Đáp án:
$\begin{array}{l}
a)2\sqrt {27} = \sqrt {4.27} = \sqrt {108} < \sqrt {147} \\
Vậy\,2\sqrt {27} < \sqrt {147} \\
b) - 3\sqrt 5 = - \sqrt {45} \\
- 5\sqrt 3 = - \sqrt {75} < - \sqrt {45} \\
\Leftrightarrow - 3\sqrt 5 > - 5\sqrt 3 \\
c)21 = \sqrt {441} \\
2\sqrt 7 = \sqrt {4.7} = \sqrt {28} \\
15\sqrt 3 = \sqrt {225.3} = \sqrt {675} \\
- \sqrt {123} \\
\Leftrightarrow - \sqrt {123} ,2\sqrt 7 ,21,15\sqrt 3 \\
d)2\sqrt {15} = \sqrt {4.15} = \sqrt {60} > \sqrt {59} \\
Vậy\,2\sqrt {15} > \sqrt {59} \\
e)2\sqrt 2 - 1 - 2\\
= 2\sqrt 2 - 3\\
= \sqrt 8 - \sqrt 9 < 0\\
\Leftrightarrow 2\sqrt 2 - 1 < 2\\
f)6 = \sqrt {36} < \sqrt {41} \\
g)\dfrac{{\sqrt 3 }}{2} = \dfrac{{\sqrt 3 }}{{\sqrt 4 }} < 1\\
h) - \dfrac{{\sqrt {10} }}{2} = - \dfrac{{\sqrt {20} }}{{2\sqrt 2 }}\\
- 2\sqrt 5 = - \sqrt {20} < - \dfrac{{\sqrt {20} }}{{2\sqrt 2 }}\\
Vậy\, - \dfrac{{\sqrt {10} }}{2} < - \dfrac{{\sqrt {20} }}{{2\sqrt 2 }}\\
i)\sqrt 6 - 1 - 3 = \sqrt 6 - 4 = \sqrt 6 - \sqrt {16} < 0\\
\Leftrightarrow \sqrt 6 - 1 < 3\\
j)2\sqrt 5 - 5\sqrt 2 = \sqrt {20} - \sqrt {50} < 0\\
\Leftrightarrow 2\sqrt 5 - 5\sqrt 2 < 1\\
k)\dfrac{{\sqrt 8 }}{3} = \sqrt {\dfrac{8}{9}} \\
\dfrac{3}{4} = \sqrt {\dfrac{9}{{16}}} \\
Do:\dfrac{8}{9} > \dfrac{9}{{16}}\\
\Leftrightarrow \dfrac{{\sqrt 8 }}{3} > \dfrac{3}{4}\\
k)6\sqrt {\dfrac{1}{4}} = 6.\dfrac{1}{2} = 3\\
4\sqrt {\dfrac{1}{2}} = 2\sqrt 2 < 3\\
2\sqrt 3 = \sqrt {12} > 3\\
\sqrt {\dfrac{{15}}{5}} = \sqrt 3 < 4\sqrt {\dfrac{1}{2}} \\
\Leftrightarrow 2\sqrt 3 ,6\sqrt {\dfrac{1}{4}} ,4\sqrt {\dfrac{1}{2}} ,\sqrt {\dfrac{{15}}{5}} ; - \sqrt {132} \\
m) - 2\sqrt 6 = - \sqrt {24} < - \sqrt {23} \\
q)\sqrt 9 = 3\\
\sqrt {25} - \sqrt {16} = 5 - 4 = 1\\
\Leftrightarrow \sqrt 9 > \sqrt {25} - \sqrt {16} \\
r)\sqrt {111} - 7 < \sqrt {121} - 7 = 11 - 7 = 4\\
\Leftrightarrow \sqrt {111} - 7 < 4\\
p) - 27\\
4\sqrt 3 = \sqrt {48} \\
16\sqrt 5 = \sqrt {256.5} = \sqrt {1280} \\
21\sqrt 2 = \sqrt {442.2} = \sqrt {884} \\
\Leftrightarrow 16\sqrt 5 ,21\sqrt 2 ,4\sqrt 3 , - 27
\end{array}$
