Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
DKXD:\,\,\,\left\{ \begin{array}{l}
x \ge 0\\
y \ge 0\\
xy \ne 1
\end{array} \right.\\
a,\\
P = \left( {\dfrac{{\sqrt x + \sqrt y }}{{1 - \sqrt {xy} }} + \dfrac{{\sqrt x - \sqrt y }}{{1 + \sqrt {xy} }}} \right):\left( {1 + \dfrac{{x + y + 2xy}}{{1 - xy}}} \right)\\
= \dfrac{{\left( {\sqrt x + \sqrt y } \right)\left( {1 + \sqrt {xy} } \right) + \left( {\sqrt x - \sqrt y } \right)\left( {1 - \sqrt {xy} } \right)}}{{\left( {1 - \sqrt {xy} } \right)\left( {1 + \sqrt {xy} } \right)}}:\dfrac{{\left( {1 - xy} \right) + x + y + 2xy}}{{1 - xy}}\\
= \dfrac{{\sqrt x + \sqrt {{x^2}y} + \sqrt y + \sqrt {x{y^2}} + \sqrt x - \sqrt {{x^2}y} - \sqrt y + \sqrt {x{y^2}} }}{{{1^2} - {{\sqrt {xy} }^2}}}:\dfrac{{1 + x + y + xy}}{{1 - xy}}\\
= \dfrac{{2\sqrt x + 2\sqrt {x{y^2}} }}{{1 - xy}}:\dfrac{{\left( {1 + x} \right) + \left( {y + xy} \right)}}{{1 - xy}}\\
= \dfrac{{2\sqrt x \left( {1 + \sqrt {{y^2}} } \right)}}{{1 - xy}}:\dfrac{{\left( {1 + x} \right) + y\left( {1 + x} \right)}}{{1 - xy}}\\
= \dfrac{{2\sqrt x \left( {1 + y} \right)}}{{1 - xy}}:\dfrac{{\left( {1 + x} \right)\left( {1 + y} \right)}}{{1 - xy}}\\
= \dfrac{{2\sqrt x \left( {1 + y} \right)}}{{1 - xy}}.\dfrac{{1 - xy}}{{\left( {1 + x} \right)\left( {1 + y} \right)}}\\
= \dfrac{{2\sqrt x }}{{1 + x}}\\
b,\\
x = \dfrac{2}{{2 + \sqrt 3 }} = \dfrac{4}{{4 + 2\sqrt 3 }} = \dfrac{4}{{{{\left( {\sqrt 3 + 1} \right)}^2}}}\\
\Rightarrow \sqrt x = \dfrac{2}{{\sqrt 3 + 1}}\\
\Rightarrow P = \dfrac{{2\sqrt x }}{{1 + x}} = \dfrac{{2.\dfrac{2}{{\sqrt 3 + 1}}}}{{1 + \dfrac{2}{{2 + \sqrt 3 }}}} = \dfrac{{\dfrac{4}{{\sqrt 3 + 1}}}}{{\dfrac{{2 + \sqrt 3 + 2}}{{2 + \sqrt 3 }}}} = \dfrac{4}{{\sqrt 3 + 1}}:\dfrac{{4 + \sqrt 3 }}{{2 + \sqrt 3 }}\\
= \dfrac{{4.\left( {2 + \sqrt 3 } \right)}}{{\left( {\sqrt 3 + 1} \right)\left( {4 + \sqrt 3 } \right)}} = \dfrac{{4\left( {2 + \sqrt 3 } \right)}}{{4\sqrt 3 + 3 + 4 + \sqrt 3 }} = \dfrac{{4\left( {2 + \sqrt 3 } \right)}}{{7 + 5\sqrt 3 }}\\
= \dfrac{{4\left( {2 + \sqrt 3 } \right)\left( {7 - 5\sqrt 3 } \right)}}{{\left( {7 + 5\sqrt 3 } \right)\left( {7 - 5\sqrt 3 } \right)}} = \dfrac{{4.\left( {14 - 10\sqrt 3 + 7\sqrt 3 - 5.3} \right)}}{{{7^2} - {{\left( {5\sqrt 3 } \right)}^2}}}\\
= \dfrac{{4.\left( { - 1 - 3\sqrt 3 } \right)}}{{49 - 75}}\\
= \dfrac{{2.\left( {3\sqrt 3 + 1} \right)}}{{13}}
\end{array}\)
