Đáp án:
$\begin{array}{l}
a)\dfrac{1}{{3{b^2}c}}\sqrt {27{a^5}{b^4}{c^2}} \left( {a,b,c > 0} \right)\\
= \dfrac{1}{{3{b^2}c}}.3{a^2}{b^2}c\sqrt {3a} \\
= 3{a^2}\sqrt {3a} \\
b)\sqrt {{a^6}{b^5}{c^7}} \left( {a < 0;b > 0;c > 0} \right)\\
= - {a^3}{b^2}{c^3}.\sqrt {bc} \\
c)\sqrt {16x} - \sqrt {225{x^3}} + \sqrt {144x} - \sqrt {49x} \\
= 4\sqrt x - 15x\sqrt x + 12\sqrt x - 7\sqrt x \\
= - 8\sqrt x \\
d)\left( {\sqrt 6 + \sqrt 2 } \right)\left( {\sqrt 3 - 2} \right)\sqrt {\sqrt 3 + 2} \\
= \left( {\sqrt 3 + 1} \right).\left( {\sqrt 3 - 2} \right).\sqrt 2 .\sqrt {\sqrt 3 + 2} \\
= \dfrac{1}{2}\left( {2\sqrt 3 - 4} \right).\left( {\sqrt 3 + 1} \right).\sqrt {4 + 2\sqrt 3 } \\
= - \dfrac{1}{2}\left( {4 - 2\sqrt 3 } \right)\left( {\sqrt 3 + 1} \right).\sqrt {{{\left( {\sqrt 3 + 1} \right)}^2}} \\
= \dfrac{{ - 1}}{2}.{\left( {\sqrt 3 - 1} \right)^2}\left( {\sqrt 3 + 1} \right)\left( {\sqrt 3 + 1} \right)\\
= \dfrac{1}{2}.{\left[ {\left( {\sqrt 3 - 1} \right)\left( {\sqrt 3 + 1} \right)} \right]^2}\\
= \dfrac{1}{2}.{\left( {3 - 1} \right)^2}\\
= 2\\
c)\dfrac{3}{{3a - 1}}\sqrt {45{a^6} - 30{a^5} + 5{a^4}} \\
= \dfrac{3}{{3a - 1}}.\sqrt {5{a^4}\left( {9{a^2} - 6a + 1} \right)} \\
= \dfrac{3}{{3a - 1}}.{a^2}.\sqrt 5 .\sqrt {{{\left( {3a - 1} \right)}^2}} \\
= \left[ \begin{array}{l}
3{a^2}\sqrt 5 \left( {khi:a > \dfrac{1}{3}} \right)\\
- 3{a^2}\sqrt 5 \left( {khi:a < \dfrac{1}{3}} \right)
\end{array} \right.\\
e)\dfrac{{x - 5\sqrt x + 6}}{{x - 4}} = \dfrac{{\left( {\sqrt x - 2} \right)\left( {\sqrt x - 3} \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}\\
= \dfrac{{\sqrt x - 3}}{{\sqrt x + 2}}
\end{array}$
