Đáp án:
\(\begin{array}{l}
a)A = 1\\
b)B = \sqrt 3 + 1\\
c)C = 2
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)A = \sqrt {2 + \sqrt 3 } .\sqrt {2 + \sqrt {2 + \sqrt 3 } } .\sqrt {\left( {2 + \sqrt {2 + \sqrt {2 + \sqrt 3 } } } \right).\left( {2 - \sqrt {2 + \sqrt {2 + \sqrt 3 } } } \right)} \\
= \sqrt {2 + \sqrt 3 } .\sqrt {2 + \sqrt {2 + \sqrt 3 } } .\sqrt {\left( {4 - \left( {2 + \sqrt {2 + \sqrt 3 } } \right)} \right)} \\
= \sqrt {2 + \sqrt 3 } .\sqrt {2 + \sqrt {2 + \sqrt 3 } } .\sqrt {2 - \sqrt {2 + \sqrt 3 } } \\
= \sqrt {2 + \sqrt 3 } .\sqrt {\left( {2 + \sqrt {2 + \sqrt 3 } } \right)\left( {2 - \sqrt {2 + \sqrt 3 } } \right)} \\
= \sqrt {2 + \sqrt 3 } .\sqrt {4 - \left( {2 + \sqrt 3 } \right)} \\
= \sqrt {2 + \sqrt 3 } .\sqrt {2 - \sqrt 3 } \\
= \sqrt {4 - 3} = 1\\
b)B = \sqrt {2\sqrt 3 + \sqrt {\dfrac{{\sqrt {16 + 2.4.\sqrt 5 + 5} }}{{4 + \sqrt 5 }}.\sqrt {5 - 2.2.\sqrt 5 + 4} - \sqrt 5 + 18} } \\
= \sqrt {2\sqrt 3 + \sqrt {\dfrac{{\sqrt {{{\left( {4 + \sqrt 5 } \right)}^2}} }}{{4 + \sqrt 5 }}.\sqrt {{{\left( {\sqrt 5 - 2} \right)}^2}} - \sqrt 5 + 18} } \\
= \sqrt {2\sqrt 3 + \sqrt {\dfrac{{4 + \sqrt 5 }}{{4 + \sqrt 5 }}.\left( {\sqrt 5 - 2} \right) - \sqrt 5 + 18} } \\
= \sqrt {2\sqrt 3 + \sqrt {\sqrt 5 - 2 - \sqrt 5 + 18} } \\
= \sqrt {2\sqrt 3 + \sqrt {16} } \\
= \sqrt {4 + 2\sqrt 3 } = \sqrt {3 + 2\sqrt 3 .1 + 1} \\
= \sqrt {{{\left( {\sqrt 3 + 1} \right)}^2}} = \sqrt 3 + 1\\
c)C = \sqrt {2 + \sqrt 3 + \sqrt {4 - 2\sqrt 3 - \sqrt {12 - 2.2\sqrt 3 .3 + 9} } } \\
= \sqrt {2 + \sqrt 3 + \sqrt {4 - 2\sqrt 3 - \sqrt {{{\left( {2\sqrt 3 - 3} \right)}^2}} } } \\
= \sqrt {2 + \sqrt 3 + \sqrt {4 - 2\sqrt 3 - \left( {2\sqrt 3 - 3} \right)} } \\
= \sqrt {2 + \sqrt 3 + \sqrt {4 - 2\sqrt 3 - 2\sqrt 3 + 3} } \\
= \sqrt {2 + \sqrt 3 + \sqrt {7 - 4\sqrt 3 } } \\
= \sqrt {2 + \sqrt 3 + \sqrt {4 - 2.2\sqrt 3 + 3} } \\
= \sqrt {2 + \sqrt 3 + \sqrt {{{\left( {2 - \sqrt 3 } \right)}^2}} } \\
= \sqrt {2 + \sqrt 3 + 2 - \sqrt 3 } = \sqrt 4 = 2
\end{array}\)
