Đáp án:
$\begin{array}{l}
a)\sqrt {{a^2}{{\left( {a + 1} \right)}^2}} \left( {a > 0} \right)\\
= a.\left( {a + 1} \right)\\
= {a^2} + a\left( {do:a > 0} \right)\\
b)\dfrac{{\sqrt {16{a^4}{b^6}} }}{{\sqrt {128{a^6}{b^6}} }} = \dfrac{{4{a^2}\left| {{b^3}} \right|}}{{8\sqrt 2 .\left( { - {a^3}} \right).\left| {{b^3}} \right|}} = \dfrac{{ - \sqrt 2 }}{{4a}}\\
c)\sqrt {9{{\left( {x - 5} \right)}^2}} \left( {x \ge 5} \right)\\
= 3\left( {x - 5} \right)\\
= 3x - 15\\
b)\sqrt {{x^2} + 6x + 9} \\
= \sqrt {{{\left( {x + 3} \right)}^2}} \\
= \left| {x + 3} \right|\\
c)\dfrac{{\sqrt {108{x^3}} }}{{\sqrt {12x} }} = \sqrt {9{x^2}} = 3x\left( {do:x > 0} \right)\\
d)\dfrac{{\sqrt {13{x^4}{y^6}} }}{{\sqrt {208{x^6}{y^6}} }}\left( {x < 0;y\# 0} \right)\\
= \sqrt {\dfrac{1}{{16{x^2}}}} = \dfrac{{ - 1}}{{4x}}\left( {do:x < 0} \right)
\end{array}$
