Đáp án:
a. \(\left[ \begin{array}{l}
x = - 1\\
x = 3
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a.DK:x \ge - 1\\
\sqrt {x + 1} \left( {\sqrt {x + 1} - 2} \right) = 0\\
\to \left[ \begin{array}{l}
x + 1 = 0\\
\sqrt {x + 1} - 2 = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = - 1\\
\sqrt {x + 1} = 2
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = - 1\\
x + 1 = 4
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = - 1\\
x = 3
\end{array} \right.\left( {TM} \right)\\
b.DK:x \ge 2\\
2\left( {x - 2} \right) - \sqrt {x - 2} = 0\\
\to \sqrt {x - 2} \left( {2\sqrt {x - 2} - 1} \right) = 0\\
\to \left[ \begin{array}{l}
x - 2 = 0\\
\sqrt {x - 2} = \dfrac{1}{2}
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 2\\
x - 2 = \dfrac{1}{4}
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 2\\
x = \dfrac{9}{4}
\end{array} \right.\left( {TM} \right)\\
c.DK:{x^2} - 9 \ge 0 \to \left[ \begin{array}{l}
x \ge 3\\
x \le - 3
\end{array} \right.\\
\sqrt {x - 3} - 2\sqrt {\left( {x - 3} \right)\left( {x + 3} \right)} = 0\\
\to \sqrt {x - 3} \left( {1 - 2\sqrt {x + 3} } \right) = 0\\
\to \left[ \begin{array}{l}
x - 3 = 0\\
\sqrt {x + 3} = \dfrac{1}{2}
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 3\\
x + 3 = \dfrac{1}{4}
\end{array} \right. \to \left[ \begin{array}{l}
x = 3\left( {TM} \right)\\
x = - \dfrac{{11}}{4}\left( l \right)
\end{array} \right.
\end{array}\)
