Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
\sqrt {{{\left( {1 + \sqrt 2 } \right)}^2}} + \sqrt {{{\left( {\sqrt 3 - \sqrt 2 } \right)}^2}} \\
= \left| {1 + \sqrt 2 } \right| + \left| {\sqrt 3 - \sqrt 2 } \right|\\
= \left( {1 + \sqrt 2 } \right) + \left( {\sqrt 3 - \sqrt 2 } \right)\\
= 1 + \sqrt 3 \\
b,\\
\sqrt {7 - 2\sqrt {10} } - \sqrt 5 \\
= \sqrt {5 - 2.\sqrt 5 .\sqrt 2 + 2} - \sqrt 5 \\
= \sqrt {{{\left( {\sqrt 5 - \sqrt 2 } \right)}^2}} - \sqrt 5 \\
= \left| {\sqrt 5 - \sqrt 2 } \right| - \sqrt 5 \\
= \sqrt 5 - \sqrt 2 - \sqrt 5 \\
= - \sqrt 2 \\
c,\\
\sqrt {11 + 2\sqrt {18} } + \sqrt {11 - 2\sqrt {18} } \\
= \sqrt {9 + 2.\sqrt 9 .\sqrt 2 + 2} + \sqrt {9 - 2.\sqrt 9 .\sqrt 2 + 2} \\
= \sqrt {{{\left( {\sqrt 9 + \sqrt 2 } \right)}^2}} + \sqrt {{{\left( {\sqrt 9 - \sqrt 2 } \right)}^2}} \\
= \left| {3 + \sqrt 2 } \right| + \left| {3 - \sqrt 2 } \right|\\
= 3 + \sqrt 2 + 3 - \sqrt 2 \\
= 6\\
d,\\
\sqrt {3 + 2\sqrt 2 } - \sqrt {6 - 2\sqrt 8 } \\
= \sqrt {2 + 2\sqrt 2 .1 + 1} - \sqrt {4 - 2.\sqrt 4 .\sqrt 2 + 2} \\
= \sqrt {{{\left( {\sqrt 2 + 1} \right)}^2}} - \sqrt {{{\left( {\sqrt 4 - \sqrt 2 } \right)}^2}} \\
= \left| {\sqrt 2 + 1} \right| - \left| {2 - \sqrt 2 } \right|\\
= \sqrt 2 + 1 - \left( {2 - \sqrt 2 } \right)\\
= 2\sqrt 2 - 1
\end{array}\)
