Đáp án:
Giải thích các bước giải:
\(\begin{array}{l}
\sqrt {2{x^2} - 2x + m} = x + 1\left( 1 \right)\\
\to \left\{ \begin{array}{l}
x \ge - 1\\
2{x^2} - 2x + m = {x^2} + 2x + 1
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x \ge - 1\\
{x^2} - 4x + m - 1 = 0\left( * \right)
\end{array} \right.\\
\end{array}\)
Để (1) có nghiệm
⇔ (*) có nghiệm
⇔Δ'≥0
\(\begin{array}{l}
\to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
4 - m + 1 \ge 0\\
{x_1} = 2 + \sqrt {5 - m} \ge - 1
\end{array} \right.\\
\left\{ \begin{array}{l}
4 - m + 1 \ge 0\\
{x_2} = 2 - \sqrt {5 - m} \ge - 1
\end{array} \right.
\end{array} \right.\\
\to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
m \le 5\\
\sqrt {5 - m} \ge - 3\left( {ld} \right)
\end{array} \right.\\
\left\{ \begin{array}{l}
m \le 5\\
\sqrt {5 - m} \le 3
\end{array} \right. \to \left\{ \begin{array}{l}
m \le 5\\
5 - m \le 9
\end{array} \right. \to - 4 \le m \le 5
\end{array} \right.\\
\to \left[ \begin{array}{l}
m \le 5\\
- 4 \le m \le 5
\end{array} \right.
\end{array}\)
