Đáp án:
a. \(\dfrac{{\sqrt 5 + 1}}{2}\)
Giải thích các bước giải:
\(\begin{array}{l}
a.A = \sqrt {\dfrac{2}{{3 - \sqrt 5 }}} = \sqrt {\dfrac{4}{{6 - 2\sqrt 5 }}} \\
= \dfrac{2}{{\sqrt {5 - 2\sqrt 5 .1 + 1} }} = \dfrac{2}{{\sqrt {{{\left( {\sqrt 5 - 1} \right)}^2}} }}\\
= \dfrac{2}{{\sqrt 5 - 1}} = \dfrac{{2\left( {\sqrt 5 + 1} \right)}}{{5 - 1}} = \dfrac{{2\sqrt 5 + 2}}{4}\\
= \dfrac{{\sqrt 5 + 1}}{2}\\
b.B = \sqrt {\dfrac{{a - 4}}{{2\left( {\sqrt a - 2} \right)}}} = \sqrt {\dfrac{{\left( {\sqrt a - 2} \right)\left( {\sqrt a + 2} \right)}}{{2\left( {\sqrt a - 2} \right)}}} \\
= \sqrt {\dfrac{{\sqrt a + 2}}{2}} \\
c.C = \sqrt {\dfrac{1}{{a\left( {1 - \sqrt 3 } \right)}}} = \dfrac{{\sqrt {1 + \sqrt 3 } }}{{\sqrt {a\left( {1 - 3} \right)} }}\\
= \dfrac{{\sqrt {1 + \sqrt 3 } }}{{\sqrt { - 2a} }}\left( {DK:a < 0} \right)\\
d.D = \sqrt {\dfrac{a}{{4 - 2\sqrt 3 }}} = \sqrt {\dfrac{a}{{3 - 2\sqrt 3 .1 + 1}}} \\
= \sqrt {\dfrac{a}{{{{\left( {\sqrt 3 - 1} \right)}^2}}}} \\
= \dfrac{{\sqrt a }}{{\sqrt 3 - 1}}
\end{array}\)
