Đáp án:
\(\begin{array}{l}
8,\\
{a^2} - 2a - 2 = 0\\
9,\\
T = \dfrac{1}{2}\\
10,\\
{x_0}^4 - 16{x_0}^2 + 32 = 0
\end{array}\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
8,\\
a = \sqrt {3 + \sqrt {5 + 2\sqrt 3 } } + \sqrt {3 - \sqrt {5 + 2\sqrt 3 } } \\
\Leftrightarrow {a^2} = {\left( {\sqrt {3 + \sqrt {5 + 2\sqrt 3 } } + \sqrt {3 - \sqrt {5 + 2\sqrt 3 } } } \right)^2}\\
\Leftrightarrow {a^2} = {\sqrt {3 + \sqrt {5 + 2\sqrt 3 } } ^2} + 2.\sqrt {3 + \sqrt {5 + 2\sqrt 3 } } .\sqrt {3 - \sqrt {5 + 2\sqrt 3 } } + {\sqrt {3 - \sqrt {5 + 2\sqrt 3 } } ^2}\\
\Leftrightarrow {a^2} = \left( {3 + \sqrt {5 + 2\sqrt 3 } } \right) + 2.\sqrt {\left( {3 + \sqrt {5 + 2\sqrt 3 } } \right).\left( {3 - \sqrt {5 + 2\sqrt 3 } } \right)} + \left( {3 - \sqrt {5 + 2\sqrt 3 } } \right)\\
\Leftrightarrow {a^2} = 3 + \sqrt {5 + 2\sqrt 3 } + 2.\sqrt {{3^2} - {{\sqrt {5 + 2\sqrt 3 } }^2}} + 3 - \sqrt {5 + 2\sqrt 3 } \\
\Leftrightarrow {a^2} = 6 + 2\sqrt {9 - \left( {5 + 2\sqrt 3 } \right)} \\
\Leftrightarrow {a^2} = 6 + 2\sqrt {4 - 2\sqrt 3 } \\
\Leftrightarrow {a^2} = 6 + 2.\sqrt {3 - 2\sqrt 3 + 1} \\
\Leftrightarrow {a^2} = 6 + 2.\sqrt {{{\sqrt 3 }^2} - 2.\sqrt 3 .1 + {1^2}} \\
\Leftrightarrow {a^2} = 6 + 2.\sqrt {{{\left( {\sqrt 3 - 1} \right)}^2}} \\
\Leftrightarrow {a^2} = 6 + 2.\left| {\sqrt 3 - 1} \right|\\
\Leftrightarrow {a^2} = 6 + 2.\left( {\sqrt 3 - 1} \right)\\
\Leftrightarrow {a^2} = 4 + 2\sqrt 3 \\
\Leftrightarrow {a^2} = 3 + 2\sqrt 3 + 1\\
\Leftrightarrow {a^2} = {\sqrt 3 ^2} + 2.\sqrt 3 .1 + {1^2}\\
\Leftrightarrow {a^2} = {\left( {\sqrt 3 + 1} \right)^2}\\
\Leftrightarrow a = \left| {\sqrt 3 + 1} \right|\\
a = \sqrt {3 + \sqrt {5 + 2\sqrt 3 } } + \sqrt {3 - \sqrt {5 + 2\sqrt 3 } } \Rightarrow a > 0\\
\Rightarrow a = \sqrt 3 + 1\\
{a^2} - 2a - 2 = \left( {4 + 2\sqrt 3 } \right) - 2.\left( {\sqrt 3 + 1} \right) - 2\\
= 4 + 2\sqrt 3 - 2\sqrt 3 - 2 - 2 = 0\\
\Rightarrow {a^2} - 2a - 2 = 0\\
9,\\
a = \sqrt {4 + \sqrt {10 + 2\sqrt 5 } } + \sqrt {4 - \sqrt {10 + 2\sqrt 5 } } \\
\Leftrightarrow {a^2} = {\left( {\sqrt {4 + \sqrt {10 + 2\sqrt 5 } } + \sqrt {4 - \sqrt {10 + 2\sqrt 5 } } } \right)^2}\\
