Đáp án:
$\begin{array}{l}
1c)\left( {20\sqrt {300} - 15\sqrt {675} + 5\sqrt {75} } \right):\sqrt {15} \\
= \left( {20.10\sqrt 5 - 15.15\sqrt 5 + 5.5\sqrt 5 } \right):\sqrt {15} \\
= 0\\
d)\left( {\sqrt {325} - \sqrt {117} + 2\sqrt {208} } \right):\sqrt {13} \\
= \left( {5\sqrt {13} - 3\sqrt {13} + 2.4\sqrt {13} } \right):\sqrt {13} \\
= 8\sqrt {13} :\sqrt {13} \\
= 8\\
B3)\\
a)Dkxd:x \ge 3\\
\sqrt {x - 3} - 2\sqrt {{x^2} - 9} = 0\\
\Leftrightarrow \sqrt {x - 3} - 2.\sqrt {\left( {x - 3} \right)\left( {x + 3} \right)} = 0\\
\Leftrightarrow \sqrt {x - 3} .\left( {1 - 2\sqrt {x + 3} } \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x - 3 = 0\\
\sqrt {x + 3} = \dfrac{1}{2}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 3\left( {tm} \right)\\
x + 3 = \dfrac{1}{4} \Leftrightarrow x = \dfrac{{ - 11}}{4}\left( {ktm} \right)
\end{array} \right.\\
Vậy\,x = 3\\
b)Dkxd:y \ge 5\\
\sqrt {4y - 20} + \sqrt {y - 5} - \dfrac{1}{3}\sqrt {9y - 45} = 4\\
\Leftrightarrow 2\sqrt {y - 5} + \sqrt {y - 5} - \dfrac{1}{3}.3\sqrt {y - 5} = 4\\
\Leftrightarrow 3\sqrt {y - 5} - \sqrt {y - 5} = 4\\
\Leftrightarrow 2\sqrt {y - 5} = 4\\
\Leftrightarrow \sqrt {y - 5} = 2\\
\Leftrightarrow y - 5 = 4\\
\Leftrightarrow y = 9\left( {tmdk} \right)\\
Vậy\,y = 9
\end{array}$
