Đáp án:
\[\begin{array}{l}
a.5,\,\,\,\,\,A = 2\sqrt 6 - 3\sqrt 3 \\
a.6,\,\,\,\,A = - \sqrt 7
\end{array}\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a.5,\\
A = \sqrt {74 + 40\sqrt 3 } - \sqrt {77 + 30\sqrt 6 } \\
= \sqrt {50 + 2.20\sqrt 3 + 24} - \sqrt {50 + 2.15\sqrt 6 + 27} \\
= \sqrt {50 + 2.\sqrt {{{20}^2}.3} + 24} - \sqrt {50 + 2.\sqrt {{{15}^2}.6} + 27} \\
= \sqrt {50 + 2.\sqrt {1200} + 24} - \sqrt {50 + 2.\sqrt {1350} + 27} \\
= \sqrt {50 + 2.\sqrt {50} .\sqrt {24} + 24} - \sqrt {50 + 2.\sqrt {50} .\sqrt {27} + 27} \\
= \sqrt {{{\left( {\sqrt {50} + \sqrt {24} } \right)}^2}} - \sqrt {{{\left( {\sqrt {50} + \sqrt {27} } \right)}^2}} \\
= \left( {\sqrt {50} + \sqrt {24} } \right) - \left( {\sqrt {50} + \sqrt {27} } \right)\\
= \sqrt {24} - \sqrt {27} \\
= \sqrt {{2^2}.6} - \sqrt {{3^2}.3} \\
= 2\sqrt 6 - 3\sqrt 3 \\
a.6,\\
A = \sqrt {73 - 12\sqrt {35} } - \sqrt {52 - 6\sqrt {35} } \\
= \sqrt {45 - 2.6.\sqrt {35} + 28} - \sqrt {45 - 2.3.\sqrt {35} + 7} \\
= \sqrt {45 - 2.\sqrt {{6^2}.35} + 28} - \sqrt {45 - 2.\sqrt {{3^2}.35} + 7} \\
= \sqrt {45 - 2.\sqrt {1260} + 28} - \sqrt {45 - 2.\sqrt {315} + 7} \\
= \sqrt {45 - 2.\sqrt {45} .\sqrt {28} + 28} - \sqrt {45 - 2.\sqrt {45} .\sqrt 7 + 7} \\
= \sqrt {{{\left( {\sqrt {45} - \sqrt {28} } \right)}^2}} - \sqrt {{{\left( {\sqrt {45} - \sqrt 7 } \right)}^2}} \\
= \left( {\sqrt {45} - \sqrt {28} } \right) - \left( {\sqrt {45} - \sqrt 7 } \right)\\
= - \sqrt {28} + \sqrt 7 \\
= - \sqrt {{2^2}.7} + \sqrt 7 \\
= - 2\sqrt 7 + \sqrt 7 \\
= - \sqrt 7
\end{array}\)
