Đáp án:
$\begin{array}{l}
a)\dfrac{{a - b}}{{\sqrt a - \sqrt b }} + \dfrac{{\sqrt {{a^3}} - \sqrt {{b^3}} }}{{a - b}}\\
= \dfrac{{\left( {\sqrt a - \sqrt b } \right)\left( {\sqrt a + \sqrt b } \right)}}{{\sqrt a - \sqrt b }} + \dfrac{{a\sqrt a - b\sqrt b }}{{\left( {\sqrt a - \sqrt b } \right)\left( {\sqrt a + \sqrt b } \right)}}\\
= \sqrt a + \sqrt b + \dfrac{{\left( {\sqrt a - \sqrt b } \right)\left( {a + \sqrt {ab} + b} \right)}}{{\left( {\sqrt a - \sqrt b } \right)\left( {\sqrt a + \sqrt b } \right)}}\\
= \dfrac{{{{\left( {\sqrt a + \sqrt b } \right)}^2} + a + \sqrt {ab} + b}}{{\sqrt a + \sqrt b }}\\
= \dfrac{{a + 2\sqrt {ab} + b + a + \sqrt {ab} + b}}{{\sqrt a + \sqrt b }}\\
= \dfrac{{2a + 3\sqrt {ab} + 2b}}{{\sqrt a + \sqrt b }}\\
b)Dkxd:x \ge - 5\\
\sqrt {4x + 20} - 3\sqrt {5 + x} + \dfrac{4}{3}\sqrt {9x + 45} - 6 = 0\\
\Rightarrow 2\sqrt {x + 5} - 3\sqrt {x + 5} + \dfrac{4}{3}.3\sqrt {x + 5} = 6\\
\Rightarrow 2\sqrt {x + 5} - 3\sqrt {x + 5} + 4\sqrt {x + 5} = 6\\
\Rightarrow 3\sqrt {x + 5} = 6\\
\Rightarrow \sqrt {x + 5} = 2\\
\Rightarrow x + 5 = 4\\
\Rightarrow x = - 1\left( {tmdk} \right)\\
Vay\,x = - 1
\end{array}$
