Đáp án:
\(\begin{array}{l}
d,\,\,\,3\sqrt 3 \\
e,\,\,\,A = \sqrt 2 \\
f,\,\,\,B = \sqrt 2 \\
g,\,\,\,C = 2
\end{array}\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
d,\\
\sqrt {5 + 2\sqrt 6 } + \sqrt {14 - 4\sqrt 6 } \\
= \sqrt {3 + 2.\sqrt 3 .\sqrt 2 + 2} + \sqrt {12 - 2.2.\sqrt 2 .\sqrt 3 + 2} \\
= \sqrt {{{\sqrt 3 }^2} + 2.\sqrt 3 .\sqrt 2 + {{\sqrt 2 }^2}} + \sqrt {{{\left( {2\sqrt 3 } \right)}^2} - 2.2\sqrt 3 .\sqrt 2 + {{\sqrt 2 }^2}} \\
= \sqrt {{{\left( {\sqrt 3 + \sqrt 2 } \right)}^2}} + \sqrt {{{\left( {2\sqrt 3 - \sqrt 2 } \right)}^2}} \\
= \left| {\sqrt 3 + \sqrt 2 } \right| + \left| {2\sqrt 3 - \sqrt 2 } \right|\\
= \left( {\sqrt 3 + \sqrt 2 } \right) + \left( {2\sqrt 3 - \sqrt 2 } \right)\\
= 3\sqrt 3 \\
e,\\
A = \sqrt {6 - 3\sqrt 3 } + \sqrt {2 - \sqrt 3 } \\
= \dfrac{1}{{\sqrt 2 }}.\left( {\sqrt 2 .\sqrt {6 - 3\sqrt 3 } + \sqrt 2 .\sqrt {2 - \sqrt 3 } } \right)\\
= \dfrac{1}{{\sqrt 2 }}.\left( {\sqrt {2.\left( {6 - 3\sqrt 3 } \right)} + \sqrt {2.\left( {2 - \sqrt 3 } \right)} } \right)\\
= \dfrac{1}{{\sqrt 2 }}.\left( {\sqrt {12 - 6\sqrt 3 } + \sqrt {4 - 2\sqrt 3 } } \right)\\
= \dfrac{1}{{\sqrt 2 }}.\left( {\sqrt {9 - 2.3.\sqrt 3 + 3} + \sqrt {3 - 2.\sqrt 3 .1 + 1} } \right)\\
= \dfrac{1}{{\sqrt 2 }}.\left( {\sqrt {{{\left( {3 - \sqrt 3 } \right)}^2}} + \sqrt {{{\left( {\sqrt 3 - 1} \right)}^2}} } \right)\\
= \dfrac{1}{{\sqrt 2 }}.\left( {\left| {3 - \sqrt 3 } \right| + \left| {\sqrt 3 - 1} \right|} \right)\\
= \dfrac{1}{{\sqrt 2 }}.\left( {3 - \sqrt 3 + \sqrt 3 - 1} \right)\\
= \dfrac{1}{{\sqrt 2 }}.2\\
= \sqrt 2 \\
f,\\
B = \sqrt {3 + \sqrt 5 } - \sqrt {3 - \sqrt 5 } \\
= \dfrac{1}{{\sqrt 2 }}.\left( {\sqrt 2 .\sqrt {3 + \sqrt 5 } - \sqrt 2 .\sqrt {3 - \sqrt 5 } } \right)\\
= \dfrac{1}{{\sqrt 2 }}.\left( {\sqrt {2.\left( {3 + \sqrt 5 } \right)} - \sqrt {2.\left( {3 - \sqrt 5 } \right)} } \right)\\
= \dfrac{1}{{\sqrt 2 }}.\left( {\sqrt {6 + 2\sqrt 5 } - \sqrt {6 - 2\sqrt 5 } } \right)\\
= \dfrac{1}{{\sqrt 2 }}.\left( {\sqrt {5 + 2.\sqrt 5 .1 + 1} - \sqrt {5 - 2.\sqrt 5 .1 + 1} } \right)\\
= \dfrac{1}{{\sqrt 2 }}.\left( {\sqrt {{{\left( {\sqrt 5 + 1} \right)}^2}} - \sqrt {{{\left( {\sqrt 5 - 1} \right)}^2}} } \right)\\
= \dfrac{1}{{\sqrt 2 }}.\left( {\left| {\sqrt 5 + 1} \right| - \left| {\sqrt 5 - 1} \right|} \right)\\
= \dfrac{1}{{\sqrt 2 }}.\left( {\sqrt 5 + 1 - \left( {\sqrt 5 - 1} \right)} \right)\\
= \dfrac{1}{{\sqrt 2 }}.2\\
= \sqrt 2 \\
g,\\
C = \left( {\sqrt 6 - \sqrt 2 } \right).\sqrt {2 + \sqrt 3 } \\
= \left( {\sqrt 2 .\sqrt 3 - \sqrt 2 } \right).\sqrt {2 + \sqrt 3 } \\
= \left( {\sqrt 3 - 1} \right).\sqrt 2 .\sqrt {2 + \sqrt 3 } \\
= \left( {\sqrt 3 - 1} \right).\sqrt {2.\left( {2 + \sqrt 3 } \right)} \\
= \left( {\sqrt 3 - 1} \right).\sqrt {4 + 2\sqrt 3 } \\
= \left( {\sqrt 3 - 1} \right).\sqrt {3 + 2.\sqrt 3 .1 + 1} \\
= \left( {\sqrt 3 - 1} \right).\sqrt {{{\left( {\sqrt 3 + 1} \right)}^2}} \\
= \left( {\sqrt 3 - 1} \right).\left( {\sqrt 3 + 1} \right)\\
= {\sqrt 3 ^2} - {1^2}\\
= 2
\end{array}\)
