Đáp án:
3) \(\left[ \begin{array}{l}
x = - 2\\
x = 3
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
1)DK:{x^2} - x - 9 \ge 0\\
\to \left[ \begin{array}{l}
x \ge \dfrac{{1 + \sqrt {37} }}{2}\\
x \le \dfrac{{1 - \sqrt {37} }}{2}
\end{array} \right.\\
\sqrt {{x^2} - \left( {x + 9} \right)} = 1\\
\to \sqrt {{x^2} - x - 9} = 1\\
\to {x^2} - x - 9 = 1\\
\to {x^2} - x - 10 = 0\\
\to {x^2} - 2.x.\dfrac{1}{2} + \dfrac{1}{4} - \dfrac{{41}}{4} = 0\\
\to {\left( {x - \dfrac{1}{2}} \right)^2} = \dfrac{{41}}{4}\\
\to \left[ \begin{array}{l}
x - \dfrac{1}{2} = \dfrac{{\sqrt {41} }}{2}\\
x - \dfrac{1}{2} = - \dfrac{{\sqrt {41} }}{2}
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{{1 + \sqrt {41} }}{2}\\
x = \dfrac{{1 - \sqrt {41} }}{2}
\end{array} \right.\\
2)DK:\dfrac{4}{9} \ge x\\
\sqrt {4 - 9x} = 12\\
\to 4 - 9x = 144\\
\to 9x = - 140\\
\to x = - \dfrac{{140}}{9}\left( {TM} \right)\\
3)\sqrt {{{\left( {1 - 2x} \right)}^2}} = 5\\
\to \left| {1 - 2x} \right| = 5\\
\to \left[ \begin{array}{l}
1 - 2x = 5\\
1 - 2x = - 5
\end{array} \right.\\
\to \left[ \begin{array}{l}
2x = - 4\\
2x = 6
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = - 2\\
x = 3
\end{array} \right.
\end{array}\)
