Đáp án:
\(\left[ \begin{array}{l}
x = 4 + 2\sqrt 5 \left( {TM} \right)\\
x = \sqrt 6 \left( {TM} \right)
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:x \ne \left\{ { - 3;2} \right\}\\
1 + \dfrac{2}{{x - 2}} = \dfrac{{10}}{{x + 3}}\\
\to \dfrac{{\left( {x - 2} \right)\left( {x + 3} \right) + 2\left( {x + 3} \right) - 10\left( {x - 2} \right)}}{{\left( {x - 2} \right)\left( {x + 3} \right)}} = 0\\
\to \dfrac{{{x^2} + x - 6 + 2x + 6 - 10x + 20}}{{\left( {x - 2} \right)\left( {x + 3} \right)}} = 0\\
\to {x^2} - 7x + 20 = 0\\
Do:\Delta = 49 - 4.20 = - 31 < 0\\
\to x \in \emptyset \\
b)\left| {{x^2} - 4x - 5} \right| = 4x - 1\\
\to \left[ \begin{array}{l}
{x^2} - 4x - 5 = 4x - 1\left( {DK:\left[ \begin{array}{l}
x \ge 5\\
x \le - 1
\end{array} \right.} \right)\\
{x^2} - 4x - 5 = - 4x + 1\left( {DK: - 1 < x < 5} \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
{x^2} - 8x - 4 = 0\\
{x^2} - 6 = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 4 + 2\sqrt 5 \left( {TM} \right)\\
x = 4 - 2\sqrt 5 \left( l \right)\\
x = \sqrt 6 \left( {TM} \right)\\
x = - \sqrt 6 \left( l \right)
\end{array} \right.
\end{array}\)
