Cho \({x_1} = \sqrt {3 + \sqrt 5 } \) và \({x_2} = \sqrt {3 - \sqrt 5 } \). Khi đó:     

a) Đúng.

\[{x_1}{x_2} = \sqrt {3 + \sqrt 5 } .\sqrt {3 - \sqrt 5 } = \sqrt {\left( {3 + \sqrt 5 } \right)\left( {3 - \sqrt 5 } \right)} = \sqrt {9 - 5} = \sqrt 4 = 2\].

b) Đúng.

\(x_1^2 + x_2^2 = {\left( {\sqrt {3 + \sqrt 5 } } \right)^2} + {\left( {\sqrt {3 - \sqrt 5 } } \right)^2} = 3 + \sqrt 5 + 3 - \sqrt 5 = 6.\)

c) Sai.

Ta có: \({\left( {{x_1} + {x_2}} \right)^2} = x_1^2 + 2{x_1}{x_2} + x_2^2 = \left( {x_1^2 + x_2^2} \right) + 2{x_1}{x_2} = 6 + 2.2 = 10\).

d) Đúng.

Cách 1. \(x_1^3 + x_2^3 = \left( {{x_1} + {x_2}} \right)\left( {x_1^2 - {x_1}{x_2} + x_2^2} \right)\)

\( = \left( {\sqrt {3 + \sqrt 5 } + \sqrt {3 - \sqrt 5 } } \right)\left( {6 - 2} \right)\)

\( = 4\left( {\sqrt {3 + \sqrt 5 } + \sqrt {3 - \sqrt 5 } } \right)\)

\( = 4\sqrt {3 + \sqrt 5 } + 4\sqrt {3 - \sqrt 5 } \)

\( = \sqrt {48 + 16\sqrt 5 } + \sqrt {48 - 16\sqrt 5 } \)

\( = \sqrt {48 + 2.8\sqrt 5 } + \sqrt {48 - 2.8\sqrt 5 } \)

\( = \sqrt {48 + 2.\sqrt 8 .\sqrt 8 .\sqrt 5 }  + \sqrt {48 - 2.\sqrt 8 .\sqrt 8 .\sqrt 5 } \)

\( = \sqrt {48 + 2.\sqrt 8 .\sqrt {40} } + \sqrt {48 - 2.\sqrt 8 .\sqrt {40} } \)

\( = \sqrt {40 + 2.\sqrt 8 .\sqrt {40} + 8} + \sqrt {40 - 2.\sqrt 8 .\sqrt {40} + 8} \)

\( = \sqrt {{{\left( {\sqrt 8 + \sqrt {40} } \right)}^2}} + \sqrt {{{\left( {\sqrt {40} - \sqrt 8 } \right)}^2}} \)

\( = \sqrt 8 + \sqrt {40} + \sqrt {40} - \sqrt 8 \)

\( = 2\sqrt {40} \)

\( = 4\sqrt {10} \).

Cách 2. Có \({\left( {{x_1} + {x_2}} \right)^2} = 10\) nên \(\sqrt {{{\left( {{x_1} + {x_2}} \right)}^2}} = \sqrt {10} \) hay \({x_1} + {x_2} = \sqrt {10} \).

Do đó, \(x_1^3 + x_2^3 = \left( {{x_1} + {x_2}} \right)\left( {x_1^2 - {x_1}{x_2} + x_2^2} \right) = \sqrt {10} .\left( {6 - 2} \right) = 4\sqrt {10} \).