Đáp án:
\(\left[ \begin{array}{l}
x = 1 + \sqrt 3 \\
x = 0
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
Xét:{x^2} - 1 > 0\\
\to \left( {x - 1} \right)\left( {x + 1} \right) > 0\\
\to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x - 1 > 0\\
x + 1 > 0
\end{array} \right.\\
\left\{ \begin{array}{l}
x - 1 < 0\\
x + 1 < 0
\end{array} \right.
\end{array} \right. \to \left[ \begin{array}{l}
x > 1\\
x < - 1
\end{array} \right.\\
{x^2} - 1 \le 0 \to - 1 \le x \le 1\\
Có:\left| {{x^2} - 1} \right| = 2x + 1\\
\to \left[ \begin{array}{l}
{x^2} - 1 = 2x + 1\left( {DK:\left[ \begin{array}{l}
x > 1\\
x < - 1
\end{array} \right.} \right)\\
{x^2} - 1 = - 2x - 1\left( {DK: - 1 \le x \le 1} \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
{x^2} - 2x - 2 = 0\left( 1 \right)\\
{x^2} + 2x = 0\left( 2 \right)
\end{array} \right.\\
Pt\left( 1 \right): Δ'= 1 + 2 = 3\\
\to \left[ \begin{array}{l}
x = 1 + \sqrt 3 \left( {TM} \right)\\
x = 1 - \sqrt 3 \left( l \right)
\end{array} \right.\\
Pt\left( 2 \right):x\left( {x + 2} \right) = 0\\
\to \left[ \begin{array}{l}
x = 0\left( {TM} \right)\\
x = - 2\left( l \right)
\end{array} \right.\\
KL:\left[ \begin{array}{l}
x = 1 + \sqrt 3 \\
x = 0
\end{array} \right.
\end{array}\)
