Đáp án:
$\begin{array}{l}
1.7a)\left( {\sqrt {12} - \sqrt {75} + \sqrt {48} } \right):\sqrt 3 \\
= \left( {2\sqrt 3 - 5\sqrt 3 + 4\sqrt 3 } \right):\sqrt 3 \\
= \sqrt 3 :\sqrt 3 \\
= 1\\
b)\dfrac{{1 + \sqrt 5 }}{{\sqrt {15} - \sqrt 5 + \sqrt 3 - 1}}\\
= \dfrac{{1 + \sqrt 5 }}{{\sqrt 5 \left( {\sqrt 3 - 1} \right) + \sqrt 3 - 1}}\\
= \dfrac{{1 + \sqrt 5 }}{{\left( {\sqrt 3 - 1} \right)\left( {\sqrt 5 + 1} \right)}}\\
= \dfrac{1}{{\sqrt 3 - 1}}\\
= \dfrac{{\sqrt 3 + 1}}{{3 - 1}}\\
= \dfrac{{\sqrt 3 + 1}}{2}\\
1.8a)A = \sqrt 3 .\sqrt {27} - \sqrt {144} :\sqrt {36} \\
= \sqrt 3 .3\sqrt 3 - 12:4\\
= 9 - 3\\
= 6\\
b)B = \left( {\dfrac{{a + 3\sqrt a }}{{\sqrt a + 3}} - 2} \right).\left( {\dfrac{{a - 1}}{{\sqrt a - 1}} + 1} \right)\\
= \left( {\dfrac{{\sqrt a \left( {\sqrt a + 3} \right)}}{{\sqrt a + 3}} - 2} \right).\left( {\dfrac{{\left( {\sqrt a - 1} \right)\left( {\sqrt a + 1} \right)}}{{\sqrt a - 1}} + 1} \right)\\
= \left( {\sqrt a - 2} \right)\left( {\sqrt a + 1 + 1} \right)\\
= \left( {\sqrt a - 2} \right)\left( {\sqrt a + 2} \right)\\
= a - 4
\end{array}$
