x1^2 + 2x1x2 +x2 = 1 ( Biết x2 = -1 -1×1)

$\begin{array}{l} x_1^2 + 2{x_1}{x_2} + {x_2} = 1\\  \Leftrightarrow x_1^2 + 2{x_1}\left( { – 1 – {x_1}} \right) + \left( { – 1 – {x_1}} \right) = 1\\  \Leftrightarrow x_1^2 – 2{x_1} – 2x_1^2 – 1 – {x_1} – 1 = 0\\  \Leftrightarrow x_1^2 + 3{x_1} + 2 = 0\\  \Leftrightarrow \left( {{x_1} + 1} \right)\left( {{x_1} + 2} \right) = 0\\  \Leftrightarrow \left[ \begin{array}{l} {x_1} =  – 1 \Rightarrow {x_2} = 0\\ {x_1} =  – 2 \Rightarrow {x_2} = 1 \end{array} \right. \end{array}$