Đáp án:
\(\left[ \begin{array}{l}
x = 6\\
x = - \sqrt {22}
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
\left| {4x - 17} \right| = {x^2} - 4x - 5\\
\to \left[ \begin{array}{l}
4x - 17 = {x^2} - 4x - 5\left( {DK:x \ge \dfrac{{17}}{4}} \right)\\
- 4x + 17 = {x^2} - 4x - 5\left( {DK:x < \dfrac{{17}}{4}} \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
{x^2} - 8x + 12 = 0\\
{x^2} - 22 = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
{x^2} - 6x - 2x + 12 = 0\\
{x^2} = 22
\end{array} \right.\\
\to \left[ \begin{array}{l}
x\left( {x - 6} \right) - 2\left( {x - 6} \right) = 0\\
x = \sqrt {22} \left( l \right)\\
x = - \sqrt {22}
\end{array} \right.\\
\to \left[ \begin{array}{l}
\left( {x - 2} \right)\left( {x - 6} \right) = 0\\
x = - \sqrt {22}
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 2\left( l \right)\\
x = 6\\
x = - \sqrt {22}
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 6\\
x = - \sqrt {22}
\end{array} \right.
\end{array}\)
