Đáp án:
$\begin{array}{l}
3B\\
a)\left( {\sqrt {\dfrac{3}{4}} - \sqrt 3 + 5\sqrt {\dfrac{4}{3}} } \right).\sqrt {12} \\
= \left( {\dfrac{{\sqrt 3 }}{2} - \sqrt 3 + \dfrac{{5.2\sqrt 3 }}{3}} \right).2\sqrt 3 \\
= \left( {\dfrac{{\sqrt 3 }}{2} - \sqrt 3 + \dfrac{{10\sqrt 3 }}{3}} \right).2\sqrt 3 \\
= \dfrac{{17\sqrt 3 }}{3}.2\sqrt 3 \\
= 34\\
b)\sqrt {3 - \sqrt 5 } .\sqrt 8 \\
= \sqrt 4 .\sqrt {6 - 2\sqrt 5 } \\
= 2.\sqrt {{{\left( {\sqrt 5 - 1} \right)}^2}} \\
= 2.\left( {\sqrt 5 - 1} \right)\\
= 2\sqrt 5 - 2\\
4A)\\
a)\left( {\sqrt {\dfrac{1}{7}} - \sqrt {\dfrac{{16}}{7}} + \sqrt 7 } \right):\sqrt 7 \\
= \left( {\dfrac{{\sqrt 7 }}{7} - \dfrac{{4\sqrt 7 }}{7} + \sqrt 7 } \right):\sqrt 7 \\
= \dfrac{1}{7} - \dfrac{4}{7} + 1\\
= \dfrac{3}{7}\\
b)\sqrt {36 - 12\sqrt 5 } :\sqrt 6 \\
= \sqrt 6 .\sqrt {6 - 2\sqrt 5 } :\sqrt 6 \\
= \sqrt {{{\left( {\sqrt 5 - 1} \right)}^2}} \\
= \sqrt 5 - 1\\
4B)\\
a)\left( {\sqrt {\dfrac{1}{3}} - \sqrt {\dfrac{4}{3}} + \sqrt 3 } \right):\sqrt 3 \\
= \left( {\dfrac{{\sqrt 3 }}{3} - \dfrac{{2\sqrt 3 }}{3} + \sqrt 3 } \right):\sqrt 3 \\
= \dfrac{{2\sqrt 3 }}{3}:\sqrt 3 \\
= \dfrac{2}{3}\\
b)\sqrt {3 - \sqrt 5 } :\sqrt 2 \\
= \dfrac{{\sqrt {6 - 2\sqrt 5 } }}{2}\\
= \dfrac{{\sqrt {{{\left( {\sqrt 5 - 1} \right)}^2}} }}{2}\\
= \dfrac{{\sqrt 5 - 1}}{2}
\end{array}$
