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