Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
\sqrt {6 + 2\sqrt 5 } + \sqrt {6 - 2\sqrt 5 } \\
= \sqrt {5 + 2\sqrt 5 + 1} + \sqrt {5 - 2\sqrt 5 + 1} \\
= \sqrt {{{\sqrt 5 }^2} + 2.\sqrt 5 .1 + {1^2}} + \sqrt {{{\sqrt 5 }^2} - 2.\sqrt 5 .1 + {1^2}} \\
= \sqrt {{{\left( {\sqrt 5 + 1} \right)}^2}} + \sqrt {{{\left( {\sqrt 5 - 1} \right)}^2}} \\
= \left| {\sqrt 5 + 1} \right| + \left| {\sqrt 5 - 1} \right|\\
= \left( {\sqrt 5 + 1} \right) + \left( {\sqrt 5 - 1} \right)\\
= 2\sqrt 5 \\
b,\\
\sqrt {8 - 2\sqrt 7 } - \sqrt {8 + 2\sqrt 7 } \\
= \sqrt {7 - 2.\sqrt 7 .1 + 1} - \sqrt {7 + 2.\sqrt 7 .1 + 1} \\
= \sqrt {{{\left( {\sqrt 7 - 1} \right)}^2}} - \sqrt {{{\left( {\sqrt 7 + 1} \right)}^2}} \\
= \left| {\sqrt 7 - 1} \right| - \left| {\sqrt 7 + 1} \right|\\
= \left( {\sqrt 7 - 1} \right) - \left( {\sqrt 7 + 1} \right)\\
= - 2\\
c,\\
\sqrt {11 + 6\sqrt 2 } - \sqrt {11 - 6\sqrt 2 } \\
= \sqrt {9 + 2.3\sqrt 2 + 2} - \sqrt {9 - 2.3.\sqrt 2 + 2} \\
= \sqrt {{{\left( {3 + \sqrt 2 } \right)}^2}} - \sqrt {{{\left( {3 - \sqrt 2 } \right)}^2}} \\
= \left| {3 + \sqrt 2 } \right| - \left| {3 - \sqrt 2 } \right|\\
= \left( {3 + \sqrt 2 } \right) - \left( {3 - \sqrt 2 } \right)\\
= 2\sqrt 2 \\
d,\\
\sqrt {17 + 12\sqrt 2 } + \sqrt {17 - 12\sqrt 2 } \\
= \sqrt {9 + 2.3.2\sqrt 2 + 8} + \sqrt {9 - 2.3.2\sqrt 2 + 8} \\
= \sqrt {{3^2} + 2.3.2\sqrt 2 + {{\left( {2\sqrt 2 } \right)}^2}} + \sqrt {{3^2} - 2.3.2\sqrt 2 .{{\left( {2\sqrt 2 } \right)}^2}} \\
= \sqrt {{{\left( {3 + 2\sqrt 2 } \right)}^2}} + \sqrt {{{\left( {3 - 2\sqrt 2 } \right)}^2}} \\
= \left| {3 + 2\sqrt 2 } \right| + \left| {3 - 2\sqrt 2 } \right|\\
= \left( {3 + 2\sqrt 2 } \right) + \left( {3 - 2\sqrt 2 } \right)\\
= 6
\end{array}\)
