Question

Đã trả lời bởi chuyên gia

Answer 1

\[\begin{array}{l}
\sqrt {5 + 2\sqrt 6 } - \sqrt {5 - 2\sqrt 6 } \\
= \left( {\sqrt {3 + 2\sqrt 3 .\sqrt 2 + 2} } \right) - \sqrt {3 - 2\sqrt 6 + 2} \\
= \sqrt {{{\left( {\sqrt 3 + \sqrt 2 } \right)}^2}} - \sqrt {{{\left( {\sqrt 3 - \sqrt 2 } \right)}^2}} \\
= |\sqrt 3 + \sqrt 2 | - |\sqrt 3 - \sqrt 2 |\\
= \sqrt 3 + \sqrt 2 - \sqrt 3 + \sqrt 2 \left( {do:\sqrt 3 + \sqrt 2 > 0;\sqrt 3 - \sqrt 2 > 0} \right)\\
= 2\sqrt 2 \\
\sqrt {7 - 2\sqrt {10} } - \sqrt {7 + 2\sqrt {10} } \\
= \sqrt {5 - 2\sqrt {10} + 2} - \sqrt {5 + 2\sqrt {10} + 2} \\
= \sqrt {{{\left( {\sqrt 5 - \sqrt 2 } \right)}^2}} - \sqrt {{{\left( {\sqrt 5 + \sqrt 2 } \right)}^2}} \\
= |\sqrt 5 - \sqrt 2 | - |\sqrt 5 + \sqrt 2 |\\
= \sqrt 5 - \sqrt 2 - \sqrt 5 - \sqrt 2 \left( {do:\sqrt 5 - \sqrt 2 > 0;\sqrt 5 + \sqrt 2 > 0} \right)\\
= - 2\sqrt 2 \\
\sqrt {4 - 2\sqrt 3 } + \sqrt {4 + 2\sqrt 3 } \\
= \sqrt {3 - 2\sqrt 3 + 1} - \sqrt {3 + 2\sqrt 3 + 1} \\
= \sqrt {{{\left( {\sqrt 3 - 1} \right)}^2}} - \sqrt {{{\left( {\sqrt 3 + 1} \right)}^2}} \\
= |\sqrt 3 - 1| - |\sqrt 3 + 1|\\
= \sqrt 3 - 1 - \sqrt 3 - 1\left( {do:\sqrt 3 - 1 > 0;\sqrt 3 + 1 > 0} \right)\\
= - 2\\
\sqrt {24 + 8\sqrt 5 } + \sqrt {9 - 4\sqrt 5 } \\
= \sqrt {20 + 2.2\sqrt 5 .2 + 4} - \sqrt {5 - 4\sqrt 5 + 4} \\
= \sqrt {{{\left( {2\sqrt 5 + 2} \right)}^2}} - \sqrt {{{\left( {\sqrt 5 - 2} \right)}^2}} \\
= |2\sqrt 5 + 2| - |\sqrt 5 - 2|\\
= 2\sqrt 5 + 2 - \sqrt 5 + 2\\
= \sqrt 5 + 4\\
\sqrt {17 - 12\sqrt 2 } + \sqrt {9 + 4\sqrt 2 } \\
= \sqrt {9 - 2.3.2\sqrt 2 + 8} - \sqrt {8 + 2.2\sqrt 2 + 1} \\
= \sqrt {{{\left( {3 - 2\sqrt 2 } \right)}^2}} - \sqrt {{{\left( {2\sqrt 2 - 1} \right)}^2}} \\
= |3 - 2\sqrt 2 | - |2\sqrt 2 - 1|\\
= 3 - 2\sqrt 2 - 2\sqrt 2 + 1\\
= 4\\
\sqrt {6 + 4\sqrt 2 } + \sqrt {22 - 12\sqrt 2 } \\
= \sqrt {4 + 2.2.\sqrt 2 + 2} - \sqrt {18 - 2.3\sqrt 2 .2 + 4} \\
= \sqrt {{{\left( {2 + \sqrt 2 } \right)}^2}} - \sqrt {{{\left( {3\sqrt 2 - 2} \right)}^2}} \\
= |2 + \sqrt 2 | - |3\sqrt 2 - 2|\\
= 2 + \sqrt 2 - 3\sqrt 2 + 2\\
= 4 - 2\sqrt 2
\end{array}\]

