Đáp án:
\(f.2\sqrt 3 \)
Giải thích các bước giải:
\(\begin{array}{l}
2)\\
a.\left| {2\sqrt 2 - 3} \right| = - \left( {2\sqrt 2 - 3} \right)\left( {do:2\sqrt 2 < 3} \right)\\
= 3 - 2\sqrt 2 \\
b.\left| {3 + \sqrt 2 } \right| - \left| {1 - \sqrt 2 } \right|\\
= 3 + \sqrt 2 - \left( { - 1 + \sqrt 2 } \right)\left( {do:\left( \begin{array}{l}
3 > \sqrt 2 \\
1 < \sqrt 2
\end{array} \right.} \right)\\
= 3 + \sqrt 2 + 1 - \sqrt 2 = 4\\
c.\left| {\sqrt 5 + \sqrt 2 } \right| + \left| {\sqrt 5 - \sqrt 2 } \right|\\
= \sqrt 5 + \sqrt 2 + \sqrt 5 - \sqrt 2 \left( {do:\sqrt 5 > \sqrt 2 } \right)\\
= 2\sqrt 5 \\
d.\left| {\dfrac{1}{{\sqrt 2 }} - \dfrac{1}{2}} \right| = - \left( {\dfrac{1}{{\sqrt 2 }} - \dfrac{1}{2}} \right)\\
= \dfrac{1}{2} - \dfrac{1}{{\sqrt 2 }}\\
e.\sqrt {5 + 2\sqrt 6 } - \sqrt {5 - 2\sqrt 6 } \\
= \sqrt {3 + 2\sqrt 3 .\sqrt 2 + 2} - \sqrt {3 - 2\sqrt 3 .\sqrt 2 + 2} \\
= \sqrt {{{\left( {\sqrt 3 + \sqrt 2 } \right)}^2}} - \sqrt {{{\left( {\sqrt 3 - \sqrt 2 } \right)}^2}} \\
= \sqrt 3 + \sqrt 2 - \sqrt 3 + \sqrt 2 = 2\sqrt 2 \\
f.\sqrt {3 + 2\sqrt 3 .1 + 1} + \sqrt {3 - 2\sqrt 3 .1 + 1} \\
= \sqrt {{{\left( {\sqrt 3 + 1} \right)}^2}} + \sqrt {{{\left( {\sqrt 3 - 1} \right)}^2}} \\
= \sqrt 3 + 1 + \sqrt 3 - 1 = 2\sqrt 3
\end{array}\)
