Đáp án:
$\begin{array}{l}
1)3 + 2\sqrt 2 = \sqrt {9 + 2.3.2\sqrt 2 + 8} \\
= \sqrt {17 + 12\sqrt 2 } = \sqrt {17 + 6.2\sqrt 2 } \\
= \sqrt {17 + 6.\sqrt 8 } \\
\sqrt {35} = \sqrt {17 + 18} = \sqrt {17 + 6.3} = \sqrt {17 + 6\sqrt 9 } \\
\Leftrightarrow 3 + 2\sqrt 2 < \sqrt {35} \\
2)\sqrt 5 + \sqrt 6 \\
= \sqrt {5 + 2\sqrt 5 .\sqrt 6 + 6} \\
= \sqrt {11 + \sqrt {120} } \\
\sqrt {22} = \sqrt {11 + 11} = \sqrt {11 + \sqrt {121} } \\
\Leftrightarrow \sqrt 5 + \sqrt 6 < \sqrt {22} \\
3)\sqrt {5 - \sqrt {21 - 4\sqrt 5 } } \\
= \sqrt {5 - \sqrt {20 - 2.2\sqrt 5 + 1} } \\
= \sqrt {5 - \sqrt {{{\left( {2\sqrt 5 - 1} \right)}^2}} } \\
= \sqrt {5 - 2\sqrt 5 + 1} \\
= \sqrt {6 - 2\sqrt 5 } \\
= \sqrt {{{\left( {\sqrt 5 - 1} \right)}^2}} \\
= \sqrt 5 - 1 > \sqrt 3 - 1\\
Vậy\,\sqrt {5 - \sqrt {21 - 4\sqrt 5 } } > \sqrt 3 - 1\\
4)\sqrt {4 + \sqrt 7 } - \sqrt {4 - \sqrt 7 } > 0\\
= \dfrac{1}{{\sqrt 2 }}.\left( {\sqrt {8 + 2\sqrt 7 } - \sqrt {8 - 2\sqrt 7 } } \right)\\
= \dfrac{{\sqrt 2 }}{2}.\left( {\sqrt {{{\left( {\sqrt 7 + 1} \right)}^2}} - \sqrt {{{\left( {\sqrt 7 - 1} \right)}^2}} } \right)\\
= \dfrac{{\sqrt 2 }}{2}.\left( {\sqrt 7 + 1 - \sqrt 7 + 1} \right)\\
= \sqrt 2 > - \sqrt 2 \\
5)\sqrt 7 + \sqrt {15} = \sqrt {7 + 2\sqrt 7 .\sqrt {15} + 15} \\
= \sqrt {22 + \sqrt {420} } \\
7 = \sqrt {49} = \sqrt {22 + 27} = \sqrt {22 + 729} \\
\Leftrightarrow \sqrt 7 + \sqrt {15} < 7\\
6)\sqrt 2 + \sqrt {15} \\
= \sqrt {2 + 2\sqrt 2 .\sqrt {15} + 15} \\
= \sqrt {17 + 2\sqrt {30} } \\
\sqrt 3 + \sqrt 5 = \sqrt {8 + 2\sqrt {15} } < \sqrt {17 + 2\sqrt {30} } \\
\Leftrightarrow \sqrt 2 + \sqrt {15} > \sqrt 3 + \sqrt 5 \\
7)\\
3\sqrt {26} \\
15 = 3.5 = 3.\sqrt {25} < 3.\sqrt {26} \\
\Leftrightarrow 3\sqrt {26} > 15\\
8)\sqrt {5\sqrt 3 } = \sqrt {\sqrt {25.3} } = \sqrt {\sqrt {75} } \\
\sqrt {3\sqrt 5 } = \sqrt {\sqrt {9.5} } = \sqrt {\sqrt {45} } < \sqrt {\sqrt {75} } \\
\Leftrightarrow \sqrt {5\sqrt 3 } > \sqrt {3\sqrt 5 } \\
9(\sqrt {37} .\sqrt {39} = \sqrt {\left( {38 - 1} \right)\left( {38 + 1} \right)} = \sqrt {{{38}^2} - 1} \\
\Leftrightarrow \sqrt {37} .\sqrt {39} < 38\\
10)\\
\sqrt {15} - \sqrt {10} = \dfrac{{15 - 10}}{{\sqrt {15} + \sqrt {10} }} = \dfrac{5}{{\sqrt {15} + \sqrt {10} }}\\
\sqrt {26} - \sqrt {21} = \dfrac{{26 - 21}}{{\sqrt {26} + \sqrt {21} }} = \dfrac{5}{{\sqrt {26} + \sqrt {21} }}\\
Do:\sqrt {15} + \sqrt {10} < \sqrt {26} + \sqrt {21} \\
\Leftrightarrow \dfrac{5}{{\sqrt {15} + \sqrt {10} }} > \dfrac{5}{{\sqrt {26} + \sqrt {21} }}\\
\Leftrightarrow \sqrt {15} - \sqrt {10} > \sqrt {26} - \sqrt {21}
\end{array}$