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Answer 2

\[\begin{array}{l}
\sqrt {5 + 2\sqrt 6 } - \sqrt {5 - 2\sqrt 6 } \\
= \left( {\sqrt {3 + 2\sqrt 3 .\sqrt 2 + 2} } \right) - \sqrt {3 - 2\sqrt 6 + 2} \\
= \sqrt {{{\left( {\sqrt 3 + \sqrt 2 } \right)}^2}} - \sqrt {{{\left( {\sqrt 3 - \sqrt 2 } \right)}^2}} \\
= |\sqrt 3 + \sqrt 2 | - |\sqrt 3 - \sqrt 2 |\\
= \sqrt 3 + \sqrt 2 - \sqrt 3 + \sqrt 2 \left( {do:\sqrt 3 + \sqrt 2 > 0;\sqrt 3 - \sqrt 2 > 0} \right)\\
= 2\sqrt 2 \\
\sqrt {7 - 2\sqrt {10} } - \sqrt {7 + 2\sqrt {10} } \\
= \sqrt {5 - 2\sqrt {10} + 2} - \sqrt {5 + 2\sqrt {10} + 2} \\
= \sqrt {{{\left( {\sqrt 5 - \sqrt 2 } \right)}^2}} - \sqrt {{{\left( {\sqrt 5 + \sqrt 2 } \right)}^2}} \\
= |\sqrt 5 - \sqrt 2 | - |\sqrt 5 + \sqrt 2 |\\
= \sqrt 5 - \sqrt 2 - \sqrt 5 - \sqrt 2 \left( {do:\sqrt 5 - \sqrt 2 > 0;\sqrt 5 + \sqrt 2 > 0} \right)\\
= - 2\sqrt 2 \\
\sqrt {4 - 2\sqrt 3 } + \sqrt {4 + 2\sqrt 3 } \\
= \sqrt {3 - 2\sqrt 3 + 1} - \sqrt {3 + 2\sqrt 3 + 1} \\
= \sqrt {{{\left( {\sqrt 3 - 1} \right)}^2}} - \sqrt {{{\left( {\sqrt 3 + 1} \right)}^2}} \\
= |\sqrt 3 - 1| - |\sqrt 3 + 1|\\
= \sqrt 3 - 1 - \sqrt 3 - 1\left( {do:\sqrt 3 - 1 > 0;\sqrt 3 + 1 > 0} \right)\\
= - 2\\
\sqrt {24 + 8\sqrt 5 } + \sqrt {9 - 4\sqrt 5 } \\
= \sqrt {20 + 2.2\sqrt 5 .2 + 4} - \sqrt {5 - 4\sqrt 5 + 4} \\
= \sqrt {{{\left( {2\sqrt 5 + 2} \right)}^2}} - \sqrt {{{\left( {\sqrt 5 - 2} \right)}^2}} \\
= |2\sqrt 5 + 2| - |\sqrt 5 - 2|\\
= 2\sqrt 5 + 2 - \sqrt 5 + 2\\
= \sqrt 5 + 4\\
\sqrt {17 - 12\sqrt 2 } + \sqrt {9 + 4\sqrt 2 } \\
= \sqrt {9 - 2.3.2\sqrt 2 + 8} - \sqrt {8 + 2.2\sqrt 2 + 1} \\
= \sqrt {{{\left( {3 - 2\sqrt 2 } \right)}^2}} - \sqrt {{{\left( {2\sqrt 2 - 1} \right)}^2}} \\
= |3 - 2\sqrt 2 | - |2\sqrt 2 - 1|\\
= 3 - 2\sqrt 2 - 2\sqrt 2 + 1\\
= 4\\
\sqrt {6 + 4\sqrt 2 } + \sqrt {22 - 12\sqrt 2 } \\
= \sqrt {4 + 2.2.\sqrt 2 + 2} - \sqrt {18 - 2.3\sqrt 2 .2 + 4} \\
= \sqrt {{{\left( {2 + \sqrt 2 } \right)}^2}} - \sqrt {{{\left( {3\sqrt 2 - 2} \right)}^2}} \\
= |2 + \sqrt 2 | - |3\sqrt 2 - 2|\\
= 2 + \sqrt 2 - 3\sqrt 2 + 2\\
= 4 - 2\sqrt 2
\end{array}\]

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