Đáp án:
\(\begin{array}{l}
3a,\\
- 3\sqrt 2 < - \sqrt {14} < - 2\sqrt 3 < - 3 < 0 < 2\sqrt 2 \\
4b,\\
B = \sqrt 2 + \sqrt 5
\end{array}\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
3a,\\
2\sqrt 3 = \sqrt {{2^2}.3} = \sqrt {12} \\
3\sqrt 2 = \sqrt {{3^2}.2} = \sqrt {18} \\
3 = \sqrt {{3^2}} = \sqrt 9 \\
18 > 14 > 12 > 9\\
\Rightarrow \sqrt {18} > \sqrt {14} > \sqrt {12} > \sqrt 9 \\
\Rightarrow - \sqrt {18} < - \sqrt {14} < - \sqrt {12} < - \sqrt 9 < 0 < 2\sqrt 2 \\
\Leftrightarrow - 3\sqrt 2 < - \sqrt {14} < - 2\sqrt 3 < - 3 < 0 < 2\sqrt 2 \\
4b,\\
B = \sqrt {29 + 6\sqrt 6 } - \sqrt {32 - 6\sqrt {15} } \\
= \sqrt {27 + 6\sqrt 6 + 2} - \sqrt {27 - 6\sqrt {15} + 5} \\
= \sqrt {{{\left( {3\sqrt 3 } \right)}^2} + 2.3\sqrt 3 .\sqrt 2 + {{\left( {\sqrt 2 } \right)}^2}} - \sqrt {{{\left( {3\sqrt 3 } \right)}^2} - 2.3\sqrt 3 .\sqrt 5 + {{\sqrt 5 }^2}} \\
= \sqrt {{{\left( {3\sqrt 3 + \sqrt 2 } \right)}^2}} - \sqrt {{{\left( {3\sqrt 3 - \sqrt 5 } \right)}^2}} \\
= \left| {3\sqrt 3 + \sqrt 2 } \right| - \left| {3\sqrt 3 - \sqrt 5 } \right|\\
= \left( {3\sqrt 3 + \sqrt 2 } \right) - \left( {3\sqrt 3 - \sqrt 5 } \right)\\
= \sqrt 2 + \sqrt 5
\end{array}\)
