Đáp án:
2) \(\dfrac{{2x - 3\sqrt x + 2}}{{x - 4}}\)
Giải thích các bước giải:
\(\begin{array}{l}
DK:x \ge - 5\\
\sqrt {4x + 20} - 3\sqrt {5 + x} + \dfrac{1}{3}.\sqrt {9x + 45} = 6\\
\to \sqrt {4\left( {x + 5} \right)} - 3\sqrt {x + 5} + \dfrac{1}{3}.\sqrt {9\left( {x + 5} \right)} = 6\\
\to 2\sqrt {x + 5} - 3\sqrt {x + 5} + \dfrac{1}{3}.3\sqrt {x + 5} = 6\\
\to \left( {2 - 3 + 1} \right)\sqrt {x + 5} = 6\\
\to 0.\sqrt {x + 5} = 6\\
\to 0 = 6\left( {vô lý} \right)\\
\to x \in \emptyset \\
2)DK:x \ge 0;x \ne 4\\
\dfrac{{\sqrt x + 2 + 2\sqrt x \left( {\sqrt x - 2} \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}\\
= \dfrac{{\sqrt x + 2 + 2x - 4\sqrt x }}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}\\
= \dfrac{{2x - 3\sqrt x + 2}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}\\
= \dfrac{{2x - 3\sqrt x + 2}}{{x - 4}}
\end{array}\)