\Leftrightarrow {a^2} = {\sqrt {4 + \sqrt {10 + 2\sqrt 5 } } ^2} + 2.\sqrt {4 + \sqrt {10 + 2\sqrt 5 } } .\sqrt {4 - \sqrt {10 + 2\sqrt 5 } } + {\sqrt {4 - \sqrt {10 + 2\sqrt 5 } } ^2}\\
\Leftrightarrow {a^2} = \left( {4 + \sqrt {10 + 2\sqrt 5 } } \right) + 2.\sqrt {\left( {4 + \sqrt {10 + 2\sqrt 5 } } \right).\left( {4 - \sqrt {10 + 2\sqrt 5 } } \right)} + \left( {4 - \sqrt {10 + 2\sqrt 5 } } \right)\\
\Leftrightarrow {a^2} = 4 + \sqrt {10 + 2\sqrt 5 } + 2.\sqrt {{4^2} - {{\sqrt {10 + 2\sqrt 5 } }^2}} + 4 - \sqrt {10 + 2\sqrt 5 } \\
\Leftrightarrow {a^2} = 8 + 2.\sqrt {16 - \left( {10 + 2\sqrt 5 } \right)} \\
\Leftrightarrow {a^2} = 8 + 2\sqrt {6 - 2\sqrt 5 } \\
\Leftrightarrow {a^2} = 8 + 2\sqrt {5 - 2\sqrt 5 + 1} \\
\Leftrightarrow {a^2} = 8 + 2\sqrt {{{\sqrt 5 }^2} - 2.\sqrt 5 .1 + {1^2}} \\
\Leftrightarrow {a^2} = 8 + 2.\sqrt {{{\left( {\sqrt 5 - 1} \right)}^2}} \\
\Leftrightarrow {a^2} = 8 + 2.\left| {\sqrt 5 - 1} \right|\\
\Leftrightarrow {a^2} = 8 + 2\left( {\sqrt 5 - 1} \right)\\
\Leftrightarrow {a^2} = 6 + 2\sqrt 5 \\
\Leftrightarrow {a^2} = 5 + 2\sqrt 5 + 1\\
\Leftrightarrow {a^2} = {\sqrt 5 ^2} + 2.\sqrt 5 .1 + {1^2}\\
\Leftrightarrow {a^2} = {\left( {\sqrt 5 + 1} \right)^2}\\
\Leftrightarrow a = \left| {\sqrt 5 + 1} \right|\\
a = \sqrt {4 + \sqrt {10 + 2\sqrt 5 } } + \sqrt {4 - \sqrt {10 + 2\sqrt 5 } } \Rightarrow a > 0\\
\Rightarrow a = \sqrt 5 + 1\\
{a^2} - 2a - 4 = \left( {6 + 2\sqrt 5 } \right) - 2.\left( {\sqrt 5 + 1} \right) - 4\\
= 6 + 2\sqrt 5 - 2\sqrt 5 - 2 - 4 = 0\\
\Rightarrow {a^2} - 2a - 4 = 0\\
T = \dfrac{{{a^4} - 4{a^3} + {a^2} + 6a + 4}}{{{a^2} - 2a + 12}}\\
= \dfrac{{\left( {{a^4} - 2{a^3} - 4{a^2}} \right) + \left( { - 2{a^3} + 4{a^2} + 8a} \right) + \left( {{a^2} - 2a - 4} \right) + 8}}{{{a^2} - 2a + 12}}\\
= \dfrac{{{a^2}.\left( {{a^2} - 2a - 4} \right) - 2a.\left( {{a^2} - 2a - 4} \right) + \left( {{a^2} - 2a - 4} \right) + 8}}{{{a^2} - 2a + 12}}\\
= \dfrac{{\left( {{a^2} - 2a - 4} \right).\left( {{a^2} - 2a + 1} \right) + 8}}{{\left( {{a^2} - 2a - 4} \right) + 16}}\\
= \dfrac{{0.\left( {{a^2} - 2a + 1} \right) + 8}}{{0 + 16}}\\
= \dfrac{8}{{16}}\\
= \dfrac{1}{2}\\
10,\\
{x_0} = \sqrt {2 + \sqrt {2 + \sqrt 3 } } - \sqrt {6 - 3\sqrt {2 + \sqrt 3 } } \\
