Đáp án:
\(N = \dfrac{{2\sqrt x }}{{x + 1}}\)
Giải thích các bước giải:
\(\begin{array}{l}
DK:x \ge 0;y \ge 0;x \ne y;xy \ne 1\\
N = \left[ {\dfrac{{\left( {\sqrt x + \sqrt y } \right)\left( {1 + \sqrt {xy} } \right) + \left( {\sqrt x - \sqrt y } \right)\left( {1 - \sqrt {xy} } \right)}}{{1 - xy}}} \right]:\dfrac{{1 - xy + x + y + 2xy}}{{1 - xy}}\\
= \dfrac{{\sqrt x + \sqrt y + x\sqrt y + y\sqrt x + \sqrt x - \sqrt y - x\sqrt y + y\sqrt x }}{{1 - xy}}.\dfrac{{1 - xy}}{{x + y + xy + 1}}\\
= \dfrac{{2\sqrt x + 2y\sqrt x }}{{x\left( {y + 1} \right) + \left( {y + 1} \right)}} = \dfrac{{2\sqrt x \left( {y + 1} \right)}}{{\left( {y + 1} \right)\left( {x + 1} \right)}} = \dfrac{{2\sqrt x }}{{x + 1}}\\
Thay:x = \dfrac{2}{{2 + \sqrt 3 }} = \dfrac{4}{{4 + 2\sqrt 3 }} = \dfrac{4}{{3 + 2\sqrt 3 .1 + 1}}\\
= \dfrac{4}{{{{\left( {\sqrt 3 + 1} \right)}^2}}}\\
\to N = 2\left( {\sqrt {\dfrac{4}{{{{\left( {\sqrt 3 + 1} \right)}^2}}}} } \right):\left( {\dfrac{2}{{2 + \sqrt 3 }} + 1} \right)\\
= \dfrac{4}{{\sqrt 3 + 1}}.\dfrac{1}{{5 - 2\sqrt 3 }} = \dfrac{{2 + 6\sqrt 3 }}{{13}}
\end{array}\)
