x^2 – 2 x – 2|x-1| + a = 0 .tìm a để pt có 4 nghiệm

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

\(\begin{array}{l} \Leftrightarrow \left\{ \begin{array}{l}1 + \sqrt {2 – a}  \ne 1 – \sqrt {2 – a} \\1 – \sqrt {2 – a}  > 0\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}a < 2\\\sqrt {2 – a}  < 1\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}a < 2\\2 – a < 1\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}a < 2\\a > 1\end{array} \right. \Leftrightarrow 1 < a < 2\end{array}\)