Đáp án:
$\begin{array}{l}
b)D = \dfrac{1}{{2a - 1}}.\sqrt {25{a^4} - 100{a^5} + 100{a^6}} \\
= \dfrac{1}{{2a - 1}}.\sqrt {25{a^4}\left( {1 - 4a + 4{a^2}} \right)} \\
= \dfrac{{5{a^2}\sqrt {{{\left( {2a - 1} \right)}^2}} }}{{2a - 1}}\\
= \left[ \begin{array}{l}
5{a^2}\left( {khi:a > \dfrac{1}{2}} \right)\\
- 5{a^2}\left( {khi:a < \dfrac{1}{2}} \right)
\end{array} \right.\\
3)a)\\
A = \sqrt {\dfrac{{{{\left( {x - 5} \right)}^4}}}{{{{\left( {4 - x} \right)}^2}}}} - \dfrac{{{x^2} - 25}}{{x - 4}}\\
= \dfrac{{{{\left( {x - 5} \right)}^2}}}{{4 - x}} - \dfrac{{{x^2} - 25}}{{x - 4}}\\
= \dfrac{{{x^2} - 10x + 25 + {x^2} - 25}}{{4 - x}}\\
= \dfrac{{2{x^2} - 10x}}{{4 - x}}\\
= \dfrac{{{{2.3}^2} - 10.3}}{{4 - 3}}\left( {x = 3} \right)\\
= - 12\\
b)\\
B = 3x - \sqrt {27} + \dfrac{{\sqrt {{x^3} + 3{x^2}} }}{{\sqrt {x + 3} }}\\
= 3x - 3\sqrt 3 + \dfrac{{\sqrt {{x^2}\left( {x + 3} \right)} }}{{\sqrt {x + 3} }}\\
= 3x - 3\sqrt 3 + x\\
= 2x - 3\sqrt 3 \\
= 2.\sqrt 3 - 3\sqrt 3 \\
= - \sqrt 3
\end{array}$
