Đưa thừa số vào trong dấu căn\(\begin{array}{l}a)\,\,\frac{1}{{xy}}\sqrt {\frac{{{x^2}{y^2}}}{2}}  &  &  & b)\,\,a\sqrt 2  &  &  & c)\,\, - \frac{a}{b}\sqrt...

A \(\begin{array}{l}
a)\,\,\left\{ \begin{array}{l}
\frac{{\sqrt 2 }}{2}\,\,\,khi\,\,\,xy > 0\\
- \frac{{\sqrt 2 }}{2}\,\,\,khi\,\,\,xy < 0
\end{array} \right.\\
b)\,\,\left\{ \begin{array}{l}
\sqrt {2{a^2}} \,\,\,\,khi\,\,\,a \ge 0\\
- \sqrt {2{a^2}} \,\,\,\,khi\,\,\,a < 0
\end{array} \right.\\
c)\,\,\sqrt {\frac{a}{b}} \\d)\,\, \sqrt{3} a\\
e)\,\,\left\{ \begin{array}{l}
\sqrt 5 \,\,\,khi\,\,\,x > \frac{1}{2}\\
- \sqrt 5 \,\,\,khi\,\,\,x < \frac{1}{2}
\end{array} \right.
\end{array}\)

B \(\begin{array}{l}
a)\,\,\left\{ \begin{array}{l}
\frac{{\sqrt 2 }}{2}\,\,\,khi\,\,\,xy > 0\\
- \frac{{\sqrt 2 }}{2}\,\,\,khi\,\,\,xy < 0
\end{array} \right.\\
b)\,\,\left\{ \begin{array}{l}
\sqrt {2{a^2}} \,\,\,\,khi\,\,\,a \ge 0\\
- \sqrt {2{a^2}} \,\,\,\,khi\,\,\,a < 0
\end{array} \right.\\
c)\,\, - \sqrt {\frac{a}{b}} \\d)\,\, \sqrt{3a} \\
e)\,\, \sqrt{3} a\\e)\,\,\left\{ \begin{array}{l}\sqrt 5 \,\,\,khi\,\,\,x \ge \frac{1}{2}\\- \sqrt 5 \,\,\,khi\,\,\,x < \frac{1}{2}\end{array} \right.\end{array}\)

C \(\begin{array}{l}
a)\,\,\frac{{\sqrt 2 }}{2}\\
b)\,\,\left\{ \begin{array}{l}
\sqrt {2{a^2}} \,\,\,\,khi\,\,\,a \ge 0\\
- \sqrt {2{a^2}} \,\,\,\,khi\,\,\,a < 0
\end{array} \right.\\
c)\,\,\sqrt {\frac{a}{b}} \\d)\,\, \sqrt{3a} \\
e)\,\,\left\{ \begin{array}{l}
\sqrt 5 \,\,\,khi\,\,\,x \ge \frac{1}{2}\\
- \sqrt 5 \,\,\,khi\,\,\,x < \frac{1}{2}
\end{array} \right.
\end{array}\)

D \(\begin{array}{l}
a)\,\,\frac{{\sqrt 2 }}{2}\\
b)\,\,\left\{ \begin{array}{l}
\sqrt {2{a^2}} \,\,\,\,khi\,\,\,a \ge 0\\
- \sqrt {2{a^2}} \,\,\,\,khi\,\,\,a < 0
\end{array} \right.\\
c)\,\,\sqrt {\frac{a}{b}} \\d)\,\, \sqrt{3} a\\
e)\,\,\sqrt 5
\end{array}\)