\Leftrightarrow {x_0}^2 = {\left( {\sqrt {2 + \sqrt {2 + \sqrt 3 } } - \sqrt {6 - 3\sqrt {2 + \sqrt 3 } } } \right)^2}\\
\Leftrightarrow {x_0}^2 = {\sqrt {2 + \sqrt {2 + \sqrt 3 } } ^2} - 2.\sqrt {2 + \sqrt {2 + \sqrt 3 } } .\sqrt {6 - 3\sqrt {2 + \sqrt 3 } } + {\sqrt {6 - 3\sqrt {2 + \sqrt 3 } } ^2}\\
\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)} + \left( {6 - 3\sqrt {2 + \sqrt 3 } } \right)\\
\Leftrightarrow {x_0}^2 = 2 + \sqrt {2 + \sqrt 3 } - 2.\sqrt {3.\left( {2 + \sqrt {2 + \sqrt 3 } } \right)\left( {2 - \sqrt {2 + \sqrt 3 } } \right)} + 6 - 3\sqrt {2 + \sqrt 3 } \\
\Leftrightarrow {x_0}^2 = 8 - 2\sqrt {2 + \sqrt 3 } - 2.\sqrt {3.\left( {{2^2} - {{\sqrt {2 + \sqrt 3 } }^2}} \right)} \\
\Leftrightarrow {x_0}^2 = 8 - 2\sqrt {2 + \sqrt 3 } - 2.\sqrt {3.\left( {4 - \left( {2 + \sqrt 3 } \right)} \right)} \\
\Leftrightarrow {x_0}^2 = 8 - 2\sqrt {2 + \sqrt 3 } - 2\sqrt {3.\left( {2 - \sqrt 3 } \right)} \\
\Leftrightarrow {x_0}^2 = 8 - \sqrt 2 .\sqrt 2 .\sqrt {2 + \sqrt 3 } - \sqrt 2 .\sqrt 2 .\sqrt 3 .\sqrt {2 - \sqrt 3 } \\
\Leftrightarrow {x_0}^2 = 8 - \sqrt 2 .\sqrt {2.\left( {2 + \sqrt 3 } \right)} - \left( {\sqrt 2 .\sqrt 3 } \right).\sqrt {2.\left( {2 - \sqrt 3 } \right)} \\
\Leftrightarrow {x_0}^2 = 8 - \sqrt 2 .\sqrt {4 + 2\sqrt 3 } - \sqrt 6 .\sqrt {4 - 2\sqrt 3 } \\
\Leftrightarrow {x_0}^2 = 8 - \sqrt 2 .\sqrt {3 + 2.\sqrt 3 .1 + {1^2}} - \sqrt 6 .\sqrt {3 - 2.\sqrt 3 .1 + {1^2}} \\
\Leftrightarrow {x_0}^2 = 8 - \sqrt 2 .\sqrt {{{\left( {\sqrt 3 + 1} \right)}^2}} - \sqrt 6 .\sqrt {{{\left( {\sqrt 3 - 1} \right)}^2}} \\
\Leftrightarrow {x_0}^2 = 8 - \sqrt 2 .\left| {\sqrt 3 + 1} \right| - \sqrt 6 .\left| {\sqrt 3 - 1} \right|\\
\Leftrightarrow {x_0}^2 = 8 - \sqrt 2 .\left( {\sqrt 3 + 1} \right) - \sqrt 6 .\left( {\sqrt 3 - 1} \right)\\
\Leftrightarrow {x_0}^2 = 8 - \sqrt 6 - \sqrt 2 - \sqrt {18} + \sqrt 6 \\
\Leftrightarrow {x_0}^2 = 8 - \sqrt 2 - \sqrt {18} \\
\Leftrightarrow {x_0}^2 = 8 - \sqrt 2 - 3\sqrt 2 \\
\Leftrightarrow {x_0}^2 = 8 - 4\sqrt 2 \\
\Leftrightarrow 8 - {x_0}^2 = 4\sqrt 2 \\
\Leftrightarrow {\left( {8 - {x_0}^2} \right)^2} = {\left( {4\sqrt 2 } \right)^2}\\
\Leftrightarrow 64 - 16{x_0}^2 + {\left( {{x_0}^2} \right)^2} = 32\\
\Leftrightarrow {x_0}^4 - 16{x_0}^2 + 32 = 0
\end{array}\)
