Đặt$\dfrac{3}{\sqrt{x-1}}=a(a>0), x+3y=b$
Hệ phương trình trở thành:
$\begin{array}{l} \left\{ \begin{array}{l} \dfrac{3}{{\sqrt {x - 1} }} + x - 3y = 1\\ \dfrac{1}{{\sqrt {x - 1} }} - 3x + 9y = 7 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} 3a + b = 1\\ a - 3b = 7 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} a = 1\\ b = - 2 \end{array} \right.\\ \to \left\{ \begin{array}{l} \dfrac{1}{{\sqrt {x - 1} }} = 1\\ x - 3y = - 2 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} x = 2\\ y = \dfrac{4}{3} \end{array} \right. \Rightarrow \left( {x;y} \right) = \left( {2;\dfrac{4}{3}} \right) \end{array}$
