Đáp án:
1) \(\left\{ \begin{array}{l}
x = 4\\
y = 1
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
1)DK:x \ne 3;y \ge - 3\\
\left\{ \begin{array}{l}
\dfrac{1}{{x - 3}} + 3\sqrt {y + 3} = 7\\
\dfrac{5}{{x - 3}} - 2\sqrt {y + 3} = 1
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\dfrac{2}{{x - 3}} + 6\sqrt {y + 3} = 14\\
\dfrac{{15}}{{x - 3}} - 6\sqrt {y + 3} = 3
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\dfrac{{17}}{{x - 3}} = 17\\
\dfrac{1}{{x - 3}} + 3\sqrt {y + 3} = 7
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\dfrac{1}{{x - 3}} = 1\\
1 + 3\sqrt {y + 3} = 7
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x - 3 = 1\\
\sqrt {y + 3} = 2
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = 4\\
y + 3 = 4
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = 4\\
y = 1
\end{array} \right.
\end{array}\)
2) Để phương trình có 2 nghiệm phân biệt
\(\begin{array}{l}
\to \Delta > 0\\
\to 25 - 4\left( {m - 1} \right) > 0\\
\to \dfrac{{25}}{4} > m - 1\\
\to \dfrac{{29}}{4} > m\\
Vi - et:\left\{ \begin{array}{l}
{x_1} + {x_2} = 5\\
{x_1}{x_2} = m - 1
\end{array} \right.\\
\sqrt {{x_1}} + \sqrt {{x_2}} = 3\\
\to {x_1} + 2\sqrt {{x_1}{x_2}} + {x_2} = 9\\
\to \left( {{x_1} + {x_2}} \right) + 2\sqrt {{x_1}{x_2}} = 9\\
\to 5 + 2\sqrt {m - 1} = 9\\
\to 2\sqrt {m - 1} = 4\\
\to m - 1 = 4\\
\to m = 5
\end{array}\)
