Đáp án:
$\begin{array}{l}
b)5\sqrt a - 4b\sqrt {25{a^3}} + 5a\sqrt {16a{b^2}} - 2\sqrt {9a} \\
= 5\sqrt a - 4b.5a\sqrt a + 5a.4b\sqrt a - 2.3\sqrt a \\
= 5\sqrt a - 20ab\sqrt a + 20ab\sqrt a - 6\sqrt a \\
= - \sqrt a \\
2)Dkxd:x \ge 1\\
\sqrt {4x - 4} + 2\sqrt {16x - 16} - \dfrac{2}{5} = \dfrac{3}{5} + \sqrt {9x - 9} \\
\Leftrightarrow 2\sqrt {x - 1} + 2.4\sqrt {x - 1} - 3\sqrt {x - 1} = 1\\
\Leftrightarrow 2\sqrt {x - 1} + 8\sqrt {x - 1} - 3\sqrt {x - 1} = 1\\
\Leftrightarrow 7\sqrt {x - 1} = 1\\
\Leftrightarrow \sqrt {x - 1} = \dfrac{1}{7}\\
\Leftrightarrow x - 1 = \dfrac{1}{{49}}\\
\Leftrightarrow x = \dfrac{{50}}{{49}}\\
Vay\,x = \dfrac{{50}}{{49}}\\
3)Dkxd:x > 0\\
\left( {x\sqrt {\dfrac{6}{x}} + \sqrt {\dfrac{{2x}}{3}} + \sqrt {6x} } \right):\sqrt {6x} \\
= \left( {\sqrt x .\sqrt 6 + \dfrac{{\sqrt {6x} }}{3} + \sqrt {6x} } \right):\sqrt {6x} \\
= \left( {1 + \dfrac{1}{3} + 1} \right).\sqrt {6x} :\sqrt {6x} \\
= 2\dfrac{1}{3}\\
4)a + b + c - \sqrt {ab} - \sqrt {bc} - \sqrt {ca} \\
= \dfrac{1}{2}.\left( {2a + 2b + 2c - 2\sqrt {ab} - 2\sqrt {bc} - 2\sqrt {ca} } \right)\\
= \dfrac{1}{2}.\left( \begin{array}{l}
a - 2\sqrt {ab} + b + b - 2\sqrt {bc} + c\\
+ c - 2\sqrt {ca} + a
\end{array} \right)\\
= \dfrac{1}{2}.\left[ {{{\left( {\sqrt a - \sqrt b } \right)}^2} + {{\left( {\sqrt b - \sqrt c } \right)}^2} + {{\left( {\sqrt c - \sqrt a } \right)}^2}} \right] \ge 0\\
\Leftrightarrow a + b + c \ge \sqrt {ab} + \sqrt {bc} + \sqrt {ca}
\end{array}$
