Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
T = \sqrt {\sqrt 5 - \sqrt {3 - \sqrt {29 - 12\sqrt 5 } } } \\
= \sqrt {\sqrt 5 - \sqrt {3 - \sqrt {20 - 2.2\sqrt 5 .3 + 9} } } \\
= \sqrt {\sqrt 5 - \sqrt {3 - \sqrt {{{\left( {2\sqrt 5 } \right)}^2} - 2.2\sqrt 5 .3 + {3^2}} } } \\
= \sqrt {\sqrt 5 - \sqrt {3 - \sqrt {{{\left( {2\sqrt 5 - 3} \right)}^2}} } } \\
= \sqrt {\sqrt 5 - \sqrt {3 - \left( {2\sqrt 5 - 3} \right)} } \\
= \sqrt {\sqrt 5 - \sqrt {6 - 2\sqrt 5 } } \\
= \sqrt {\sqrt 5 - \sqrt {5 - 2.\sqrt 5 .1 + 1} } \\
= \sqrt {\sqrt 5 - \sqrt {{{\left( {\sqrt 5 - 1} \right)}^2}} } \\
= \sqrt {\sqrt 5 - \left( {\sqrt 5 - 1} \right)} \\
= \sqrt 1 \\
= 1\\
b,\\
A = \dfrac{{\sqrt {\sqrt 5 + 2} + \sqrt {\sqrt 5 - 2} }}{{\sqrt {\sqrt 5 + 1} }}\\
\Rightarrow {A^2} = \dfrac{{{{\left( {\sqrt {\sqrt 5 + 2} + \sqrt {\sqrt 5 - 2} } \right)}^2}}}{{\sqrt 5 + 1}}\\
\Leftrightarrow {A^2} = \dfrac{{\left( {\sqrt 5 + 2} \right) + 2.\sqrt {\left( {\sqrt 5 - 2} \right)\left( {\sqrt 5 + 2} \right)} + \left( {\sqrt 5 - 2} \right)}}{{\sqrt 5 + 1}}\\
\Leftrightarrow {A^2} = \dfrac{{2\sqrt 5 + 2.\sqrt {{{\sqrt 5 }^2} - {2^2}} }}{{\sqrt 5 + 1}}\\
\Leftrightarrow {A^2} = \dfrac{{2\sqrt 5 + 2}}{{\sqrt 5 + 1}}\\
\Leftrightarrow {A^2} = 2\\
\Leftrightarrow A = \sqrt 2 \,\,\,\,\,\,\left( {A > 0} \right)
\end{array}\)
