Đáp án:
c) \(\left[ \begin{array}{l}
x = 3\\
x = - 2
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:x \ge 5\\
\sqrt {x - 5} = x - 4\\
\to x - 5 = {x^2} - 8x + 16\\
\to {x^2} - 9x + 21 = 0\\
Do:\Delta = 81 - 4.21 = - 3 < 0\\
\to x \in \emptyset \\
b)\left| {3x - 2} \right| = 2x - 1\\
\to \left[ \begin{array}{l}
3x - 2 = 2x - 1\left( {DK:x \ge \dfrac{2}{3}} \right)\\
3x - 2 = - 2x + 1\left( {DK:x < \dfrac{2}{3}} \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 1\\
5x = 3
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 1\\
x = \dfrac{3}{5}
\end{array} \right.\left( {TM} \right)\\
c)DK:x \ne \left\{ { - 6;0} \right\}\\
\dfrac{1}{{x + 6}} - \dfrac{1}{{{x^2}}} = 0\\
\to \dfrac{{{x^2} - x - 6}}{{{x^2}\left( {x + 6} \right)}} = 0\\
\to {x^2} - x - 6 = 0\\
\to \left[ \begin{array}{l}
x = 3\\
x = - 2
\end{array} \right.\left( {TM} \right)
\end{array}\)
