$\left ( x+y \right )+\left ( x+\frac{1}{x} \right )=3+1=x^{2}+xy+\frac{y}{x}+1=\left ( x+y \right )\left ( x+\frac{1}{x} \right )$nên $\left ( \left ( x+y \right )+\left ( x+\frac{1}{x} \right ) \right )^{2}=4\left ( x+y \right )\left ( x+\frac{1}{x} \right )$$\Rightarrow x+y=x+\frac{1}{x}\Rightarrow y=\frac{1}{x}$.Thay vào hệ đc$\begin{cases}x+\frac{1}{x}=2 \\ x^{2}+\frac{1}{x^{2}}=2 \end{cases}\Leftrightarrow x=1\Rightarrow y=1$
$\left ( x+y \right )+\left ( x+\frac{1}{x} \right )=3+1=x^{2}+xy+\frac{y}{x}+1=\left ( x+y \right )\left ( x+\frac{1}{x} \right )$$\Rightarrow x+y=x+\frac{1}{x}\Rightarrow y=\frac{1}{x}$.Thay vào hệ đc$\begin{cases}x+\frac{1}{x}=2 \\ x^{2}+\frac{1}{x^{2}}=2 \end{cases}\Leftrightarrow x=1\Rightarrow y=1$
$\left ( x+y \right )+\left ( x+\frac{1}{x} \right )=3+1=x^{2}+xy+\frac{y}{x}+1=\left ( x+y \right )\left ( x+\frac{1}{x} \right )$
nên $\left ( \left ( x+y \right )+\left ( x+\frac{1}{x} \right ) \right )^{2}=4\left ( x+y \right )\left ( x+\frac{1}{x} \right )$$\Rightarrow x+y=x+\frac{1}{x}\Rightarrow y=\frac{1}{x}$.Thay vào hệ đc$\begin{cases}x+\frac{1}{x}=2 \\ x^{2}+\frac{1}{x^{2}}=2 \end{cases}\Leftrightarrow x=1\Rightarrow y=1$