Đáp án:
\(\left\{ \begin{array}{l}
x = 1\\
y = 7
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
1)DK:x \ne - 1;y > 3\\
\left\{ \begin{array}{l}
\dfrac{x}{{x + 1}} + \dfrac{5}{{\sqrt {y - 3} }} = 3\\
\dfrac{{3x}}{{x + 1}} - \dfrac{1}{{\sqrt {y - 3} }} = 1
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\dfrac{x}{{x + 1}} + \dfrac{5}{{\sqrt {y - 3} }} = 3\\
\dfrac{{15x}}{{x + 1}} - \dfrac{5}{{\sqrt {y - 3} }} = 5
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\dfrac{{16x}}{{x + 1}} = 8\\
\dfrac{{3x}}{{x + 1}} - \dfrac{1}{{\sqrt {y - 3} }} = 1
\end{array} \right.\\
\to \left\{ \begin{array}{l}
16x = 8x + 8\\
\dfrac{{3x}}{{x + 1}} - \dfrac{1}{{\sqrt {y - 3} }} = 1
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = 1\\
\dfrac{1}{{\sqrt {y - 3} }} = \dfrac{1}{2}
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = 1\\
y - 3 = 4
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = 1\\
y = 7
\end{array} \right.
\end{array}\)
