Biểu thức $\sqrt {x - 3} $  có nghĩa khi

$\sqrt {{A^2}B}  = A\sqrt B $

$\sqrt {{A^2}B}  =  - A\sqrt B $

$\sqrt {{A^2}B}  = -B\sqrt A $

$\sqrt {{A^2}B}  = B\sqrt A $

$\dfrac{9}{{13}}$

$\dfrac{9}{{169}}$

$\dfrac{3}{{13}}$

$\dfrac{{13}}{9}$

\(14a\)

\(20a\)

\(9a\)

\( - 8a\)

$\sqrt {ab}  = a\sqrt b $

$\sqrt a \sqrt b  = b\sqrt a $

$\sqrt a .\sqrt b  = \sqrt {ab} $

$\sqrt {ab}  = \dfrac{{\sqrt a }}{{\sqrt b }}$

$5\sqrt 3  > 4\sqrt 5 $

$5\sqrt 3  = 4\sqrt 5 $

$5\sqrt 3  \ge 4\sqrt 5 $

$5\sqrt 3  < 4\sqrt 5 $

\(\sqrt[3]{{ab}} = \sqrt a .\sqrt b \)

\(\dfrac{{\sqrt[3]{a}}}{{\sqrt[3]{b}}} = \dfrac{a}{b}\) với \(b \ne 0\)

\({\left( {\sqrt[3]{a}} \right)^3} =  - a\, khi\,a < 0\)

\(\dfrac{{\sqrt[3]{a}}}{{\sqrt[3]{b}}} = \sqrt[3]{{\dfrac{a}{b}}}\) với \(b \ne 0\)