Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
9,\\
A = \sqrt {2 + \sqrt 3 + \sqrt {4 - 2\sqrt 3 - \sqrt {21 - 12\sqrt 3 } } } \\
= \sqrt {2 + \sqrt 3 + \sqrt {4 - 2\sqrt 3 - \sqrt {12 - 2.2\sqrt 3 .3 + 9} } } \\
= \sqrt {2 + \sqrt 3 + \sqrt {4 - 2\sqrt 3 - \sqrt {{{\left( {2\sqrt 3 - 3} \right)}^2}} } } \\
= \sqrt {2 + \sqrt 3 + \sqrt {4 - 2\sqrt 3 - \left( {2\sqrt 3 - 3} \right)} } \\
= \sqrt {2 + \sqrt 3 + \sqrt {7 - 4\sqrt 3 } } \\
= \sqrt {2 + \sqrt 3 + \sqrt {4 - 2.2.\sqrt 3 + 3} } \\
= \sqrt {2 + \sqrt 3 + \sqrt {{{\left( {2 - \sqrt 3 } \right)}^2}} } \\
= \sqrt {2 + \sqrt 3 + 2 - \sqrt 3 } \\
= \sqrt 4 = 2\\
10,\\
{x_0} = \sqrt {2 + \sqrt {2 + \sqrt 3 } } - \sqrt {6 - 3\sqrt {2 + \sqrt 3 } } \\
\Leftrightarrow {x_0}^2 = \left( {2 + \sqrt {2 + \sqrt 3 } } \right) - 2.\sqrt {\left( {2 + \sqrt {2 + \sqrt 3 } } \right)\left( {6 - 3\sqrt {2 + \sqrt 3 } } \right)} + 6 - 3\sqrt {2 + \sqrt 3 } \\
\Leftrightarrow {x_0}^2 = 8 - 2\sqrt {2 + \sqrt 3 } - 2.\sqrt {12 - 3.{{\sqrt {2 + \sqrt 3 } }^2}} \\
\Leftrightarrow {x_0}^2 = 8 - 2\sqrt {2 + \sqrt 3 } - 2.\sqrt {12 - 3.\left( {2 + \sqrt 3 } \right)} \\
\Leftrightarrow {x_0}^2 = 8 - 2\sqrt {2 + \sqrt 3 } - 2\sqrt {6 - 3\sqrt 3 } \\
\Leftrightarrow 8 - {x_0}^2 = 2.\left( {\sqrt {2 + \sqrt 3 } + \sqrt {6 - 3\sqrt 3 } } \right)\\
\Leftrightarrow 64 - 16{x_0}^2 + {x_0}^4 = 4.\left( {2 + \sqrt 3 + 2.\sqrt {\left( {2 + \sqrt 3 } \right).\left( {6 - 3\sqrt 3 } \right)} + 6 - 3\sqrt 3 } \right)\\
\Leftrightarrow 64 - 16{x_0}^2 + {x_0}^4 = 4.\left( {8 - 2\sqrt 3 + 2.\sqrt {12 - 3.3} } \right)\\
\Leftrightarrow 64 - 16{x_0}^2 + {x_0}^4 = 4.\left( {8 - 2\sqrt 3 + 2\sqrt 3 } \right)\\
\Leftrightarrow {x_0}^4 - 16{x_0}^2 + 32 = 0
\end{array}\)
