Giải thích các bước giải:
ĐKXĐ: x, y≥0
Với m=1 thì hệ trở thành:
$\begin{array}{l} \left\{ \begin{array}{l} 2\sqrt x + 3\sqrt y = 5\\ 3\sqrt x - \sqrt y = 2 \end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l} 2\sqrt x + 3\sqrt y = 5\\ 9\sqrt x - 3\sqrt y = 6 \end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l} 11\sqrt x = 11\\ 3\sqrt x - \sqrt y = 2 \end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l} x = 1\\ 3\sqrt x - 2 = \sqrt y \end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l} x = 1\\ 1 = \sqrt y \end{array} \right.\\ \Leftrightarrow x = y = 1 \end{array}$
$\begin{array}{l} \left\{ \begin{array}{l} 2\sqrt x + 3\sqrt y = m + 4\\ 3\sqrt x - \sqrt y = 7m - 5 \end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l} 2\sqrt x + 3\sqrt y = m + 4\\ 9\sqrt x - 3\sqrt y = 21m - 15 \end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l} 2\sqrt x + 3\sqrt y = m + 4\\ 9\sqrt x - 3\sqrt y = 21m - 15 \end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l} 11\sqrt x = 22m - 11\\ 3\sqrt x - \sqrt y = 7m - 5 \end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l} \sqrt x = 2m - 1\\ \sqrt y = - m + 2 \end{array} \right.\\ \end{array}$
Để pt có nghiệm thì:
$\begin{array}{l} \left\{ \begin{array}{l} - m + 2 \ge 0\\ 2m - 1 \ge 0 \end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l} 2 \ge m\\ m \ge \frac{1}{2} \end{array} \right.\\ \Leftrightarrow \frac{1}{2} \le m \le 2 \end{array}$
