right) + y - \sqrt {xy} }}{{\sqrt x  + \sqrt y }}} \right):\left( {\frac{x}{{\sqrt y \left( {\sqrt x  + \sqrt y } \right)}} + \frac{y}{{\sqrt x \left( {\sqrt y  - \sqrt x } \right)}} - \frac{{x + y}}{{\sqrt {xy} }}} \right)\\
 = \left( {\frac{{x + y}}{{\sqrt x  + \sqrt y }}} \right):\left( {\frac{{x\sqrt x \left( {\sqrt y  - \sqrt x } \right) + y\sqrt y \left( {\sqrt x  + \sqrt y } \right)}}{{\sqrt {xy} \left( {\sqrt x  + \sqrt y } \right)\left( {\sqrt y  - \sqrt x } \right)}} - \frac{{x + y}}{{\sqrt {xy} }}} \right)\\
 = \left( {\frac{{x + y}}{{\sqrt x  + \sqrt y }}} \right):\left( {\frac{{x\sqrt {xy}  - {x^2} + y\sqrt {xy}  + {y^2} - \left( {x + y} \right)\left( {y - x} \right)}}{{\sqrt {xy} \left( {y - x} \right)}}} \right)\\
 = \left( {\frac{{x + y}}{{\sqrt x  + \sqrt y }}} \right):\left( {\frac{{\sqrt {xy} \left( {x + y} \right) - {x^2} + {y^2} - \left( {{y^2} - {x^2}} \right)}}{{\sqrt {xy} \left( {y - x} \right)}}} \right)\\
 = \left( {\frac{{x + y}}{{\sqrt x  + \sqrt y }}} \right):\left( {\frac{{\sqrt {xy} \left( {x + y} \right)}}{{\sqrt {xy} \left( {y - x} \right)}}} \right)\\
 = \frac{{x + y}}{{\sqrt x  + \sqrt y }}.\frac{{y - x}}{{x + y}}\\
 = \frac{{y - x}}{{\sqrt x  + \sqrt y }} = \sqrt y  - \sqrt x 
\end{array}\)

b,

\(\begin{array}{l}
x = 3 \Rightarrow \sqrt x  = \sqrt 3 \\
y = 4 + 2\sqrt 3  \Rightarrow \sqrt y  = \sqrt {4 + 2\sqrt 3 }  = \sqrt {{{\left( {\sqrt 3  + 1} \right)}^2}}  = \sqrt 3  + 1\\
{B_1} = \sqrt y  - \sqrt x  = 1
\end{array}\)