Đáp án:
\(\begin{array}{l}
a)2\\
b)\sqrt {10} \\
c)2\\
d)8
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\sqrt {4 + \sqrt 8 } .\sqrt {{2^2} - {{\left( {\sqrt {2 + \sqrt 2 } } \right)}^2}} \\
= \sqrt {4 + \sqrt 8 } .\sqrt {4 - 2 - \sqrt 2 } \\
= \sqrt {4 + 2\sqrt 2 } .\sqrt {2 - \sqrt 2 } \\
= \sqrt 2 .\sqrt {2 + \sqrt 2 } .\sqrt {2 - \sqrt 2 } \\
= \sqrt 2 .\sqrt {4 - 2} = \sqrt 2 .\sqrt 2 = 2\\
b)\dfrac{{\sqrt {6 + 2\sqrt 5 } + \sqrt {6 - 2\sqrt 5 } }}{{\sqrt 2 }}\\
= \dfrac{{\sqrt {5 + 2.\sqrt 5 .1 + 1} + \sqrt {5 - 2.\sqrt 5 .1 + 1} }}{{\sqrt 2 }}\\
= \dfrac{{\sqrt {{{\left( {\sqrt 5 + 1} \right)}^2}} + \sqrt {{{\left( {\sqrt 5 - 1} \right)}^2}} }}{{\sqrt 2 }}\\
= \dfrac{{\sqrt 5 + 1 + \sqrt 5 - 1}}{{\sqrt 2 }}\\
= \dfrac{{2\sqrt 5 }}{{\sqrt 2 }} = \sqrt {10} \\
c)\left( {4 + \sqrt {15} } \right).\left( {\sqrt 5 - \sqrt 3 } \right)\sqrt 2 .\sqrt {4 - \sqrt {15} } \\
= \left( {4 + \sqrt {15} } \right).\left( {\sqrt 5 - \sqrt 3 } \right)\sqrt {8 - 2\sqrt {15} } \\
= \left( {4 + \sqrt {15} } \right).\left( {\sqrt 5 - \sqrt 3 } \right)\sqrt {5 - 2.\sqrt 5 .\sqrt 3 + 3} \\
= \left( {4 + \sqrt {15} } \right).\left( {\sqrt 5 - \sqrt 3 } \right)\sqrt {{{\left( {\sqrt 5 - \sqrt 3 } \right)}^2}} \\
= \left( {4 + \sqrt {15} } \right).{\left( {\sqrt 5 - \sqrt 3 } \right)^2}\\
= \left( {4 + \sqrt {15} } \right).\left( {8 - 2\sqrt {15} } \right)\\
= \left( {4 + \sqrt {15} } \right).2\left( {4 - \sqrt {15} } \right)\\
= 2\left( {16 - 15} \right) = 2\\
d)\left( {3 + \sqrt 5 } \right).\left( {\sqrt 5 - 1} \right)\sqrt 2 .\sqrt {3 - \sqrt 5 } \\
= \left( {3 + \sqrt 5 } \right).\left( {\sqrt 5 - 1} \right)\sqrt {6 - 2\sqrt 5 } \\
= \left( {3 + \sqrt 5 } \right).\left( {\sqrt 5 - 1} \right)\sqrt {5 - 2.\sqrt 5 .1 + 1} \\
= \left( {3 + \sqrt 5 } \right).\left( {\sqrt 5 - 1} \right)\sqrt {{{\left( {\sqrt 5 - 1} \right)}^2}} \\
= \left( {3 + \sqrt 5 } \right).{\left( {\sqrt 5 - 1} \right)^2}\\
= \left( {3 + \sqrt 5 } \right)\left( {6 - 2\sqrt 5 } \right)\\
= 2\left( {3 + \sqrt 5 } \right)\left( {3 - \sqrt 5 } \right)\\
= 2.\left( {9 - 5} \right) = 2.4 = 8
\end{array}\)
