Điều kiện xác định $x>2y; x\ne 0$
$\begin{array}{l} \left\{ \begin{array}{l} \sqrt {x - 2y} = 4 - x\\ \frac{1}{{\sqrt {x - 2y} }} + \frac{1}{x} = 1 \end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l} \sqrt {x - 2y} + x = 4\\ \frac{1}{{\sqrt {x - 2y} }} + \frac{1}{x} = 1 \end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l} \sqrt {x - 2y} + x = 4\\ \frac{{x + \sqrt {x - 2y} }}{{x\sqrt {x - 2y} }} = 1 \end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l} \sqrt {x - 2y} + x = 4\\ \frac{4}{{x\sqrt {x - 2y} }} = 1 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} \sqrt {x - 2y} + x = 4\\ x\sqrt {x - 2y} = 4 \end{array} \right. \end{array}$
$\sqrt{x-2y}, x$ là nghiệm của phương trình $t^2-4t+4=0\Rightarrow (t-2)^2=0\Rightarrow t=2$
$\begin{array}{l} \Rightarrow \left\{ \begin{array}{l} \sqrt {x - 2y} = 2\\ x = 2 \end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l} x = 2\\ \sqrt {2 - 2y} = 2 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} x = 2\\ 2 - 2y = 4 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} x = 2\\ y = - 1 \end{array} \right.\\ \Rightarrow \left( {x;y} \right) = \left( {2; - 1} \right) \end{array}$
