Đưa thừa số vào trong dấu căn
\(\begin{array}{l}a)\,\,\frac{1}{{xy}}\sqrt {\frac{{{x^2}{y^2}}}{2}} & & & b)\,\,a\sqrt 2 & & & c)\,\, - \frac{a}{b}\sqrt {\frac{b}{a}} \,\,\,\left( {a > 0,\,\,\,b > 0} \right)\\d)\,\,a\sqrt {\frac{3}{a}} & & & e)\,\,\frac{1}{{2x - 1}}\sqrt {5\left( {1 - 4x + 4{x^2}} \right)} .\end{array}\)
A.\(\begin{array}{l}a)\,\,\left\{ \begin{array}{l}\frac{{\sqrt 2 }}{2}\,\,\,khi\,\,\,xy > 0\\ - \frac{{\sqrt 2 }}{2}\,\,\,khi\,\,\,xy < 0\end{array} \right.\\b)\,\,\left\{ \begin{array}{l}\sqrt {2{a^2}} \,\,\,\,khi\,\,\,a \ge 0\\ - \sqrt {2{a^2}} \,\,\,\,khi\,\,\,a \frac{1}{2}\\ - \sqrt 5 \,\,\,khi\,\,\,x < \frac{1}{2}\end{array} \right.\end{array}\)
B.\(\begin{array}{l}a)\,\,\left\{ \begin{array}{l}\frac{{\sqrt 2 }}{2}\,\,\,khi\,\,\,xy > 0\\ - \frac{{\sqrt 2 }}{2}\,\,\,khi\,\,\,xy < 0\end{array} \right.\\b)\,\,\left\{ \begin{array}{l}\sqrt {2{a^2}} \,\,\,\,khi\,\,\,a \ge 0\\ - \sqrt {2{a^2}} \,\,\,\,khi\,\,\,a < 0\end{array} \right.\\c)\,\, - \sqrt {\frac{a}{b}} \\d)\,\,\left\{ \begin{array}{l}\sqrt 5 \,\,\,khi\,\,\,x \ge \frac{1}{2}\\ - \sqrt 5 \,\,\,khi\,\,\,x < \frac{1}{2}\end{array} \right.\end{array}\)
C.\(\begin{array}{l}a)\,\,\frac{{\sqrt 2 }}{2}\\b)\,\,\left\{ \begin{array}{l}\sqrt {2{a^2}} \,\,\,\,khi\,\,\,a \ge 0\\ - \sqrt {2{a^2}} \,\,\,\,khi\,\,\,a < 0\end{array} \right.\\c)\,\,\sqrt {\frac{a}{b}} \\d)\,\,\left\{ \begin{array}{l}\sqrt 5 \,\,\,khi\,\,\,x \ge \frac{1}{2}\\ - \sqrt 5 \,\,\,khi\,\,\,x < \frac{1}{2}\end{array} \right.\end{array}\)
D.\(\begin{array}{l}a)\,\,\frac{{\sqrt 2 }}{2}\\b)\,\,\left\{ \begin{array}{l}\sqrt {2{a^2}} \,\,\,\,khi\,\,\,a \ge 0\\ - \sqrt {2{a^2}} \,\,\,\,khi\,\,\,a < 0\end{array} \right.\\c)\,\,\sqrt {\frac{a}{b}} \\d)\,\,\sqrt 5 \end{array}\)
