Đáp án:
a. x=1
b. \( - 1 < m \le 1\)
Giải thích các bước giải:
a. Thay m=1 vào phương trình
\(\begin{array}{l}
\to \sqrt {{x^2} + 2x - 3} = x - 1\\
\leftrightarrow \left\{ \begin{array}{l}
x - 1 \ge 0\\
{x^2} + 2x - 3 = {x^2} - 2x + 1
\end{array} \right. \leftrightarrow \left\{ \begin{array}{l}
x \ge 1\\
4x = 4
\end{array} \right. \leftrightarrow \left\{ \begin{array}{l}
x \ge 1\\
x = 1
\end{array} \right. \leftrightarrow x = 1
\end{array}\)
\(\begin{array}{l}
b.\sqrt {{x^2} + 2mx - 3} = x - 1\\
\leftrightarrow \left\{ \begin{array}{l}
x - 1 \ge 0\\
{x^2} + 2mx - 3 = {x^2} - 2x + 1
\end{array} \right. \leftrightarrow \left\{ \begin{array}{l}
x \ge 1\\
x(m + 1) = 2
\end{array} \right. \leftrightarrow \left\{ \begin{array}{l}
x \ge 1\\
x = \frac{2}{{m + 1}}(m \ne - 1)
\end{array} \right.\\
\to \frac{2}{{m + 1}} \ge 1\\
\leftrightarrow \frac{{2 - m - 1}}{{m + 1}} \ge 0 \leftrightarrow \frac{{1 - m}}{{m + 1}} \ge 0\\
\leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
1 - m \ge 0\\
m + 1 > 0
\end{array} \right.\\
\left\{ \begin{array}{l}
1 - m \le 0\\
m + 1 < 0
\end{array} \right.
\end{array} \right. \leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
m \le 1\\
m > - 1
\end{array} \right.\\
\left\{ \begin{array}{l}
m \ge 1\\
m < - 1
\end{array} \right.(VN)
\end{array} \right. \leftrightarrow - 1 < m \le 1
\end{array}\)
