Đáp án:
\(\begin{array}{l}
2B.\\
a.\sqrt {3a} \\
b.\sqrt {2{a^2}}
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
2A\\
a.Do:a \ge 0\\
\to a\sqrt {13} = \sqrt {13{a^2}} \\
b.Do:a < 0\\
\to a\sqrt {\dfrac{{ - 15}}{a}} = \sqrt {{{\left( { - a} \right)}^2}.\left( {\dfrac{{ - 15}}{a}} \right)} = \sqrt {{a^2}.\left( {\dfrac{{ - 15}}{a}} \right)} \\
= \sqrt { - 15a} \\
2B.\\
a.Do:a > 0\\
\to \dfrac{a}{2}.\sqrt {\dfrac{{12}}{a}} = \sqrt {\dfrac{{{a^2}}}{4}.\dfrac{{12}}{a}} = \sqrt {\dfrac{{a.3}}{1}} = \sqrt {3a} \\
b.Do:a \le 0\\
\to a\sqrt 2 = \sqrt {2{a^2}}
\end{array}\)
