Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
\dfrac{{3 + \sqrt 3 }}{{1 + \sqrt 3 }} = \dfrac{{\sqrt 3 \left( {\sqrt 3 + 1} \right)}}{{1 + \sqrt 3 }} = \sqrt 3 \\
b,\\
\dfrac{{\sqrt {7 - 4\sqrt 3 } }}{{\sqrt 3 - 2}} = \dfrac{{\sqrt {4 - 2.2.\sqrt 3 + 3} }}{{\sqrt 3 - 2}} = \dfrac{{\sqrt {{{\left( {2 - \sqrt 3 } \right)}^2}} }}{{\sqrt 3 - 2}} = \dfrac{{\left| {2 - \sqrt 3 } \right|}}{{\sqrt 3 - 2}} = \dfrac{{2 - \sqrt 3 }}{{\sqrt 3 - 2}} = - 1\\
c,\\
\dfrac{{x - y - \sqrt x - \sqrt y }}{{\sqrt x + \sqrt y }} = \dfrac{{\left( {x - y} \right) - \left( {\sqrt x + \sqrt y } \right)}}{{\sqrt x + \sqrt y }}\\
= \dfrac{{\left( {\sqrt x - \sqrt y } \right)\left( {\sqrt x + \sqrt y } \right) - \left( {\sqrt x + \sqrt y } \right)}}{{\sqrt x + \sqrt y }}\\
= \dfrac{{\left( {\sqrt x + \sqrt y } \right)\left( {\sqrt x - \sqrt y - 1} \right)}}{{\sqrt x + \sqrt y }}\\
= \sqrt x - \sqrt y - 1
\end{array}\)
