Đáp án:
$\begin{array}{l}
B1)\\
a)\sqrt {600} = \sqrt {6.100} = 10\sqrt 6 \\
- 2\sqrt {108} = - 2.\sqrt {36.3} = - 2.6\sqrt 3 = - 12\sqrt 3 \\
\sqrt {48{x^6}{y^4}} = \sqrt {3.16.{{\left( {{x^3}} \right)}^2}.{{\left( {{y^2}} \right)}^2}} = - 4{x^3}{y^2}\sqrt 3 \\
\left( {do:x < 0} \right)\\
b) - 7\sqrt 3 = - \sqrt {49.3} = - \sqrt {147} \\
6\sqrt {\dfrac{2}{3}} = \sqrt {36.\dfrac{2}{3}} = \sqrt {24} \\
B2)\\
a)\sqrt {\dfrac{5}{{98}}} = \dfrac{{\sqrt {5.98} }}{{98}} = \dfrac{{7\sqrt {10} }}{{98}} = \dfrac{{\sqrt {10} }}{{14}}\\
\sqrt {\dfrac{{{{\left( {1 - \sqrt 3 } \right)}^2}}}{{27}}} = \dfrac{{\sqrt 3 - 1}}{{3\sqrt 3 }} = \dfrac{{3 - \sqrt 3 }}{{27}}\\
\sqrt {\dfrac{7}{{12}}} = \dfrac{{\sqrt 7 .\sqrt {12} }}{{12}} = \dfrac{{2\sqrt 3 .\sqrt 7 }}{{12}} = \dfrac{{\sqrt {21} }}{6}\\
\sqrt {\dfrac{{{x^2}}}{5}} = \dfrac{{x\sqrt 5 }}{5}\\
\sqrt {{y^2} - \dfrac{{{y^2}}}{7}} = \dfrac{{ - y\sqrt {{y^2} - 1} }}{{\sqrt 7 }} = \dfrac{{ - y\sqrt {7{y^2} - 7} }}{7}\\
b)\dfrac{{\sqrt 5 - \sqrt 3 }}{{\sqrt 2 }} = \dfrac{{\sqrt {10} - \sqrt 6 }}{2}\\
\dfrac{{2\sqrt {10} - 5}}{{4 - \sqrt {10} }} = \dfrac{{\sqrt 5 \left( {2\sqrt 2 - \sqrt 5 } \right)}}{{\sqrt 2 \left( {2\sqrt 2 - \sqrt 5 } \right)}} = \dfrac{{\sqrt 5 }}{{\sqrt 2 }} = \dfrac{{\sqrt {10} }}{2}\\
\dfrac{{a + \sqrt {ab} }}{{\sqrt a + \sqrt b }} = \dfrac{{\sqrt a \left( {\sqrt a + \sqrt b } \right)}}{{\sqrt a + \sqrt b }} = \sqrt a \\
\dfrac{{a - \sqrt a }}{{1 - \sqrt a }} = \dfrac{{ - \sqrt a \left( {1 - \sqrt a } \right)}}{{1 - \sqrt a }} = - \sqrt a \\
B3)\\
a)\sqrt {98} - 3\sqrt 2 + 2\sqrt {50} \\
= 7\sqrt 2 - 3\sqrt 2 + 2.5\sqrt 2 \\
= 14\sqrt 2 \\
b)\dfrac{3}{{2 - x}}\sqrt {\dfrac{{7{x^4}\left( {{x^2} - 4x + 4} \right)}}{{36}}} \left( {x < 2} \right)\\
= \dfrac{3}{{2 - x}}.\dfrac{{{x^2}\left| {x - 2} \right|\sqrt 7 }}{6}\\
= \dfrac{3}{{2 - x}}.\dfrac{{{x^2}\left( {2 - x} \right).\sqrt 7 }}{6}\\
= \dfrac{{{x^2}\sqrt 7 }}{2}
\end{array}$
