Đáp án:
$\begin{array}{l}
25)\sqrt 6 \\
26)\sqrt {10} \\
27) - \sqrt 7 \\
28) - \sqrt 5 \\
29) - 3\sqrt 2 \\
30)\sqrt 3 \\
31) - \sqrt 5 \\
32)\sqrt 5 \\
33)\sqrt 6 \\
34) - 2\sqrt 2 \\
35)\sqrt {30} \\
36)2\sqrt 3 \\
37)\dfrac{2}{{3\sqrt 5 }}\\
38)2\sqrt 2 \\
39)3\sqrt 3
\end{array}$
Giải thích các bước giải:
$\begin{array}{l}
25)\dfrac{{\sqrt 6 - 6}}{{1 - \sqrt 6 }} = \dfrac{{\sqrt 6 \left( {1 - \sqrt 6 } \right)}}{{1 - \sqrt 6 }} = \sqrt 6 \\
26)\dfrac{{5\sqrt 2 - 2\sqrt 5 }}{{\sqrt 5 - \sqrt 2 }} = \dfrac{{\sqrt {10} \left( {\sqrt 5 - \sqrt 2 } \right)}}{{\sqrt 5 - \sqrt 2 }} = \sqrt {10} \\
27)\dfrac{{\sqrt {14} - \sqrt 7 }}{{1 - \sqrt 2 }} = \dfrac{{\sqrt 7 \left( {\sqrt 2 - 1} \right)}}{{1 - \sqrt 2 }} = - \sqrt 7 \\
28)\dfrac{{\sqrt {15} - \sqrt 5 }}{{1 - \sqrt 3 }} = \dfrac{{\sqrt 5 \left( {\sqrt 3 - 1} \right)}}{{1 - \sqrt 3 }} = - \sqrt 5 \\
29)\dfrac{{3\sqrt 2 - 6}}{{\sqrt 2 - 1}} = \dfrac{{3\sqrt 2 \left( {1 - \sqrt 2 } \right)}}{{\sqrt 2 - 1}} = - 3\sqrt 2 \\
30)\dfrac{{\sqrt {15} - \sqrt {12} }}{{\sqrt 5 - 2}} = \dfrac{{\sqrt 3 \left( {\sqrt 5 - \sqrt 4 } \right)}}{{\sqrt 5 - 2}} = \dfrac{{\sqrt 3 \left( {\sqrt 5 - 2} \right)}}{{\sqrt 5 - 2}} = \sqrt 3 \\
31)\dfrac{{5 - 2\sqrt 5 }}{{2 - \sqrt 5 }} = \dfrac{{\sqrt 5 \left( {\sqrt 5 - 2} \right)}}{{2 - \sqrt 5 }} = - \sqrt 5 \\
32)\dfrac{{5 + 3\sqrt 5 }}{{3 + \sqrt 5 }} = \dfrac{{\sqrt 5 \left( {\sqrt 5 + 3} \right)}}{{3 + \sqrt 5 }} = \sqrt 5 \\
33)\dfrac{{3\sqrt 2 - 2\sqrt 3 }}{{\sqrt 3 - \sqrt 2 }} = \dfrac{{\sqrt 6 \left( {\sqrt 3 - \sqrt 2 } \right)}}{{\sqrt 3 - \sqrt 2 }} = \sqrt 6 \\
34)\dfrac{{6\sqrt 2 - 4}}{{\sqrt 2 - 3}} = \dfrac{{2\sqrt 2 \left( {3 - \sqrt 2 } \right)}}{{\sqrt 2 - 3}} = - 2\sqrt 2 \\
35)\dfrac{{5\sqrt 6 + 6\sqrt 5 }}{{\sqrt 5 + \sqrt 6 }} = \dfrac{{\sqrt {30} \left( {\sqrt 5 + \sqrt 6 } \right)}}{{\sqrt 5 + \sqrt 6 }} = \sqrt {30} \\
36)\dfrac{{6 + 2\sqrt 6 }}{{\sqrt 3 + \sqrt 2 }} = \dfrac{{\sqrt {12} \left( {\sqrt 3 + \sqrt 2 } \right)}}{{\sqrt 3 + \sqrt 2 }} = \sqrt {12} = 2\sqrt 3 \\
37)\dfrac{{2\sqrt 7 - 4\sqrt 3 }}{{3\sqrt {35} - 6\sqrt {15} }} = \dfrac{{2\left( {\sqrt 7 - 2\sqrt 3 } \right)}}{{3\sqrt 5 \left( {\sqrt 7 - 2\sqrt 3 } \right)}} = \dfrac{2}{{3\sqrt 5 }}\\
38)\dfrac{{12\sqrt {10} - 16\sqrt {14} }}{{6\sqrt 5 - 8\sqrt 7 }} = \dfrac{{2\sqrt 2 \left( {6\sqrt 5 - 8\sqrt 7 } \right)}}{{6\sqrt 5 - 8\sqrt 7 }} = 2\sqrt 2 \\
39)\dfrac{{6\sqrt 6 - 27}}{{2\sqrt 2 - 3\sqrt 3 }} = \dfrac{{3\sqrt 3 \left( {2\sqrt 2 - 3\sqrt 3 } \right)}}{{2\sqrt 2 - 3\sqrt 3 }} = 3\sqrt 3
\end{array}$