Đáp án:
$\begin{array}{l}
3b)\dfrac{1}{4}\sqrt {82} = \sqrt {\dfrac{{82}}{{16}}} = \sqrt {\dfrac{{41}}{4}} \\
6\sqrt {\dfrac{1}{7}} = \sqrt {\dfrac{{36}}{7}} \\
Xet:\dfrac{{41}}{4} - \dfrac{{36}}{7} = \dfrac{{143}}{{28}} > 0\\
\Leftrightarrow \sqrt {\dfrac{{41}}{4}} > \sqrt {\dfrac{{36}}{7}} \\
\Leftrightarrow \dfrac{1}{4}\sqrt {82} > 6\sqrt {\dfrac{1}{7}} \\
B4)a)\sqrt {98} + \sqrt {72} - 0,5\sqrt 8 \\
= \sqrt {49.2} + \sqrt {36.2} - 0,5.\sqrt {4.2} \\
= 7\sqrt 2 + 6\sqrt 2 - 0,5.2.\sqrt 2 \\
= 13\sqrt 2 - \sqrt 2 \\
= 12\sqrt 2 \\
b)\sqrt {32a} + 2\sqrt {18a} - \sqrt {2a} \left( {dkxd:a \ge 0} \right)\\
= \sqrt {16.2a} + 2.\sqrt {9.2a} - \sqrt {2a} \\
= 4\sqrt {2a} + 2.3\sqrt {2a} - \sqrt {2a} \\
= 9\sqrt {2a}
\end{array}$
