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