Đáp án:
1<a<2
Giải thích các bước giải:
\(\begin{array}{l}{x^2} - 2x - 2\left| {x - 1} \right| + a = 0\\ \Leftrightarrow {x^2} - 2x + 1 - 2\left| {x - 1} \right| + a - 1 = 0\\ \Leftrightarrow {\left( {x - 1} \right)^2} - 2\left| {x - 1} \right| + a - 1 = 0\\ \Leftrightarrow {\left| {x - 1} \right|^2} - 2\left| {x - 1} \right| + 1 + a - 2 = 0\\ \Leftrightarrow {\left( {\left| {x - 1} \right| - 1} \right)^2} = 2 - a\\ \Leftrightarrow \left\{ \begin{array}{l}2 - a \ge 0\\\left[ \begin{array}{l}\left| {x - 1} \right| - 1 = \sqrt {2 - a} \\\left| {x - 1} \right| - 1 = - \sqrt {2 - a} \end{array} \right.\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}a \le 2\\\left[ \begin{array}{l}\left| {x - 1} \right| = 1 + \sqrt {2 - a} \\\left| {x - 1} \right| = 1 - \sqrt {2 - a} \end{array} \right.\end{array} \right.\end{array}\)
Phương trình có 4 nghiệm phân biệt
\(\begin{array}{l} \Leftrightarrow \left\{ \begin{array}{l}1 + \sqrt {2 - a} \ne 1 - \sqrt {2 - a} \\1 - \sqrt {2 - a} > 0\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}a < 2\\\sqrt {2 - a} < 1\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}a < 2\\2 - a < 1\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}a < 2\\a > 1\end{array} \right. \Leftrightarrow 1 < a < 2\end{array}\)